20.5: Wound Dressings
Wound dressings should be selected based on the type of the wound, the cause of the wound, and the characteristics of the wound. A specially-trained wound care nurse should be consulted, when possible, for appropriate selection of dressings for chronic wounds. See Table \(\PageIndex{1}\) for commonly used wound dressings and associated nursing considerations. [1]
| Type of Dressing | Description | Nursing Considerations |
|---|---|---|
| Sterile gauze | Nonadherent dressing used on moderately to highly exudative wounds. Nontraumatic to wound bed and promotes a moist wound environment. |
|
| Foam (see Figure \(\PageIndex{7}\) [9] ) | Nonadherent and absorptive. |
|
| Alginate/hydrofibers (see Figure \(\PageIndex{8}\) [10] ) | Nonadherent and highly absorptive for highly exudative wounds. Used for wound packing in full-thickness wounds such as Stage 3 or 4 pressure injuries. |
|
Types of Tape
There are several types of tape that can be used to secure dressings. The most commonly used types of tape are medical transpore, micropore paper, cloth, and waterproof tape.
- Medical transpore tape (often referred to as “medi-pore”) is inexpensive, durable, and very sticky. It has tiny holes in it that allow air to reach the skin underneath and sweat and body fluid to pass through it without causing it to come off. However, it leaves residue and can damage sensitive skin.
- Micropore paper tape is gentle on skin and doesn’t leave residue, but it is not waterproof and doesn’t work well on irregular areas. It allows air to reach the skin underneath.
- Cloth tape sticks well, allows air to reach the skin, and does not leave a residue. It has high strength so it can be used to secure a splint. However, it is not flexible or waterproof and can be difficult to tear.
- Waterproof tape is more expensive but it is flexible and doesn’t leave residue. It sticks well to skin but does not stick well to hair. It is waterproof when applied to dry skin.
Read Inside First Aid’s webpage about different types of medical tape:
5 Different Types of Medical Tapes and How to Use Them.
Wound Vacs
The term wound vac refers to a device used with special foam dressings and suctioning to remove fluid and decrease air pressure around a wound to assist in healing. During a wound vac procedure, the nurse applies a special foam dressing over an open wound and seals it with a thin film layer. The film has an opening that rubber tubing fits through to connect to a vacuum pump. Once connected, the vacuum pump removes fluid from the wound while also helping to pull the edges of the wound together. A person with a wound vac typically wears the device 24 hours a day while the wound is healing. [11] See Figure \(\PageIndex{9}\) [12] for an image of a wound vac foam dressing attached to suctioning by a wound vac device. Figure \(\PageIndex{10}\) [13] demonstrates the progression of a wound healing with a wound vac from image A to D.
- Cox, J. (2019). Wound care 101. Nursing, 49 (10). doi.org/10.1097/01.nurse.0000580632.58318.08↵
- Cox, J. (2019). Wound care 101. Nursing, 49 (10). doi.org/10.1097/01.nurse.0000580632.58318.08↵
- "Sterile 4x4 Dressing and Package 3I3A0330.jpg" by Deanna Hoyord, Chippewa Valley Technical College is licensed under CC BY 4.0↵
- "Kerlix with Package 3I3A0365.jpg" by Deanna Hoyord, Chippewa Valley Technical College is licensed under CC BY 4.0↵
- "Nonadherent Dressing and Packaging 3I3A0203.jpg" by Deanna Hoyord, Chippewa Valley Technical College is licensed under CC BY 4.0↵
- "Vaseline Gauze Package -1 3I3A0296.jpg" and "Vaseline Gauze Dressing 3I3A0299" by Deanna Hoyord, Chippewa Valley Technical College are licensed under CC BY 4.0 ↵
- "Transparent Film.jpg" by Deanna Hoyord, Chippewa Valley Technical College is licensed under CC BY 4.0 ↵
- "Hydrocolloid.jpg" by Deanna Hoyord, Chippewa Valley Technical College is licensed under CC BY 4.0 ↵
- "Foam Dressing 3I3A0406.jpg" by Deanna Hoyord, Chippewa Valley Technical College is licensed under CC BY 4.0 ↵
- "Alginate Dressing 3I3A0396.jpg" by Deanna Hoyord, Chippewa Valley Technical College is licensed under CC BY 4.0 ↵
- Yetman, D. (2020, March 23). What you need to know about vacuum-assisted wound closure (VAC) . Healthline. https://www.healthline.com/health/wound-vac#how-it-works ↵
- “ KCI Wound Vac01.jpg " and “ KCI Wound Vac02.jpg ” by Noles1984 at English Wikipedia are in the Public Domain ↵
- “0100-6991-rcbc-44-01-00081-gf1.gif” by unknown is licensed under CC BY 4.0 . Access for free at www.scielo.br/scielo.php?script=sci_arttext&pid=S0100-69912017000100081#f1. ↵


9.4: Strategies for Listening Well
As speaking begins, focus on individual portions of the message, rather than examining the overall message itself. Identify main ideas and supporting points. Listen for oral citations that often begin with the phrase “according to…” and evaluate the credibility of the statement. Listen analytically when piecing together all of the components of the speech individually. Consider paraphrasing the speaker’s message internally. The average person may speak conversationally at a rate of about 150 words per minute, but the average listener possesses the capability to think at the much faster rate of around 400–500 words per minute. Taking this into consideration, listeners can use “down time” in between words and sentences to paraphrase speakers’ points into their own words, adding to their ability to retain and understand them.
