5.5: Zeros of Polynomial Functions
Learning Objectives
- Evaluate a polynomial using the Remainder Theorem.
- Use the Factor Theorem to solve a polynomial equation.
- Use the Rational Zero Theorem to find rational zeros.
- Find zeros of a polynomial function.
- Use the Linear Factorization Theorem to find polynomials with given zeros.
- Use Descartes’ Rule of Signs.
- Solve real-world applications of polynomial equations
A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be?
This problem can be solved by writing a cubic function and solving a cubic equation for the volume of the cake. In this section, we will discuss a variety of tools for writing polynomial functions and solving polynomial equations.
Evaluating a Polynomial Using the Remainder Theorem
In the last section, we learned how to divide polynomials. We can now use polynomial division to evaluate polynomials using the Remainder Theorem . If the polynomial is divided by \(x–k\), the remainder may be found quickly by evaluating the polynomial function at \(k\), that is, \(f(k)\). Let’s walk through the proof of the theorem.
Recall that the Division Algorithm states that, given a polynomial dividend \(f(x)\) and a non-zero polynomial divisor \(d(x)\) where the degree of \(d(x)\) is less than or equal to the degree of \(f(x)\),there exist unique polynomials \(q(x)\) and \(r(x)\) such that
\[f(x)=d(x)q(x)+r(x) \nonumber\]
If the divisor, \(d(x)\), is \(x−k\), this takes the form
\[f(x)=(x−k)q(x)+r \nonumber\]
Since the divisor \(x−k\)
is linear, the remainder will be a constant, \(r\). And, if we evaluate this for \(x=k\), we have
\[\begin{align*} f(k)&=(k−k)q(k)+r \\[4pt] &=0{\cdot}q(k)+r \\[4pt] &=r \end{align*}\]
In other words, \(f(k)\) is the remainder obtained by dividing \(f(x)\)by \(x−k\).
The Remainder Theorem
If a polynomial \(f(x)\) is divided by \(x−k\),then the remainder is the value \(f(k)\).
Given a polynomial function \(f\), evaluate \(f(x)\) at \(x=k\) using the Remainder Theorem.
- Use synthetic division to divide the polynomial by \(x−k\).
- The remainder is the value \(f(k)\).
Example \(\PageIndex{1}\): Using the Remainder Theorem to Evaluate a Polynomial
Use the Remainder Theorem to evaluate \(f(x)=6x^4−x^3−15x^2+2x−7\) at \(x=2\).
Solution
To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by \(x−2\).
\[ 2 \begin{array}{|ccccc} \; 6 & −1 & −15 & 2 & −7 \\ \text{} & 12 & 22 & 14 & 32 \\ \hline \end{array} \\ \begin{array}{ccccc} 6 & 11 & \; 7 & \;\;16 & \;\; 25 \end{array} \]
The remainder is 25. Therefore, \(f(2)=25\).
Analysis
We can check our answer by evaluating \(f(2)\).
\[\begin{align*} f(x)&=6x^4−x^3−15x^2+2x−7 \\ f(2)&=6(2)^4−(2)^3−15(2)^2+2(2)−7 \\ &=25 \end{align*}\]
Exercise \(\PageIndex{1}\)
se the Remainder Theorem to evaluate \(f(x)=2x^5−3x^4−9x^3+8x^2+2\) at \(x=−3\).
- Answer
-
\(f(−3)=−412\)
Using the Factor Theorem to Solve a Polynomial Equation
The Factor Theorem is another theorem that helps us analyze polynomial equations. It tells us how the zeros of a polynomial are related to the factors. Recall that the Division Algorithm.
\[f(x)=(x−k)q(x)+r\]
If \(k\) is a zero, then the remainder \(r\) is \(f(k)=0\) and \(f (x)=(x−k)q(x)+0\) or \(f(x)=(x−k)q(x)\).
Notice, written in this form, \(x−k\) is a factor of \(f(x)\). We can conclude if \(k\) is a zero of \(f(x)\), then \(x−k\) is a factor of \(f(x)\).
Similarly, if \(x−k\) is a factor of \(f(x)\), then the remainder of the Division Algorithm \(f(x)=(x−k)q(x)+r\) is \(0\). This tells us that \(k\) is a zero.
This pair of implications is the Factor Theorem. As we will soon see, a polynomial of degree \(n\) in the complex number system will have \(n\) zeros. We can use the Factor Theorem to completely factor a polynomial into the product of \(n\) factors. Once the polynomial has been completely factored, we can easily determine the zeros of the polynomial.
THE FACTOR THEOREM
According to the Factor Theorem, \(k\) is a zero of \(f(x)\) if and only if \((x−k)\) is a factor of \(f(x)\).
How to: Given a factor and a third-degree polynomial, use the Factor Theorem to factor the polynomial
- Use synthetic division to divide the polynomial by \((x−k)\).
- Confirm that the remainder is \(0\).
- Write the polynomial as the product of \((x−k)\) and the quadratic quotient.
- If possible, factor the quadratic.
- Write the polynomial as the product of factors.
Example \(\PageIndex{2}\): Using the Factor Theorem to Solve a Polynomial Equation
Show that \((x+2)\) is a factor of \(x^3−6x^2−x+30\). Find the remaining factors. Use the factors to determine the zeros of the polynomial .
Solution
We can use synthetic division to show that \((x+2)\) is a factor of the polynomial.
\[ -2 \begin{array}{|cccc} \; 1 & −6 & −1 & 30 \\ \text{} & -2 & 16 & -30 \\ \hline \end{array} \\ \begin{array}{cccc} 1 & -8 & \; 15 & \;\;0 \end{array} \]
The remainder is zero, so \((x+2)\) is a factor of the polynomial. We can use the Division Algorithm to write the polynomial as the product of the divisor and the quotient:
\[(x+2)(x^2−8x+15)\]We can factor the quadratic factor to write the polynomial as
\[(x+2)(x−3)(x−5)\]By the Factor Theorem, the zeros of \(x^3−6x^2−x+30\) are –2, 3, and 5.
Exercise \(\PageIndex{2}\)
Use the Factor Theorem to find the zeros of \(f(x)=x^3+4x^2−4x−16\) given that \((x−2)\) is a factor of the polynomial.
- Answer
-
The zeros are 2, –2, and –4.
Using the Rational Zero Theorem to Find Rational Zeros
Another use for the Remainder Theorem is to test whether a rational number is a zero for a given polynomial. But first we need a pool of rational numbers to test. The Rational Zero Theorem helps us to narrow down the number of possible rational zeros using the ratio of the factors of the constant term and factors of the leading coefficient of the polynomial
Consider a quadratic function with two zeros, \(x=\frac{2}{5}\) and \(x=\frac{3}{4}\). By the Factor Theorem, these zeros have factors associated with them. Let us set each factor equal to 0, and then construct the original quadratic function absent its stretching factor.
Notice that two of the factors of the constant term, 6, are the two numerators from the original rational roots: 2 and 3. Similarly, two of the factors from the leading coefficient, 20, are the two denominators from the original rational roots: 5 and 4.
We can infer that the numerators of the rational roots will always be factors of the constant term and the denominators will be factors of the leading coefficient. This is the essence of the Rational Zero Theorem; it is a means to give us a pool of possible rational zeros.
THE RATIONAL ZERO THEOREM
The Rational Zero Theorem states that, if the polynomial \(f(x)=a_nx^n+a_{n−1}x^{n−1}+...+a_1x+a_0\) has integer coefficients, then every rational zero of \(f(x)\) has the form \(\frac{p}{q}\) where \(p\) is a factor of the constant term \(a_0\) and \(q\) is a factor of the leading coefficient \(a_n\).
When the leading coefficient is 1, the possible rational zeros are the factors of the constant term.
How to: Given a polynomial function \(f(x)\), use the Rational Zero Theorem to find rational zeros.
- Determine all factors of the constant term and all factors of the leading coefficient.
- Determine all possible values of \(\dfrac{p}{q}\), where \(p\) is a factor of the constant term and \(q\) is a factor of the leading coefficient. Be sure to include both positive and negative candidates.
- Determine which possible zeros are actual zeros by evaluating each case of \(f(\frac{p}{q})\).
Example \(\PageIndex{3}\): Listing All Possible Rational Zeros
List all possible rational zeros of \(f(x)=2x^4−5x^3+x^2−4\).
Solution
The only possible rational zeros of \(f(x)\) are the quotients of the factors of the last term, –4, and the factors of the leading coefficient, 2.
The constant term is –4; the factors of –4 are \(p=±1,±2,±4\).
The leading coefficient is 2; the factors of 2 are \(q=±1,±2\).
If any of the four real zeros are rational zeros, then they will be of one of the following factors of –4 divided by one of the factors of 2.
\[\dfrac{p}{q}=±\dfrac{1}{1},±\dfrac{1}{2} \; \; \; \; \; \; \frac{p}{q}=±\dfrac{2}{1},±\dfrac{2}{2} \; \; \; \; \; \; \dfrac{p}{q}=±\dfrac{4}{1},±\dfrac{4}{2} \nonumber\]Note that \(\frac{2}{2}=1\) and \(\frac{4}{2}=2\), which have already been listed. So we can shorten our list.
\[\dfrac{p}{q} = \dfrac{\text{Factors of the last}}{\text{Factors of the first}}=±1,±2,±4,±\dfrac{1}{2}\nonumber \]
Example \(\PageIndex{4}\): Using the Rational Zero Theorem to Find Rational Zeros
Use the Rational Zero Theorem to find the rational zeros of \(f(x)=2x^3+x^2−4x+1\).
Solution
The Rational Zero Theorem tells us that if \(\frac{p}{q}\) is a zero of \(f(x)\), then \(p\) is a factor of 1 and \(q\) is a factor of 2.
\[ \begin{align*} \dfrac{p}{q}=\dfrac{factor\space of\space constant\space term}{factor\space of\space leading\space coefficient} \\[4pt] &=\dfrac{factor\space of\space 1}{factor\space of\space 2} \end{align*}\]
The factors of 1 are ±1 and the factors of 2 are ±1 and ±2. The possible values for \(\frac{p}{q}\) are ±1 and \(±\frac{1}{2}\). These are the possible rational zeros for the function. We can determine which of the possible zeros are actual zeros by substituting these values for \(x\) in \(f(x)\).
\[f(−1)=2{(−1)}^3+{(−1)}^2−4(−1)+1=4\]
\[f(1)=1{(1)}^3+{(1)}^2−4(1)+1=0\]\[f(−\dfrac{1}{2})=2{(−\dfrac{1}{2})}^3+{(−\dfrac{1}{2})}^2−4(−\dfrac{1}{2})+1=3\]
\[f(\dfrac{1}{2})=2{(\dfrac{1}{2})}^3+{(\dfrac{1}{2})}^2−4(\dfrac{1}{2})+1=−\dfrac{1}{2}\]Of those, \(−1\),\(−\dfrac{1}{2}\), and \(\dfrac{1}{2}\) are not zeros of \(f(x)\). 1 is the only rational zero of \(f(x)\).
Exercise \(\PageIndex{3}\)
Use the Rational Zero Theorem to find the rational zeros of \(f(x)=x^3−5x^2+2x+1\).
- Answer
-
There are no rational zeros.
Finding the Zeros of Polynomial Functions
The Rational Zero Theorem helps us to narrow down the list of possible rational zeros for a polynomial function. Once we have done this, we can use synthetic division repeatedly to determine all of the zeros of a polynomial function.
How to: Given a polynomial function \(f\), use synthetic division to find its zeros.
- Use the Rational Zero Theorem to list all possible rational zeros of the function.
- Use synthetic division to evaluate a given possible zero by synthetically dividing the candidate into the polynomial. If the remainder is 0, the candidate is a zero. If the remainder is not zero, discard the candidate.
- Repeat step two using the quotient found with synthetic division. If possible, continue until the quotient is a quadratic.
- Find the zeros of the quadratic function. Two possible methods for solving quadratics are factoring and using the quadratic formula.
Example \(\PageIndex{5}\): Finding the Zeros of a Polynomial Function with Repeated Real Zeros
Find the zeros of \(f(x)=4x^3−3x−1\).
Solution
The Rational Zero Theorem tells us that if \(\dfrac{p}{q}\) is a zero of \(f(x)\), then \(p\) is a factor of –1 and \(q\) is a factor of 4.
\[\begin{align*}\dfrac{p}{q}=\dfrac{factor\space of\space constant\space term}{factor\space of\space leading\space coefficient} \\[4pt] =\dfrac{factor\space of\space -1}{factor\space of\space 4} \end{align*}\]
The factors of –1 are ±1 and the factors of 4 are ±1,±2, and ±4. The possible values for \(\dfrac{p}{q}\) are \(±1\),\(±\dfrac{1}{2}\), and \(±\dfrac{1}{4}\). These are the possible rational zeros for the function. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with 1.
Dividing by \((x−1)\) gives a remainder of 0, so 1 is a zero of the function. The polynomial can be written as
\[(x−1)(4x^2+4x+1) \nonumber\]
The quadratic is a perfect square. \(f(x)\) can be written as
\[(x−1){(2x+1)}^2\nonumber\]
We already know that 1 is a zero. The other zero will have a multiplicity of 2 because the factor is squared. To find the other zero, we can set the factor equal to 0.
\[ \begin{align*} 2x+1=0 \\[4pt] x &=−\dfrac{1}{2} \end{align*}\]
The zeros of the function are 1 and \(−\frac{1}{2}\) with multiplicity 2.
AnalysisLook at the graph of the function \(f\) in Figure \(\PageIndex{1}\). Notice, at \(x =−0.5\), the graph bounces off the x-axis, indicating the even multiplicity (2,4,6…) for the zero −0.5. At \(x=1\), the graph crosses the x-axis, indicating the odd multiplicity (1,3,5…) for the zero \(x=1\).
Using the Fundamental Theorem of Algebra
Now that we can find rational zeros for a polynomial function, we will look at a theorem that discusses the number of complex zeros of a polynomial function. The Fundamental Theorem of Algebra tells us that every polynomial function has at least one complex zero. This theorem forms the foundation for solving polynomial equations.
Suppose \(f\) is a polynomial function of degree four, and \(f (x)=0\). The Fundamental Theorem of Algebra states that there is at least one complex solution, call it \(c_1\). By the Factor Theorem, we can write \(f(x)\) as a product of \(x−c_1\) and a polynomial quotient. Since \(x−c_1\) is linear, the polynomial quotient will be of degree three. Now we apply the Fundamental Theorem of Algebra to the third-degree polynomial quotient. It will have at least one complex zero, call it \(c_2\). So we can write the polynomial quotient as a product of \(x−c_2\) and a new polynomial quotient of degree two. Continue to apply the Fundamental Theorem of Algebra until all of the zeros are found. There will be four of them and each one will yield a factor of \(f(x)\).
THE FUNDAMENTAL THEOREM OF ALGEBRA
The Fundamental Theorem of Algebra states that, if \(f(x)\) is a polynomial of degree \(n > 0\), then \(f(x)\) has at least one complex zero.
We can use this theorem to argue that, if \(f(x)\) is a polynomial of degree \(n >0\), and a is a non-zero real number, then \(f(x)\) has exactly \(n\) linear factors
\[f(x)=a(x−c_1)(x−c_2)...(x−c_n)\]where \(c_1,c_2\),...,\(c_n\) are complex numbers. Therefore, \(f(x)\) has \(n\) roots if we allow for multiplicities.
Q&A: Does every polynomial have at least one imaginary zero?
No. Real numbers are a subset of complex numbers, but not the other way around. A complex number is not necessarily imaginary. Real numbers are also complex numbers.
Example \(\PageIndex{6}\): Finding the Zeros of a Polynomial Function with Complex Zeros
Find the zeros of \(f(x)=3x^3+9x^2+x+3\).
Solution
The Rational Zero Theorem tells us that if \(\frac{p}{q}\) is a zero of \(f(x)\), then \(p\) is a factor of 3 and \(q\) is a factor of 3.
\[ \begin{align*} \dfrac{p}{q}=\dfrac{factor\space of\space constant\space term}{factor\space of\space leading\space coefficient} \\[4pt] &=\dfrac{factor\space of\space 3}{factor\space of\space 3} \end{align*}\]
The factors of 3 are ±1 and ±3. The possible values for \(\dfrac{p}{q}\), and therefore the possible rational zeros for the function, are ±3,±1, and \(±\dfrac{1}{3}\). We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with –3.
Dividing by \((x+3)\) gives a remainder of 0, so –3 is a zero of the function. The polynomial can be written as
\[(x+3)(3x^2+1) \nonumber\]We can then set the quadratic equal to 0 and solve to find the other zeros of the function.
\[ \begin{align*} 3x^2+1=0 \\[4pt] x^2 &=−\dfrac{1}{3} \\[4pt] x&=±−\sqrt{\dfrac{1}{3}} \\[4pt] &=±\dfrac{i\sqrt{3}}{3} \end{align*}\]The zeros of \(f(x)\) are \(–3\) and \(±\dfrac{i\sqrt{3}}{3}\).
AnalysisLook at the graph of the function \(f\) in Figure \(\PageIndex{2}\). Notice that, at \(x =−3\), the graph crosses the x-axis, indicating an odd multiplicity (1) for the zero \(x=–3\). Also note the presence of the two turning points. This means that, since there is a \(3^{rd}\) degree polynomial, we are looking at the maximum number of turning points. So, the end behavior of increasing without bound to the right and decreasing without bound to the left will continue. Thus, all the x-intercepts for the function are shown. So either the multiplicity of \(x=−3\) is 1 and there are two complex solutions, which is what we found, or the multiplicity at \(x =−3\) is three. Either way, our result is correct.
\(\PageIndex{4}\)
Find the zeros of \(f(x)=2x^3+5x^2−11x+4\).
Solution
The zeros are \(–4\), \(\frac{1}{2}\), and \(1\).
Using the Linear Factorization Theorem to Find Polynomials with Given Zeros
A vital implication of the Fundamental Theorem of Algebra , as we stated above, is that a polynomial function of degree n will have \(n\) zeros in the set of complex numbers, if we allow for multiplicities. This means that we can factor the polynomial function into \(n\) factors. The Linear Factorization Theorem tells us that a polynomial function will have the same number of factors as its degree, and that each factor will be in the form \((x−c)\), where c is a complex number.
Let \(f\) be a polynomial function with real coefficients, and suppose \(a +bi\), \(b≠0\), is a zero of \(f(x)\). Then, by the Factor Theorem, \(x−(a+bi)\) is a factor of \(f(x)\). For \(f\) to have real coefficients, \(x−(a−bi)\) must also be a factor of \(f(x)\). This is true because any factor other than \(x−(a−bi)\), when multiplied by \(x−(a+bi)\), will leave imaginary components in the product. Only multiplication with conjugate pairs will eliminate the imaginary parts and result in real coefficients. In other words, if a polynomial function \(f\) with real coefficients has a complex zero \(a +bi\), then the complex conjugate \(a−bi\) must also be a zero of \(f(x)\). This is called the Complex Conjugate Theorem .
COMPLEX CONJUGATE THEOREM
According to the Linear Factorization Theorem, a polynomial function will have the same number of factors as its degree, and each factor will be in the form \((x−c)\), where \(c\) is a complex number.
If the polynomial function \(f\) has real coefficients and a complex zero in the form \(a+bi\), then the complex conjugate of the zero, \(a−bi\), is also a zero.
How to
Given the zeros of a polynomial function \(f\) and a point \((c, f(c))\) on the graph of \(f\), use the Linear Factorization Theorem to find the polynomial function.
- Use the zeros to construct the linear factors of the polynomial.
- Multiply the linear factors to expand the polynomial.
- Substitute \((c,f(c))\) into the function to determine the leading coefficient.
- Simplify.
Example \(\PageIndex{7}\): Using the Linear Factorization Theorem to Find a Polynomial with Given Zeros
Find a fourth degree polynomial with real coefficients that has zeros of \(–3\), \(2\), \(i\), such that \(f(−2)=100\).