Kinesthetic or tactile learners find taking notes during the speech provides a valuable tool for retaining information, even if they never refer back to those notes again. The physical action of writing the notes helps the brain register and record the consumed information. On the other hand, a visual learner who needs to see the speaker during the presentation may find taking notes detrimental to their ability to listen.
One of the hardest things to do while listening involves resisting distractions. Auditory distractions include a loud fan or buzzing noise, people talking loudly in the hallway, or someone sitting nearby with a cold and the sniffles. Visual distractions may consist of a compelling visual aid left up by the previous speaker, someone with a visually striking outfit or item of clothing seated nearby, or a speaker who wears clothing with a distracting message emblazoned across the front. Physical distractions include hunger, leg cramps, a headache, sinus pressure, or fluctuations in temperature, not to mention the worst distraction of all: a vibrating cell phone! Lastly, mental or internal distractions could impair the ability to listen. To resist these distractions requires self-discipline and careful preparation before the speech begins. Turn off the phone. Do not arrive to the speech hungry. Set aside distracting thoughts. Use the moment to focus on the art of paying attention on purpose.
Give each and every speaker the benefit of the doubt.
Note to Self
Keep emotions in check. It is highly likely that a speaker will say something at some point that will evoke a question, and this could easily lead to temptation to dwell on that question, allowing it to ruminate, or circle round and round in the mind. If this happens, consider writing the point or question down to save it for later. Some speakers skillfully evoke questions that they end up answering later in the speech. This technique requires a lot of practice to perform successfully, but effectively keeps audiences listening closely for the answers.


2.6: Zeros of Polynomial Functions
- Refresher Topics
-
-
Simplifying Expressions
- Rationalizing Denominators and Numerators
-
Simplifying Expressions
In this section, you will:
- Evaluate a polynomial using the Remainder Theorem.
- Use the Factor Theorem to solve a polynomial equation.
- Use the Rational Zeros Theorem to find rational zeros.
- Find zeros of a polynomial function.
- Use the Linear Factorization Theorem to find polynomials with given zeros.
- Solve real-world applications of polynomial equations.
A new bakery offers decorated, multi-tiered cakes for display and cutting at Quinceañera and wedding celebrations, as well as sheet cakes to serve most of the guests. The bakery wants the volume of a small sheet 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\[\begin{array}{rcl}
f(k) & = & (k-k) q(k)+r \\[6pt]
& = & 0 \cdot q(k)+r \\[6pt]
& = & r\\[6pt]
\end{array} \nonumber\]In other words, \(f(k)\) is the remainder obtained by dividing \(f(x)\) by \(x−k\).
If a polynomial \(f(x)\) is divided by \(x−k\), then the remainder is the value \(f(k)\).
Use the Remainder Theorem to evaluate \[f(x)=6 x^4-x^3-15 x^2+2 x-7 \text { at } x=2. \nonumber\]
- Solution
-
To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by \(x−2\).\[\begin{array}{r}
\underline{2} \mid & 6 & -1 & -15 & 2 & -7 \\
& & 12 & 22 & 14 & 32 \\
\hline & 6 & 11 & 7 & 16 & 25 \\
\end{array}\nonumber \]The remainder is 25. Therefore, \(f(2)=25\).We can check our answer by evaluating \(f(2)\).\[\begin{array}{rrcl}
& f(x) & = & 6 x^4-x^3-15 x^2+2 x-7 \\[6pt]
\implies & f(2) & = & 6(2)^4-(2)^3-15(2)^2+2(2)-7 \\[6pt]
& & = & 25 \\[6pt]
\end{array} \nonumber\]
Use the Remainder Theorem to evaluate \[f(x)=2 x^5-3 x^4-9 x^3+8 x^2+2 \text { at } x=-3. \nonumber\]
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 tells us\[f(x)=(x-k) q(x)+r. \nonumber\]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 \(n\) with complex-numbered coefficients will have \(n\) complex-numbered 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.
\(k\) is a zero of \(f(x)\) if and only if \((x−k)\) is a factor of \(f(x)\).
Show that \((x+2)\) is a factor of \(x^3-6 x^2-x+30\). 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.\[\begin{array}{r}
\underline{-2} \mid & 1 & -6 & -1 & 30 \\
& & -2 & 16 & -30 \\
\hline & 1 & -8 & 15 & 0 \\
\end{array}\nonumber \]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:\[(x+2)\left(x^2-8 x+15\right). \nonumber \]We can factor the quadratic factor to write the polynomial as\[(x+2)(x−3)(x−5). \nonumber\]By the Factor Theorem, the zeros of \(x^3-6 x^2-x+30\) are \( -2 \), \( 3 \), and \( 5 \).