Solution
Because \(x =i\) is a zero, by the Complex Conjugate Theorem \(x =–i\) is also a zero. The polynomial must have factors of \((x+3),(x−2),(x−i)\), and \((x+i)\). Since we are looking for a degree 4 polynomial, and now have four zeros, we have all four factors. Let’s begin by multiplying these factors.
\[\begin{align} f(x) & =a(x+3)(x−2)(x−i)(x+i) \\ f(x) & =a(x^2+x−6)(x^2+1) \\ f(x) & =a(x^4+x^3−5x^2+x−6) \end{align} \]We need to find \(a\) to ensure \(f(–2)=100\). Substitute \(x=–2\) and \(f (-2)=100\) into \(f (x)\).
\[\begin{align} 100=a({(−2)}^4+{(−2)}^3−5{(−2)}^2+(−2)−6) \\ 100=a(−20) \\ −5=a \end{align} \]So the polynomial function is
\[f(x)=−5(x^4+x^3−5x^2+x−6)\]or
\[f(x)=−5x^4−5x^3+25x^2−5x+30\] AnalysisWe found that both \(i\) and \(−i\) were zeros, but only one of these zeros needed to be given. If \(i\) is a zero of a polynomial with real coefficients, then \(−i\) must also be a zero of the polynomial because \(−i\) is the complex conjugate of \(i\).
Q&A
If \(2+3i\) were given as a zero of a polynomial with real coefficients, would \(2−3i\) also need to be a zero?
Yes. When any complex number with an imaginary component is given as a zero of a polynomial with real coefficients, the conjugate must also be a zero of the polynomial.
\(\PageIndex{5}\)
Find a third degree polynomial with real coefficients that has zeros of \(5\) and \(−2i\) such that \(f (1)=10\).
Solution
\(f(x)=−\frac{1}{2}x^3+\frac{5}{2}x^2−2x+10\)
Using Descartes’ Rule of Signs
There is a straightforward way to determine the possible numbers of positive and negative real zeros for any polynomial function. If the polynomial is written in descending order, Descartes’ Rule of Signs tells us of a relationship between the number of sign changes in \(f(x)\) and the number of positive real zeros. For example, the polynomial function below has one sign change.
This tells us that the function must have 1 positive real zero.
There is a similar relationship between the number of sign changes in \(f(−x)\) and the number of negative real zeros.
In this case, \(f(−x)\) has 3 sign changes. This tells us that \(f(x)\) could have 3 or 1 negative real zeros.
DESCARTES’ RULE OF SIGNS
According to Descartes’ Rule of Signs , if we let \(f(x)=a_nx^n+a_{n−1}x^{n−1}+...+a_1x+a_0\) be a polynomial function with real coefficients:
- The number of positive real zeros is either equal to the number of sign changes of \(f(x)\) or is less than the number of sign changes by an even integer.
- The number of negative real zeros is either equal to the number of sign changes of \(f(−x)\) or is less than the number of sign changes by an even integer.
Example \(\PageIndex{8}\): Using Descartes’ Rule of Signs
Use Descartes’ Rule of Signs to determine the possible numbers of positive and negative real zeros for \(f(x)=−x^4−3x^3+6x^2−4x−12\).
Solution
Begin by determining the number of sign changes.
There are two sign changes, so there are either 2 or 0 positive real roots. Next, we examine \(f(−x)\) to determine the number of negative real roots.
\[ \begin{align} f(−x) & =−{(−x)}^4−3{(−x)}^3+6{(−x)}^2−4(−x)−12 \\ f(−x) & =−x^4+3x^3+6x^2+4x−12 \end{align} \]
Again, there are two sign changes, so there are either 2 or 0 negative real roots.
There are four possibilities, as we can see in Table \(\PageIndex{1}\).
| Positive Real Zeros | Negative Real Zeros | Complex Zeros | Total Zeros |
|---|---|---|---|
| 2 | 2 | 0 | 4 |
| 2 | 0 | 2 | 4 |
| 0 | 2 | 2 | 4 |
| 0 | 0 | 4 | 4 |
We can confirm the numbers of positive and negative real roots by examining a graph of the function. See Figure \(\PageIndex{3}\). We can see from the graph that the function has 0 positive real roots and 2 negative real roots.
\(\PageIndex{6}\)
Use Descartes’ Rule of Signs to determine the maximum possible numbers of positive and negative real zeros for \(f(x)=2x^4−10x^3+11x^2−15x+12\). Use a graph to verify the numbers of positive and negative real zeros for the function.
Solution
There must be 4, 2, or 0 positive real roots and 0 negative real roots. The graph shows that there are 2 positive real zeros and 0 negative real zeros.
Solving Real-World Applications
We have now introduced a variety of tools for solving polynomial equations. Let’s use these tools to solve the bakery problem from the beginning of the section.
Example \(\PageIndex{9}\)
A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be?
Solution
Begin by writing an equation for the volume of the cake. The volume of a rectangular solid is given by \(V=lwh\). We were given that the length must be four inches longer than the width, so we can express the length of the cake as \(l=w+4\). We were given that the height of the cake is one-third of the width, so we can express the height of the cake as \(h=\dfrac{1}{3}w\). Let’s write the volume of the cake in terms of width of the cake.
\[V=(w+4)(w)(\dfrac{1}{3}w)\] \[V=\dfrac{1}{3}w^3+\dfrac{4}{3}w^2\]Substitute the given volume into this equation.
\(351=13w^3+43w^2\) Substitute 351 for V. \(1053=w^3+4w^2\) Multiply both sides by 3. \(0=w^3+7w^2−1053\) Subtract 1053 from both sides.Descartes' rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are \(\pm 1,±3,±9,±13,±27,±39,±81,±117,±351,\) and \(±1053\). We can use synthetic division to test these possible zeros. Only positive numbers make sense as dimensions for a cake, so we need not test any negative values. Let’s begin by testing values that make the most sense as dimensions for a small sheet cake. Use synthetic division to check \(x=1\).
Since 1 is not a solution, we will check \(x=3\).
Since 3 is not a solution either, we will test \(x=9\).
Synthetic division gives a remainder of 0, so 9 is a solution to the equation. We can use the relationships between the width and the other dimensions to determine the length and height of the sheet cake pan.
\(l=w+4=9+4=13\) and \(h=\dfrac{1}{3}w=\dfrac{1}{3}(9)=3\)The sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches
\(\PageIndex{7}\)
A shipping container in the shape of a rectangular solid must have a volume of 84 cubic meters. The client tells the manufacturer that, because of the contents, the length of the container must be one meter longer than the width, and the height must be one meter greater than twice the width. What should the dimensions of the container be?
Solution
3 meters by 4 meters by 7 meters
Media
Access these online resources for additional instruction and practice with zeros of polynomial functions.
Key Concepts
- To find \(f(k)\), determine the remainder of the polynomial \(f(x)\) when it is divided by \(x−k\). This is known as the Remainder Theorem. See Example \(\PageIndex{1}\) .
- According to the Factor Theorem, \(k\) is a zero of \(f(x)\) if and only if \((x−k)\) is a factor of \(f(x)\). See Example\(\PageIndex{2}\) .
- According to the Rational Zero Theorem, each rational zero of a polynomial function with integer coefficients will be equal to a factor of the constant term divided by a factor of the leading coefficient. See Example \(\PageIndex{3}\) and Example \(\PageIndex{4}\) .
- When the leading coefficient is 1, the possible rational zeros are the factors of the constant term.
- Synthetic division can be used to find the zeros of a polynomial function. See Example \(\PageIndex{5}\) .
- According to the Fundamental Theorem, every polynomial function with degree greater than 0 has at least one complex zero. See Example \(\PageIndex{6}\) .
- Allowing for multiplicities, a polynomial function will have the same number of factors as its degree. Each factor will be in the form \((x−c)\), where \(c\) is a complex number. See Example \(\PageIndex{7}\) .
- The number of positive real zeros of a polynomial function is either the number of sign changes of the function or less than the number of sign changes by an even integer.
- The number of negative real zeros of a polynomial function is either the number of sign changes of \(f(−x)\) or less than the number of sign changes by an even integer. See Example \(\PageIndex{8}\) .
- Polynomial equations model many real-world scenarios. Solving the equations is easiest done by synthetic division. See Example \(\PageIndex{9}\) .
Glossary
Descartes’ Rule of Signs
a rule that determines the maximum possible numbers of positive and negative real zeros based on the number of sign changes of \(f(x)\) and \(f(−x)\)
Factor Theorem
\(k\) is a zero of polynomial function \(f(x)\) if and only if \((x−k)\) is a factor of \(f(x)\)
Fundamental Theorem of Algebra
a polynomial function with degree greater than 0 has at least one complex zero
Linear Factorization Theorem
allowing for multiplicities, a polynomial function will have the same number of factors as its degree, and each factor will be in the form \((x−c)\), where \(c\) is a complex number
Rational Zero Theorem
the possible rational zeros of a polynomial function have the form \(\frac{p}{q}\) where \(p\) is a factor of the constant term and \(q\) is a factor of the leading coefficient.
Remainder Theorem
if a polynomial \(f(x)\) is divided by \(x−k\),then the remainder is equal to the value \(f(k)\)


5: Radiation and Spectra
The nearest star is so far away that the fastest spacecraft humans have built would take almost 100,000 years to get there. Yet we very much want to know what material this neighbor star is composed of and how it differs from our own Sun. How can we learn about the chemical makeup of stars that we cannot hope to visit or sample?
In astronomy, most of the objects that we study are completely beyond our reach. The temperature of the Sun is so high that a spacecraft would be fried long before it reached it, and the stars are much too far away to visit in our lifetimes with the technology now available. Even light, which travels at a speed of 300,000 kilometers per second (km/s), takes more than 4 years to reach us from the nearest star. If we want to learn about the Sun and stars, we must rely on techniques that allow us to analyze them from a distance.
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- 5.2: The Behavior of Light
- James Clerk Maxwell showed that whenever charged particles change their motion, as they do in every atom and molecule, they give off waves of energy. Light is one form of this electromagnetic radiation. The wavelength of light determines the color of visible radiation. Wavelength (λ) is related to frequency (f) and the speed of light (c) by the equation c = λf. Electromagnetic radiation sometimes behaves like waves, but at other times, it behaves as if it were a particle- called a photon.
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- 5.3: The Electromagnetic Spectrum
- The electromagnetic spectrum consists of gamma rays, X-rays, ultraviolet radiation, visible light, infrared, and radio radiation. Many of these wavelengths cannot penetrate the layers of Earth’s atmosphere and must be observed from space, whereas others—such as visible light, FM radio and TV—can penetrate to Earth’s surface. The emission of electromagnetic radiation is intimately connected to the temperature of the source.
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- 5.4: Spectroscopy in Astronomy
- A spectrometer is a device that forms a spectrum, often utilizing the phenomenon of dispersion. The light from an astronomical source can consist of a continuous spectrum, an emission (bright line) spectrum, or an absorption (dark line) spectrum. Because each element leaves its spectral signature in the pattern of lines we observe, spectral analyses reveal the composition of the Sun and stars.
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- 5.5: The Structure of the Atom
- Atoms consist of a nucleus containing one or more positively charged protons. All atoms except hydrogen can also contain one or more neutrons in the nucleus. Negatively charged electrons orbit the nucleus. The number of protons defines an element (hydrogen has one proton, helium has two, and so on) of the atom. Nuclei with the same number of protons but different numbers of neutrons are different isotopes of the same element.
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- 5.6: Formation of Spectral Lines
- When electrons move from a higher energy level to a lower one, photons are emitted, and an emission line can be seen in the spectrum. Absorption lines are seen when electrons absorb photons and move to higher energy levels. Since each atom has its own characteristic set of energy levels, each is associated with a unique pattern of spectral lines. This allows astronomers to determine what elements are present in the stars and in the clouds of gas and dust among the stars.
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- 5.7: The Doppler Effect
- If an atom is moving toward us when an electron changes orbits and produces a spectral line, we see that line shifted slightly toward the blue of its normal wavelength in a spectrum. If the atom is moving away, we see the line shifted toward the red. This shift is known as the Doppler effect and can be used to measure the radial velocities of distant objects.
Thumbnail: This photograph of the Sun was taken at several different wavelengths of ultraviolet, which our eyes cannot see, and then color coded so it reveals activity in our Sun’s atmosphere that cannot be observed in visible light. This is why it is important to observe the Sun and other astronomical objects in wavelengths other than the visible band of the spectrum. This image was taken by a satellite from above Earth’s atmosphere, which is necessary since Earth’s atmosphere absorbs much of the ultraviolet light coming from space. (credit: modification of work by NASA).


20.3: Applying Kirchhoff’s rule to model circuits
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In this section, we show how to model a circuit using Kirchhoff’s rules. In general, one can consider a circuit to be fully modeled if one can determine the current in each segment of the circuit. We will show how one can apply the same procedure to model any circuit that contains batteries and resistors. The procedure is as follows:
- Make a good diagram of the circuit.
- Simplify any resistors that can easily be combined into effective resistors (in series or in parallel).
- Make a new diagram with the effective resistors, showing battery arrows, and labeling all of the nodes so that loops can easily be described.
- Make a guess for the directions of the current in each segment.
- Write the junction rule equations.
- Write the loop equations.
- This will lead to \(N\) independent equations that one can solve for the \(N\) different currents in the circuit.
- Once you have determined all of the currents, if some of them are negative numbers, switch the direction of those currents in the diagram (they will be negative if you guessed the direction incorrectly).
We will illustrate the procedure on the circuit shown in Figure \(\PageIndex{1}\), for which we would like to know the current through each resistor and each battery. The circuit contains 5 resistors ( \(R_1\) - \(R_5\) ), 2 real batteries (with ideal voltages \(\Delta V_1\) and \(\Delta V_2\) ), and 2 additional resistors to model the internal resistances of the real batteries ( \(r_1\) , \(r_2\) )
Exercise \(\PageIndex{1}\)
How many different currents are in the circuit shown in Figure \(\PageIndex{1}\)?
- 3
- 4
- 5
- 6
- Answer
Simplifying the resistors (step 2): In this circuit, resistors \(r_2\) , \(R_1\) and \(R_2\) are in series, so that they can be combined into an effective resistor, \(R_6\) :
\[\begin{aligned} R_6=r_2+R_1+R_2\end{aligned}\]
With this simplification, we obtain the circuit illustrated in Figure \(\PageIndex{2}\)
Next, we note that resistors \(R_4\) and \(R_5\) are in parallel and can be easily combined into a resistor, \(R_7\) :
\[\begin{aligned} R_7=\frac{R_4R_5}{R_4+R_5}\end{aligned}\]
which leads to the circuit illustrated in Figure \(\PageIndex{3}\).
Finally, we note that \(r_1\) and \(R_7\) are in series and can be combined into an effective resistor, \(R_8\) :
\[\begin{aligned} R_8=r_1+R_7=r_1+\frac{R_4R_5}{R_4+R_5}\end{aligned}\]
leading to the simplified circuit illustrated in Figure \(\PageIndex{2}\), which we have labeled with nodes and battery labels.
Guessing the directions of the currents (step 4): Before we can write the equations from Kirchhoff’s rules, we need to label the currents in the circuit diagram. In general, it is not always obvious in which way the currents will go, so we make a guess that we can fix later if we guessed wrong.
In order to guess the current directions, choose one point on the circuit and move along a segment. Label the current in that segment and continue moving through the circuit, splitting up the current when a junction is encountered. Make sure to only have one current per segment. We guess the currents as follows, referring to Figure \(\PageIndex{5}\):
- We start at point \(a\) and move upwards to point \(f\) . We will call the current in that segment, \(I_1\) .
- Since there is no junction, the current \(I_1\) continues through the resistor \(R_8\) to point \(e\) .
- There is a junction at point \(e\) , so we split the current \(I_1\) into currents \(I_2\) (towards point \(d\) ), and \(I_3\) (downwards to point \(b\) ).
- We follow current \(I_2\) first; \(I_2\) flows from \(e\) to \(d\) , then down to \(c\) , through the battery \(\Delta V_2\) , and to point \(b\) , where there is again junction.
- We follow current \(I_3\) , which just flows down to the junction at point \(b\) , where it “meets up” with current \(I_2\) .
- Currents \(I_2\) and \(I_3\) both flow into the junction at point \(b\) , and the current flowing out of the junction, through the battery \(\Delta V_1\) , and towards point \(a\) is, again, \(I_1\) , since this current then flows up to point \(f\) .
- All segments of wire have a labeled current, so we are done guessing currents.
You can proceed in an analogous way for any circuit. The final circuit, with currents labeled, is shown in Figure \(\PageIndex{5}\):
We can now proceed with using Kirchhoff’s rules to solve for the values of the currents in the circuit. It is useful to note that there are 3 unknown currents in this circuit; we thus hope that Kirchhoff’s rules will give us 3 independent equations.
Applying the junction rule (step 5): In the circuit from Figure \(\PageIndex{5}\), there are two junctions (at points \(b\) and \(e\) ), so we will get two equations from the junction rule. To apply the junction rule, the sum of the currents coming into the junction must be equal to the currents going out of the junction:
\[\begin{aligned} \text{incoming currents}&=\text{outgoing currents}&\\[4pt] I_2+I_3 &= I_1 \quad &\text{(junction $b$)}\\[4pt] I_1 &= I_2+I_3 \quad &\text{(junction $e$)}\\[4pt]\end{aligned}\]
Note that the two equations are not independent (in fact, they are the same). It is generally the case that if there \(N\) junctions, one will obtain less than \(N\) independent equations (usually, \(N-1\) equations will be independent). In this case, the two junctions only gave us one equation.
Applying the loop rule (step 6): This circuit contains 3 different loops: \(abcdefa\) , \(abefa\) , and \(bcdeb\) , which will lead to 3 equations from the loop rule. We expect that these equations will not be independent, since this would lead to 4 equations and 3 unknowns when combined with the junction rule equation. Let us start with the loop \(abcdefa\) :
- From \(a\) to \(b\) , we trace through the battery in the opposite direction from the battery arrow : \(-\Delta V_1\) .
- From \(b\) to \(c\) , we trace through the battery in the same direction as the battery arrow : \(+\Delta V_2\) .
- From \(c\) through \(d\) and through to \(e\) we go through the resistor \(R_6\) in the opposite direction from the current , \(I_2\) , in that resistor: \(+I_2R_6\) .
- From \(e\) to \(f\) , we go through the go through the resistor \(R_8\) in the opposite direction from the current , \(I_1\) , in that resistor: \(+I_1R_8\) .
- And we are back at the starting point, so the sum of the above terms is equal to zero.
which gives the equation:
\[\begin{aligned} -\Delta V_1+\Delta V_2+I_2R_6+I_1R_8=0\quad\text{(loop abcdefa)}\end{aligned}\]
Similarly, for the loop \(abefa\) , we obtain:
\[\begin{aligned} -\Delta V_1+I_3R_3+I_1R_8=0\quad\text{(loop abefa)}\end{aligned}\]
and for loop \(bcdeb\) :
\[\begin{aligned} \Delta V_2+I_2R_6-I_3R_3=0\quad\text{(loop bcdeb)}\end{aligned}\]
Although it appears that we have obtained 3 additional equations, only two of these are independent. For example, if you sum the second and third equations (loops \(abefa\) , and \(bcdeb\) ), you simply obtain the first equation (loop \(abcdefa\) ). In general, if there are \(N\) different loops, one will obtain less than \(N\) independent equations (usually \(N-1\) independent equations, as we did here).
At this point, after choosing one of the junction equations, and two of the loop equations, we have 3 independent equations that we can solve for the 3 unknown currents 1 :
\[\begin{aligned} I_1 &= I_2+I_3 \quad &\text{(junction $e$)}\\[4pt] -\Delta V_1+\Delta V_2+I_2R_6+I_1R_8&=0\quad&\text{(loop abcdefa)}\\[4pt] -\Delta V_1+I_3R_3+I_1R_8&=0\quad&\text{(loop abefa)}\end{aligned}\]
It is only a matter of some simple math to solve for the 3 unknowns from these 3 equations (which we carry out in the example below).
Example \(\PageIndex{1}\)
Referring to the circuit in Figure \(\PageIndex{6}\), what is the voltage across the real terminal of the battery with ideal voltage \(\Delta V_1\) (the voltage between points \(a\) and \(b\) )? What is the current through resistor \(R_5\) ?
Solution
Since this circuit is the same that we just analyzed, we know that it can be simplified into the circuit shown in Figure \(\PageIndex{7}\), with resistors:
\[\begin{aligned} R_6&=r_2+R_1+R_2=(1\Omega)+(3\Omega)+(4\Omega)=8\Omega\\[4pt] R_8&=r_1+\frac{R_4R_5}{R_4+R_5}=(1\Omega)+\frac{(2\Omega)(2\Omega)}{(2\Omega)+(2\Omega)}=2\Omega\end{aligned}\]
From above, we know that this leads to the following three equations:
\[\begin{aligned} I_1 &= I_2+I_3 \quad &\text{(junction $e$)}\\[4pt] -\Delta V_1+\Delta V_2+I_2R_6+I_1R_8&=0\quad&\text{(loop abcdefa)}\\[4pt] -\Delta V_1+I_3R_3+I_1R_8&=0\quad&\text{(loop abefa)}\end{aligned}\]
In order to solve these types of equations, it is usually convenient to place the battery voltages on the right hand side, and the resistor voltages on the left hand side. Although it is generally bad practice to fill numbers into the equations before solving them, it is almost always a good idea when solving the \(N\) equations for the \(N\) currents. Furthermore, in order to make the equations legible, it is also useful to not write in the units (which is very bad practice in general!). Thus, filling in the values for the resistors and the battery voltages, moving the voltages to the right hand side, we obtain the following system of equations:
\[\begin{aligned} I_1-I_2-I_3&=0 \quad &\text{(junction $e$)}\\[4pt] 2I_1+8I_2&=8 \quad&\text{(loop abcdefa)}\\[4pt] 2I_1+4I_3&=12 \quad&\text{(loop abefa)}\end{aligned}\]
Subtracting the second equation from the third equation (to eliminate \(I_1\) ):
\[\begin{aligned} 4I_3-8I_2&=4\\[4pt] \therefore I_3&=1+2I_2\end{aligned}\]
Substituting this into the junction equation:
\[\begin{aligned} I_1-I_2-I_3&=0\\[4pt] I_1-I_2-1-2I_2&=0\\[4pt] \therefore I_2=\frac{1}{3}(I_1-1)\end{aligned}\]
Finally, substituting this into the equation from loop \(abcdefa\) , allows us to determine \(I_1\) and the other two currents:
\[\begin{aligned} 2I_1+8I_2&=8\\[4pt] 2I_1+8\left(\frac{1}{3}(I_1-1) \right)&=8\\[4pt] \therefore I_1&=\frac{16}{7}=2.29\text{A}\\[4pt] \therefore I_2&=\frac{1}{3}(I_1-1)=0.43\text{A}\\[4pt] \therefore I_3&=1+2I_2=1.86\text{A}\\[4pt]\end{aligned}\]
In this case, the currents are all positive, so the diagram in Figure \(\PageIndex{7}\) is correct and we do not need to reverse the direction of any of the currents.
We can now determine the potential difference across the real terminals of the battery \(\Delta V_1\) . The current through the battery is \(I_1=2.29\text{A}\) , which cause a voltage drop, \(\Delta V_{r1}\) , across its internal resistance, \(r_1\) of:
\[\begin{aligned} \Delta V_{r1}=I_1r_1=(2.29\text{A})(1\Omega)=2.29\text{V}\end{aligned}\]
The voltage across the real terminals of the battery is then:
\[\begin{aligned} \Delta V_{real}=\Delta V_1-\Delta V_{r1}=(12\text{V})-(2.29\text{V})=9.7\text{V}\end{aligned}\]
The current through the resistor \(R_5\) (Figure \(\PageIndex{6}\)) requires a little more thought, since we calculated the current, \(I_1\) through the effective resistor \(R_8\) , which we must now “break apart”. Figure \(\PageIndex{8}\) shows the components of \(R_8\) .
The current \(I_1\) , that goes through the \(\Delta V_1\) battery also goes through the \(r_1\) internal resistance of the battery. That current then splits up into currents, \(I_4\) and \(I_5\) , to go through the resistors \(R_4\) and \(R_5\) . Although it should be obvious that half of \(I_1\) will go through each resistor (since these are equal), we can determine this from applying Kirchhoff’s rules to the combination of resistors in Figure \(\PageIndex{8}\):
\[\begin{aligned} I_1&=I_4+I_5 \quad&\text{(junction)}\\[4pt] I_5R_5-I_4R_4&=0\quad&\text{(clockwise loop)}\end{aligned}\]
From the loop equation, we have:
\[\begin{aligned} I_5=\frac{R_4}{R_5}I_4=I_4\end{aligned}\]
since \(R_4=R_5=2\Omega\) . Since \(I_4=I_5\) , the junction equation gives:
\[\begin{aligned} I_5=\frac{1}{2}I_1=1.15\text{A}\end{aligned}\]
By solving for \(I_4\) and \(I_5\) , we have now determined all of the currents through all of the segments of the original circuit in Figure \(\PageIndex{6}\).
Discussion
In this example, we showed how one can use a simplified circuit to solve the current through the effective resistors in the simplified circuit. Once those currents are known, we showed that it is straightforward to determine the currents through individual resistors that have been combined into effective resistors.
josh's thoughts
Solving a circuit can be daunting, especially if the diagram is drawn in an unfamiliar way. While the circuits in this chapter are designed to be as easy to read as possible, many circuits are much more strange. For example, here is a circuit which you may come across:
The circuit in Figure \(\PageIndex{9}\) May look like it is a difficult circuit to solve, but the diagram can be re-drawn to reveal the simplicity of the circuit:
What used to be a strange kite shape is now just a parallel circuit, which can be further simplified by calculating the effective resistance:
\[\begin{aligned} R_{eff} &= (R_1^{-1}+R_2^{-1}+(R_3+R_4)^{-1})^{-1}\end{aligned}\]
Which gives a series circuit with only one resistor:
Circuits can be drawn in many unique or potentially confusing ways, but knowing how to read the circuit and re-draw it can help make the diagram more legible and the circuit easier to solve.
Footnotes
1. The 3 unknowns do not necessarily have to be the currents, and could be any combination of the currents, battery voltage and resistors. As long as there at most 3 unknown quantities, this circuit can be solved.


16: Cardiovascular System - Blood
Single-celled organisms do not need blood. They obtain nutrients directly from and excrete wastes directly into their environment. The human organism cannot do that. Our large, complex bodies need blood to deliver nutrients to and remove wastes from our trillions of cells. The heart pumps blood throughout the body in a network of blood vessels. Together, these three components—blood, heart, and vessels—make up the cardiovascular system. (Thumbnail image credit: "Drops of Blood Medium" by unknown author is licensed under CC BY 3.0 )
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- 16.1: Introduction to Blood
- Blood serves as the main transport medium of the body, distributing oxygen and other nutrients, housekeeping and sickness fighting cells, and signalling molecules to control body functions throughout the body. This chapter covers the composition and varied functions of blood, the formation of blood cells in red bone marrow, blood typing and its implications for blood transfusions, and several disorders of blood.
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- 16.2: An Overview of Blood
- Like all connective tissues, blood is made up of cellular elements and an extracellular matrix. The cellular elements—referred to as the formed elements—include red blood cells, white blood cells, and cell fragments called platelets. The extracellular matrix, called plasma, makes blood unique among connective tissues because it is fluid. This fluid, which is mostly water, perpetually suspends the formed elements and enables them to circulate throughout the body within the cardiovascular system.
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- 16.3: Erythrocytes
- The erythrocyte, commonly known as a red blood cell (or RBC), is by far the most common formed element. A single drop of blood contains millions of erythrocytes and just thousands of leukocytes. Specifically, males have about 5.4 million erythrocytes per microliter (µL) of blood, and females have approximately 4.8 million per µL. In fact, erythrocytes are estimated to make up about 25 percent of the total cells in the body.
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- 16.4: Leukocytes and Platelets
- The leukocyte, commonly known as a white blood cell (or WBC), is a major component of the body’s defenses against disease. Leukocytes protect the body against invading microorganisms and body cells with mutated DNA, and they clean up debris. Platelets are essential for the formation of blood clots to prevent blood loss in the event of damage to the blood vessel wall.
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- 16.5: Production of the Formed Elements
- Although one type of leukocyte called memory cells can survive for years, most erythrocytes, leukocytes, and platelets normally live only a few hours to a few weeks. Thus, the body must form new blood cells and platelets quickly and continuously. Your body typically replaces the donated plasma within 24 hours and it takes about 4 to 6 weeks to replace the blood cells. The process by which this replacement occurs is called hemopoiesis, or hematopoiesis.
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- 16.6: Blood Typing
- Blood transfusions in humans were risky procedures until the discovery of the major human blood groups by Karl Landsteiner, an Austrian biologist and physician, in 1900. Until that point, physicians did not understand that death sometimes followed blood transfusions, when the type of donor blood infused into the patient was incompatible with the patient’s own blood. Blood groups are determined by the presence or absence of specific marker molecules on the plasma membranes of erythrocytes.


27.0: Introduction
At age 35, Melissa is without a home and a regular place to sleep at night. She was raised by a single mother who worked multiple waitressing jobs. Melissa worked hard in high school, earned a scholarship to a state college, and took a part-time job to pay for books, fees, and living expenses. One night she was hit by a car and suffered a traumatic back injury that propelled her into a downward spiral of trauma, chronic pain, and medical debt. The medical expenses forced her to drop out of school, and the pain made it impossible for her to keep a full-time job, without which she lacked sufficient income to pay rent. Melissa felt that the health care system stigmatized her because of her medical debt. When she sought relief from her chronic pain, she faced accusations of being a “drug seeker.” Eventually, she gave up on health care and ended up homeless and in despair.
Caring for vulnerable populations and communities, such as individuals like Melissa who are experiencing homelessness, is a key role of the nurse as these populations have multiple risk factors for negative health outcomes. Structural barriers within society and in the health care system place vulnerable populations, also known as disadvantaged groups, at higher risk for poor health.  Multiple social factors, such as living in unsafe neighborhoods, being unable to access quality education, and working in low-wage occupations, put vulnerable populations and communities at a disadvantage, perpetuating their low socioeconomic position . Nurses can meaningfully intervene and improve health outcomes by assessing these populations through the lens of the social determinants of health (SDOH).
This chapter discusses six primary types of disadvantaged populations: individuals experiencing homelessness, veterans, LGBTQIA+ populations, migrant workers, individuals with disabilities, and those who have had adverse childhood experiences. Older adults, immigrants, and refugees are also often considered vulnerable populations; Older adults are discussed in Health Promotion and Maintenance Across the Lifespan, and immigrants and refugees are discussed in Caring for Populations and Communities in Crisis. While no defined set of factors exists to determine whether a population or a community is disadvantaged, one or more of the following characteristics are often used to identify vulnerability (Bhatt & Bathija, 2018):
- Income and education
- Access to primary health care services
- Age
- Gender and sexual orientation
- Social, cultural, and linguistic needs
- Race and ethnicity
- Chronic illness or disability
- Alcohol or substance misuse
- Homelessness
- Human immunodeficiency virus (HIV), acquired immunodeficiency syndrome (AIDS)
- Veteran status
- Migrant status
- Rural residency
Nurses have a duty to ensure equitable and culturally competent care . Disadvantaged populations are often subject to significant health disparities . Recall from Social Determinants Affecting Health Outcomes that health disparities are preventable differences in health between groups of individuals, usually as a result of social or economic factors, geographic location, and environment (Centers for Disease Control and Prevention [CDC], 2022c). Health disparities often result in health inequities , “differences in health status or health resources between different population groups, arising from ... social conditions” (World Health Organization [WHO], 2018, para 2). Health inequities are a result of an unjust system. Since nursing is founded on the principle of social justice , nurses are well positioned to care for vulnerable populations and advocate for systemic changes. Ensuring access to essential health care services, advocating for culturally and linguistically appropriate care, and educating community partners on health needs are steps toward improving the health outcomes of vulnerable populations (Bhatt & Bathija, 2018).


38.20: Muscle Contraction and Locomotion - Control of Muscle Tension
- Describe the factors that control muscle tension
Control of Muscle Tension
Neural control initiates the formation of actin – myosin cross-bridges, leading to the sarcomere shortening involved in muscle contraction. These contractions extend from the muscle fiber through connective tissue to pull on bones, causing skeletal movement. The pull exerted by a muscle is called tension. The amount of force created by this tension can vary, which enables the same muscles to move very light objects and very heavy objects. In individual muscle fibers, the amount of tension produced depends primarily on the amount of cross-bridges formed, which is influenced by the cross-sectional area of the muscle fiber and the frequency of neural stimulation.
Cross-bridges and Tension
The number of cross-bridges formed between actin and myosin determine the amount of tension that a muscle fiber can produce. Cross-bridges can only form where thick and thin filaments overlap, allowing myosin to bind to actin. If more cross-bridges are formed, more myosin will pull on actin and more tension will be produced.
Maximal tension occurs when thick and thin filaments overlap to the greatest degree within a sarcomere. If a sarcomere at rest is stretched past an ideal resting length, thick and thin filaments do not overlap to the greatest degree so fewer cross-bridges can form. This results in fewer myosin heads pulling on actin and less muscle tension. As a sarcomere shortens, the zone of overlap reduces as the thin filaments reach the H zone, which is composed of myosin tails. Because myosin heads form cross-bridges, actin will not bind to myosin in this zone, reducing the tension produced by the myofiber. If the sarcomere is shortened even more, thin filaments begin to overlap with each other, reducing cross-bridge formation even further, and producing even less tension. Conversely, if the sarcomere is stretched to the point at which thick and thin filaments do not overlap at all, no cross-bridges are formed and no tension is produced. This amount of stretching does not usually occur because accessory proteins, internal sensory nerves, and connective tissue oppose extreme stretching.
The primary variable determining force production is the number of myofibers (long muscle cells) within the muscle that receive an action potential from the neuron that controls that fiber. When using the biceps to pick up a pencil, for example, the motor cortex of the brain only signals a few neurons of the biceps so only a few myofibers respond. In vertebrates, each myofiber responds fully if stimulated. On the other hand, when picking up a piano, the motor cortex signals all of the neurons in the biceps so that every myofiber participates. This is close to the maximum force the muscle can produce. As mentioned above, increasing the frequency of action potentials (the number of signals per second) can increase the force a bit more because the tropomyosin is flooded with calcium.
Contributions and Attributions
- intercalated disc. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/intercalated%20disc . License : CC BY-SA: Attribution-ShareAlike
- OpenStax College, Biology. October 23, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44788/latest...ol11448/latest . License : CC BY: Attribution
- OpenStax College, Biology. December 5, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44788/latest...ol11448/latest . License : CC BY: Attribution
- Muscle. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Muscle . License : CC BY-SA: Attribution-ShareAlike
- Animal locomotion. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Animal_locomotion . License : CC BY-SA: Attribution-ShareAlike
- actin. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/actin . License : CC BY-SA: Attribution-ShareAlike
- myosin. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/myosin . License : CC BY-SA: Attribution-ShareAlike
- OpenStax College, Biology. December 5, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44788/latest...ol11448/latest . License : CC BY: Attribution
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Key Points
- The more cross-bridges that are formed, the more tension in the muscle.
- The amount of tension produced depends on the cross-sectional area of the muscle fiber and the frequency of neural stimulation.
- Maximal tension occurs when thick and thin filaments overlap to the greatest degree within a sarcomere; less tension is produced when the sarcomere is stretched.
- If more motor neurons are stimulated, more myofibers contract, and there is greater tension in the muscle.
Key Terms
- tension : condition of being held in a state between two or more forces, which are acting in opposition to each other


6.6: Taking a “Second Look” at Prisoners’ Property
“[T]he items in question have been exposed to police view under unobjectionable circumstances, so that no reasonable expectation of privacy is breached by an officer’s taking a second look . . . ” U.S. v. Grill 1
When an arrestee is booked into jail, officers or jail staff will routinely examine and inventory most or all of his personal property. If they happen to find evidence in the process, it will ordinarily be given to investigators or stored in an evidence room. 2 The rest will be kept in a property room for safekeeping. It might sit there for days, often months or longer.
At some point, investigators might want to take a second look at it. In many cases, they will be looking for something specific, whether it pertains to the crime for which the prisoner was arrested or some other crime. Oftentimes they just want to see if there is anything with evidentiary value that was overlooked when the prisoner was booked. In either case, the question arises: Is a warrant required?
At first glance, it might seem that a warrant would never be necessary because the property is in the lawful possession of a law enforcement agency or detention facility. Thus, the prisoner cannot reasonably expect his property is protected. This may, in fact, be the view of the United States Supreme Court which made the following observation in
U.S. v. Edwards :
[I]t is difficult to perceive what is unreasonable about the police’s examining and holding as evidence those personal effects of the accused that they already have in their lawful custody as the result of a lawful arrest. 3
While this language seems to indicate that a warrant will never be required, elsewhere in Edwards the Court indicated that a warrant might be required in some situations, although it did not elaborate. 4
So, what’s the law? As we will now explain, an analysis of Edwards and other cases leads to the conclusion that a warrant is not required to search an item if there is probable cause to believe it is evidence of a crime. If probable cause does not exist, a warrant is unnecessary if, (1) the item was “subject to search” during booking, and (2) the search was conducted in a reasonable manner.
IF PROBABLE CAUSE EXISTS
If officers have probable cause to believe that an item taken from a prisoner for safekeeping is evidence of a crime, they may seize it without a warrant. This is essentially because a prisoner does not enjoy a reasonable expectation of privacy as to seizable evidence of a crime that is in the lawful possession of a law enforcement agency.
The controlling case on this issue is Edwards 5 which resulted from the defendant’s arrest late one night as he was attempting to break into a post office in Ohio. When Edwards was booked, he was allowed to keep his clothing. Meanwhile, officers at the post office determined that the burglar had unfastened a window with a pry bar and, in the process, left “paint chips on the window sill and wire mesh screen.” Figuring that some of the chips would probably have stuck to the perpetrator’s clothing, they gave Edwards some jail garb and sent his clothes to the lab for analysis. As expected, the lab found bits of paint that matched the paint at the post office.
The Ohio Court of Appeals ruled the seizure of Edwards’ clothing was unlawful because it occurred after he had been booked. The Ohio Court of Appeals ruled the seizure of Edwards’ clothing was unlawful because it occurred without a warrant after the booking process had been completed. The United States Supreme Court disagreed, ruling the search was lawful regardless of when it occurred because the officers had probable cause. Said the Court:
It must be remembered that . . . the police had lawful custody of Edwards and necessarily of the clothing he wore. When it became apparent that the articles of clothing were evidence of the crime for which Edwards was being held, the police were entitled to take, examine, and preserve them for use as evidence, just as they are normally permitted to seize evidence of crime when it is lawfully encountered.
Similarly, in U.S. v. Oaxaca 6 two men were arrested shortly after they robbed a bank in the City of Commerce. When booked into the Los Angeles County Jail, they were allowed to keep their shoes. About six weeks later, investigators took the shoes from them without a warrant in order to compare them with the perpetrators’ shoes as shown in a surveillance video. They matched.
On appeal, the defendants argued the shoes should have been suppressed because the investigators did not have a warrant. The Ninth Circuit disagreed, saying:
Both the defendants and their shoes remained in lawful custody until the time when the shoes were taken for use as evidence. To require a warrant under these circumstances would be to require a useless and meaningless formality.
IF NO PROBABLE CAUSE
If investigators lack probable cause to conduct a warrantless search of a prisoner’s stored property, they may search it nevertheless if both of the following requirements are met: (1) they searched only those items that were actually searched or “subject to search” during booking or arrest; and (2) the search was conducted in a reasonable manner.
THE “SUBJECT TO SEARCH” TEST
Under the “subject to search” test, a warrant is not required to take a second look at items that were actually observed during a search incident to arrest or during booking, or items that could have been lawfully observed at either time. 7 As the U.S. Supreme Court observed in Edwards:
[M]ost cases in the courts of appeals . . . have long since concluded that once the accused is lawfully arrested and is in custody, the effects in his possession at the place of detention that were subject to search at the time and place of his arrest may lawfully be searched and seized without a warrant even though a substantial period of time has elapsed between the arrest and subsequent administrative processing, on the one hand, and the taking of the property for use as evidence, on the other. 8
Three things should be noted about the “subject to search” test. First, a second look is permitted even though there was no “first look,” so long as officers could have taken a first look when the suspect was booked or arrested. 9 Second, because officers can lawfully look at virtually everything in an arrestee’s possession, 10 the “subject to search” requirement is seldom an obstacle. 11
Third, the “subject to search” test is actually more protective of the prisoners’ privacy than a rule permitting a second look only if officers did, in fact, see the item during booking or arrest. This is because such a rule would give the arresting officers and jail staff a perverse incentive to open every container and search everything they find to make sure that all of the prisoner’s property would be subject to a second look. As the Ninth Circuit noted in U.S. v. Burnette.
It is likely that, were we to require warrants for subsequent searches, police officers would routinely remove all items from containers seized at the time of the initial search and thereby insure that all items were discovered at that time. Thus, requiring a warrant for subsequent searches would be unlikely to provide any additional protection for individual privacy. 12
Because the courts employ a “subject to search” test, the following items are subject to a second look without a warrant.
PROPERTY NOT INSIDE A CONTAINER
Articles that were not inside a container may be inspected because they were not only “subject to search,” they were actually seen— at least briefly—when they were seized, inventoried, or stored; e.g., clothing, rings, watches, keys. 13 As the United States Supreme Court observed, “The seizure of an item whose identity is already known occasions no further invasion of privacy.” 14
PROPERTY INSIDE A CONTAINER
Items that were inside a container may be searched if, (1) officers actually opened the container during booking or arrest and saw the contents, or (2) the contents of the container were subject to search when the prisoner was booked. 15 As a practical matter, all property inside a container is subject to a second look because, as the United States Supreme Court observed, “[I]t is not unreasonable for police, as part of the routine procedure incident to incarcerating an arrested person, to search any container or article in his possession, in accordance with established inventory procedures.” 16
MANNER OF SEARCHING
As noted, a “second look” search must be conducted in a reasonable manner. In determining whether such a search was reasonable, the courts look at, (1) whether the property was searched more than once and, if so, whether there was a good reason for conducting multiple searches; and (2) whether officers damaged or destroyed property in conducting the search. 17
EXAMPLES
The following are examples of situations in which the courts ruled that a warrant was not required to take a second look at a prisoner’s property:
ROBBER’S RING : After the defendant was arrested for robbing a cab driver, investigators learned from the victim that the perpetrator was wearing a certain kind of ring. A booking inventory showed that the defendant was wearing a ring when he was arrested. Investigators retrieved the ring from property and showed it to the victim, who identified it. The ring, said the court, “did not have, nor can it acquire after booking, a vestige of privacy requiring a search warrant.” 18
ROBBER’S RING : A Sacramento bank teller noticed that the man who was robbing her was wearing a “gold nugget ring.” When Davis was arrested for the robbery a few days later, he had two rings in his possession. During booking, the rings were put in a nylon bag. When an FBI agent learned that the teller had noticed that the robber was wearing a gold ring, he asked a police detective to see if there were any rings in Davis’s property. Checking the inventory sheet, the detective saw that the bag contained two rings, so he opened it and seized the rings—one of which was identified by the teller. The court ruled the warrantless seizure was lawful because the detective “did not conduct a search but merely retrieved items, lawfully obtained, that law enforcement knew were in its possession.” 19
RAPIST’S SHOES : After Cheatham was arrested for rape, his clothes and shoes were “inventoried and stored in the jail’s property room.” Cheatham was also a suspect in another rape case in which the perpetrator left shoe prints at the scene. When investigators learned Cheatham was in custody, they obtained the shoes from the property room without a warrant, examined the tread, and determined they matched. In rejecting Cheatham’s argument that the shoes should have been suppressed because the officer did not obtain a warrant, the Washington Supreme Court said, “[O]nce an inmate’s personal effects have been exposed to police view in a lawful inventory search and stored in the continuous custody of the police, the inmate no longer has a legitimate expectation of privacy in the items free of further government intrusion.” 20
KEYS : Thompson was arrested in Texas on drug charges. During booking, officers seized some keys, among other things. Several days later, an FBI agent went to the jail and arrested Thompson for stealing dynamite. The agent was aware that some keys had been booked into property, so he inspected them and, as the result, determined they opened a storage unit in which the dynamite had been found. The court ruled the agent did not need a warrant to inspect the keys, noting, “[The FBI agent was] not searching personal effects based on mere hunches that something of evidentiary value might be found. The police officer who had arrested Thompson had already informed the federal agent about the keys. The agent’s particularized search for the keys did not require a warrant.” 21
KEYS : Two Symbionese Liberation Army members, Little and Remiro were arrested by Concord police for the murder of Oakland Schools Superintendent Marcus Foster. During booking, officers removed a set of keys from each of them. Later that day, the keys were turned over to an Oakland police officer who determined they opened the locks on some buildings connected to the SLA. In ruling the keys were seized lawfully, the Court of Appeal noted that an arrestee’s personal effects “like his person itself, are subject to reasonable inspection, examination, and test.” 22
BAIT MONEY : After arresting Westover, detectives in Kansas City searched him and found $621 which was later put into an envelope and stored in the police property room. Because Westover was also a suspect in two Sacramento bank robberies, officers later examined the money without a warrant and determined that some of the bills had been taken in the Sacramento holdup. In ruling a warrant was not required, the Ninth Circuit observed, “In taking the money, no one would suggest that at that instant a search warrant would be required to list the numbers on the bills. Thus, a search warrant to again look at the money already in police custody does not make sense.” 23
ROBBER’S CLOTHES : Earls was arrested on an unspecified Vehicle Code violation and booked into jail. During booking, his clothing “was confiscated.” Several days later, FBI agents determined that Earls was a suspect in a Sacramento bank robbery. An agent obtained Earls’ clothing and sent it to the FBI lab for analysis. The lab found fibers that linked Earls to the robbery. Court: “During their period of police custody an arrested person’s personal effects, like the person itself, are subject to reasonable inspection, examination, and test.” 24
MURDERER’S RING : LAPD detectives had probable cause to arrest Phillip Gunn for murder. When they learned that a man named Phillip Gunn was in jail on a cocaine possession charge they went there to see if the prisoner was the Gunn they were looking for. Gunn’s property had been stored in a transparent plastic bag. Inside the bag, they could see a ring which they apparently realized was similar to the ring worn by the murder victim. Before confirming that the prisoner was the murder suspect, they opened the bag and seized the ring. Later, they showed it to the victim’s wife who positively identified it. Said the court: “What the homicide investigators did in this case cannot be classified as either a search or a seizure within the meaning of the Fourth Amendment. The ring was lawfully in the custody of the police. Its storage in the plastic property bag was purely for convenience and safekeeping. No expectation of privacy was involved. The ring was no more in a place of privacy than if the booking officer had left it on the counter of the booking desk.” 25
ADDRESS BOOK : A Scottsdale police officer arrested Holzman for using a stolen credit card in a department store. During a search incident to the arrest, the officer found an address book. He opened the book and noted it contained “a bunch of names and numbers,” but he did not read any of the entries. The address book was subsequently placed with Holzman’s other property in the jail property room. As the investigation continued, the officer developed probable cause to believe that Holzman was involved in widespread credit card scam. Consequently, he went back to the jail and took a closer look at the entries in the address book and discovered incriminating evidence. Said the court, “[T]he arresting officer legitimately examined the address book during the valid arrest of Holzman, and determined that it contained ‘a bunch of names and numbers.’ At that point appellant’s expectation of privacy in the contents of the book was significantly diminished.” 26
References
- (5th Cir. 1973) 484 F.2d 990, 991
- See United States v. Edwards (1974) 415 U.S. 800, 804-5 [“The police were also entitled to take from Edwards any evidence of the crime in his immediate possession, including his clothing.”].
- (1974) 415 U.S. 800, 806.
- See United States v. Edwards (1974) 415 U.S. 800, 808 [“In upholding this search and seizure, we do not conclude that the Warrant Clause of the Fourth Amendment is never applicable to post arrest seizures of the effects of the arrestee.”].
- (1974) 415 U.S. 800.
- (9th Cir. 1978) 569 F.2d 518. ALSO SEE Jackson v. State (Del. Supreme 1994) 643 A.2d 1360, 1364-5 [“When the evidentiary value of the sneakers was realized, they were properly seized as evidence.”].
- See United States v. Edwards (1974) 415 U.S. 800, 807 [“Caruso [U.S. v. Caruso (2nd Cir. 1966) 358 F.2d 184] is typical of most cases in the courts of appeals that have long since concluded that once the accused is lawfully arrested and is in custody, the effects in his possession at the place of detention that were subject to search at the time and place of his arrest may lawfully be searched and seized without a warrant . . . ” Emphasis added. Citations omitted.]; Illinois v. Andreas (1983) 463 U.S. 765, 771 [“[O]nce police are lawfully in a position to observe an item first-hand, its owner’s privacy interest in that item is lost . . . ” Emphasis added.]; U.S. v. Grill (5th Cir. 1973) 484 F.2d 990, 991 [“[T]he items in question have been exposed to police view under unobjectionable circumstances, so that no reasonable expectation of privacy is breached by an officer’s taking a second look . . . ”]; U.S. v. Johnson (9th Cir. 1987) 820 F.2d 1065, 1072 [“Even though the officer did not in fact at first record the serial numbers of the bills, he could have done so legitimately without a warrant. Accordingly, we find that appellant’s expectation of privacy was significantly reduced, and that the information obtained during the second search was admissible.”]; U.S. v. Burnette (9th Cir. 1983) 698 F.2d 1038, 1049 [“The contents of the purse had been fully exposed to the police and, consequently, her expectation of privacy in the purse was necessarily reduced by a significant degree. . . . [so that] the subsequent warrantless search at the police station was valid.”]; Lockhart v. McCotter (5th Cir. 1986) 782 F.2d 1275, 1280 [“The police had earlier, at the time of inventory, lawfully viewed the wallet contained in Lockhart’s envelope. Because of this earlier police inspection of his personal property, Lockhart had only a diminished expectation of privacy with respect to the items contained in the envelope. By taking a ‘second look’ at the wallet without first obtaining a search warrant, the police did not unduly intrude upon whatever remaining expectation of privacy Lockhart had.”]; State v. Williams (Kan. 1991) 807 P.2d 1292, 1315 [“[The issue is] whether the law enforcement officers can go through the personal possessions of the accused that were being held for safekeeping and seize evidence without a warrant under circumstances in which the officers could have and should have examined the item seized when the defendant was first booked into jail.”]. NOTE: There is language in Edwards indicating a warrantless search is permitted regardless of whether there was probable cause. The Court noted that the prisoner’s clothes could have been searched without a warrant after booking, “particularly in view of [not because of] the existence of probable cause linking the clothes to the crime.” At p. 806. Emphasis added.
- (1974) 415 U.S. 800, 807. Emphasis added. Citations omitted.]. NOTE: Elsewhere in Edwards, the Court applied the “subject to search” principle when it noted, “Edwards was no more imposed upon than he could have been at the time and place of the arrest or immediately upon arrival at the place of detention.” At p. 805. Emphasis added.
- See State v. William (Kan. Supreme 1991) 807 P.2d 1292, 1318 [although there was no “first look” of documents, a “second look” was permitted because they could have been read when they were booked into a property locker].
- See Illinois v. Lafayette (1983) 462 U.S. 640, 648 [“[I]t is not unreasonable for police, as part of the routine procedure incident to incarcerating an arrested person, to search any container or article in his possession, in accordance with established inventory procedures.”] People v. Panfili (1983) 145 Cal.App.3d 387, 393 [“Police have not merely a right but an affirmative duty, statutorily defined, to safeguard the property of a prisoner.” Citing Penal Code § 1412.].
- NOTE: Defense attorneys sometimes cite two cases which they contend prohibit a warrantless second look unless officers took a first look. One of the cases, U.S. v. Brett (5th Cir. 1969) 412 F.2d 401, 405-6 can be disposed of quickly—it is a pre-Edwards case that is contrary to Edwards. The other case is People v. Smith (1980) 103 Cal.App.3d 840. Although Smith is still occasionally cited by defendants, we are not aware of any case in which it was followed. There are two good reasons for this. First, the court did not engage in any meaningful analysis of the central issue; i.e., whether prisoners enjoy a reasonable expectation of privacy as to items that have been taken from them and stored in a property room for safekeeping. Second, if Smith were the law, officers could comply with it by simply making it a practice to conduct highly intensive booking searches of all property—looking at everything. This would not only result in a waste of police resources, it would, as noted, result in less privacy for the prisoners. Consequently, Smith is usually either distinguished or ignored. See People v. Bradley (1981) 115 Cal.App.3d 744, 751; People v. Davis (2000) 84 Cal.App.4th 390, 394 [“Smith does not stand for the broad proposition that jail inmates retain a Fourth Amendment privacy interest in property seized upon arrest and stored in the jail property room. On the contrary, [Smith] was based on the fact that the officers searched through a purse and wallet in the defendant’s mother’s property for items which had not previously been noted or whose evidentiary value had not previously been appreciated.”]; People v. Superior Court (Gunn)(1981) 112 Cal.App.3d 970, 978, fn.2 [court notes that Smith was inconsistent with Edwards]; People v. Panfili (1983) 145 Cal.App.3d 387, 383 [unlike Smith, officers isolated the defendant’s property—they did not complete the booking process].
- (9th Cir. 1983) 698 F.2d 1038, 1049, fn.25.
- See United States v. Edwards (1974) 415 U.S. 800, 806 [clothing]; People v. Davis (2000) 84 Cal.App.4th 390 [ring]; People v. Bradley (1981) 115 Cal.App.3d 744, 750-1 [ring]; People v. Remiro (1979) 89 Cal.App.3d 809, 835 [keys]; People v. Superior Court (Gunn) (1980) 112 Cal.App.3d 970, 977-8 [ring]; U.S. v. Thompson (5th Cir. 1988) 837 F.2d 673, 675-6; U.S. v. Oaxaca (9th Cir. 1978) 569 F.2d 518, 524 [“Both the defendants and their shoes remained in lawful custody until the time when the shoes were taken for use as evidence. To require a warrant under these circumstances would be to require a useless and meaningless formality.”]; U.S. v. Caruso (2nd Cir. 1966) 358 F.2d 184; U.S. v. Turner (9th Cir. 1994) 28 F.3d 981, 983 [“We have held that if an initial seizure of clothing of the defendant is incident to a lawful arrest and therefore proper, once the clothes were properly in the custody of the sheriff’s office, the clothing could be removed or transferred without benefit of official process.”]; U.S. v. Bomengo (5th Cir. 1978) 580 F.2d 173, 175 [“In United States v. Blanton (5 Cir. 1973, 479 F.2d 327) and recently in United States v. McDaniel (5 Cir. 1978) 574 F.2d 1224, 1226, we rejected the argument that for Fourth Amendment purposes a governmental view subsequent to a private search constituted a ‘new search.’”]; United States v. Burnette (9th Cir. 1983) 698 F.2d 1038, 1049 [“[O]nce an item in an individual’s possession has been lawfully seized and searched, subsequent searches of that item, so long as it remains in the legitimate uninterrupted possession of the police, may be conducted without a warrant. . . . ¶ Requiring police to procure a warrant for subsequent searches of an item already lawfully searched would in no way provide additional protection for an individual’s legitimate privacy interests. The contents of an item previously searched are simply no longer private.”]; Hell’s Angels Motorcycle Corp. v. McKinley (9th Cir. 2004) 360 F.3d 930, 933 [“[P]ersonal items seized and examined by police during searches incident to a lawful arrest are not protected from further warrantless searches by police.” Emphasis added.]. ALSO SEE Arizona v. Hicks (1987) 480 U.S. 321, 325 [“Merely inspecting those parts of the turntable that came into view during the latter search would not have constituted an independent search, because it would have produced no additional invasion of respondent’s privacy interest.”].
- Minnesota v. Dickerson (1993) 508 U.S. 366, 377.
- See United States v. Edwards (1974) 415 U.S. 800, 807; Illinois v. Andreas (1983) 463 U.S. 765, 771 [“It is obvious that the privacy interest in the contents of a container diminishes with respect to a container that law enforcement authorities have already lawfully opened and found to contain illicit drugs.”]; United States v. Burnette (9th Cir. 1983) 698 F.2d 1038, 1049 [“The contents of an item previously searched are simply no longer private.”]. ALSO SEE the cases cited in section entitled “If no probable cause,” supra.
- Illinois v. Lafayette (1983) 462 U.S. 640, 648. ALSO SEE People v. Panfili (1983) 145 Cal.App.3d 387, 393 [“Police have not merely a right but an affirmative duty, statutorily defined, to safeguard the property of a prisoner.” Citing Penal Code § 1412.]; People v. Bradley (1981) 115 Cal.App.3d 744, 751 [“Whatever segregation the police make as a matter of internal police administration of articles taken from a prisoner at the time of his arrest and booking does not derogate the fact of their continued custody and possession of such articles.”].
- See United States v. Edwards (1974) 415 U.S. 800, 808, fn.9 [Court notes that a second look searches might be deemed unreasonable “because of their number or their manner of perpetration.”].
- People v. Bradley (1981) 115 Cal.App.3d 744, 751. ALSO SEE People v. Rivard (Mich.App. 1975) 230 N.W.2d 6, 8 [“Once the ring had been exposed to police view under unobjectionable circumstances and lawfully taken by the police for safekeeping, any expectation of privacy with respect to that item had at least partially dissipated so that no reasonable expectation of privacy was breached by Detective Van Alstine taking a ‘second look.’”].
- People v. Davis (2000) 84 Cal.App.4th 390, 394-5.
- State v. Cheatham (Wash. 2003) 81 P.3d 830, 836. ALSO SEE State v. Jellison (Mont. Supreme 1989) 769 P.2d 711 [robbery suspect’s shoes that were booked into property were taken to robbery scene and compared with a shoe print on the counter].
- U.S. v. Thompson (5th Cir. 1988) 837 F.2d 673. ALSO SEE U.S. v. Grill (5th Cir. 1973) 484 F.2d 990.
- People v. Remiro (1979) 89 Cal.App.3d 809, 835 [quoting from People v. Rogers (1966) 241 Cal.App.2d 384, 389-90.
- U.S. v. Westover (9th Cir. 1968) 394 F.2d 164, 165. ALSO SEE U.S. v. Jenkins (2nd Cir. 1974) 496 F.2d 57, 73 [“[O]nce the money had been lawfully taken by the police for safekeeping Wilcox no longer could reasonably expect any privacy with respect to the serial numbers.”]; U.S. v. Johnson (9th Cir. 1987) 820 F.2d 1065, 1072 [bait money]; U.S. v. Burnette (9th Cir. 1983) 698 F.2d 1038, 1049 [bait money]; Evalt v. U.S. (9th Cir. 1967) 382 F.2d 424, 427 [bait money]; People v. Panfili (1983) 145 Cal.App.3d 387, 393-4 [bait money].
- People v. Earls (1980) 109 Cal.App.3d 1009, 1012.
- People v. Superior Court (Gunn)(1981) 112 Cal.App.3d 970, 977. ALSO SEE People v. Richards (Ill. Supreme 1983) 445 N.E.2d 319 [officers lawfully seized necklace, having probable cause to believe the defendant had taken it in a burglary].
- U.S. v. Holzman (9th Cir. 1989) 871 F.2d 1496. ALSO SEE State v. William (Kan. 1991) 807 P.2d 1292, 425-6 [officers lawfully seized and read a document taken from the defendant and placed in storage even though the document was not read when the defendant was booked].