Use the Factor Theorem to find the zeros of \(f(x)=x^3+4 x^2-4 x-16\) given that \((x−2)\) is a factor of the polynomial.
Using the Rational Zeros 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 Zeros Theorem (also called the Rational Roots 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.\[\begin{array}{rccccl}
& x-\dfrac{2}{5} = 0 & \text{or} & x - \dfrac{3}{4}=0 & \quad & \left(\text{setting each factor equal to } 0\right) \\[6pt]
\implies & 5 x - 2 = 0 & \text{or} & 4 x-3=0 & \quad & \left(\text{multiplying both sides of the equation to eliminate fractions}\right) \\[6pt]
\end{array} \nonumber \]\[ \begin{array}{rrclcl} \implies & f(x) & = & (5 x-2)(4 x-3) & \quad & \left(\text{creating the quadratic function, multiplying the factors}\right) \\[6pt]
\implies & f(x) & = & 20 x^2-23 x+6 & \quad & \left(\text{expanding the polynomial}\right) \\[6pt]
\implies & f(x) & = & (5 \cdot 4) x^2-23 x+(2 \cdot 3) & & \\[6pt]
\end{array} \nonumber\]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 Zeros Theorem; it is a means to give us a pool of possible rational zeros.
If the polynomial \(f(x)=a_n x^n+a_{n-1} x^{n-1}+\ldots+a_1 x+a_0\) 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 Rational Zeros Theorem simplifies to stating that the possible rational zeros are the factors of the constant term.
An often overlooked requirement to using the Rational Zeros Theorem is that the coefficients of the polynomial must be integers. Do not make the mistake of trying to use the Rational Roots Theorem on polynomials with coefficients of the form \( \frac{3}{7} \), \( \sqrt{2} \), or \( 3 + 2i \).
List all possible rational zeros of \(f(x)=2 x^4-5 x^3+x^2-4\)
- 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 factors of \( -4 \) are \(p= \pm 1, \pm 2, \pm 4\).
The factors of \( 2 \) are \(q= \pm 1, \pm 2\).
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 \( 2 \).\[\dfrac{p}{q}= \pm \dfrac{1}{1}, \pm \dfrac{1}{2} \quad \dfrac{p}{q}= \pm \dfrac{2}{1}, \pm \dfrac{2}{2} \quad \dfrac{p}{q}= \pm \dfrac{4}{1}, \pm \dfrac{4}{2}. \nonumber\]Note that \(\frac{2}{2}=1\) and \(\frac{4}{2}=2\), which have already been listed. So we can shorten our list.\[\dfrac{p}{q} = \dfrac{\text {Factors of the last}}{\text {Factors of the first}}= \pm 1, \pm 2, \pm 4, \pm \dfrac{1}{2}. \nonumber \]
Use the Rational Zeros Theorem to find the rational zeros of \(f(x)=2 x^3+x^2-4 x+1\).
- Solution
-
The Rational Zeros 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 \( 2 \).\[\begin{array}{rcl}
\dfrac{p}{q} & = & \dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}} \\[6pt]
& = & \dfrac{\text{factor of } 1}{\text{factor of } 2} \\[6pt]
\end{array} \nonumber\]The factors of \( 1 \) are \(\pm 1\) and the factors of \( 2 \) are \(\pm 1\) and \(\pm 2\). The possible values for \(\frac{p}{q}\) are \(\pm 1\) and \(\pm \frac{1}{2}\). 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 \(f(x)\).\[\begin{array}{rcl}
f(-1) & = & 2(-1)^3+(-1)^2-4(-1)+1=4 \\[6pt]
f(1) & = & 2(1)^3+(1)^2-4(1)+1=0 \\[6pt]
f\left(-\dfrac{1}{2}\right) & = & 2\left(-\dfrac{1}{2}\right)^3+\left(-\dfrac{1}{2}\right)^2-4\left(-\dfrac{1}{2}\right)+1=3 \\[6pt]
f\left(\dfrac{1}{2}\right) & = & 2\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^2-4\left(\dfrac{1}{2}\right)+1=-\dfrac{1}{2} \\[6pt]
\end{array} \nonumber\]Of those, \( −1 \), \(-\frac{1}{2}\), and \(\frac{1}{2}\) are not zeros of \(f(x)\). \( 1 \) is the only rational zero of \(f(x)\).
Use the Rational Zeros Theorem to find the rational zeros of\[f(x)=x^3-5 x^2+2 x+1. \nonumber \]
Finding the Zeros of Polynomial Functions
The Rational Zeros 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.
Find the zeros of \(f(x)=4 x^3-3 x-1\).
- Solution
-
First, note that the coefficients of the given polynomial function are all integers - remember, we cannot use the Rational Zeros Theorem if this is not the case. The Rational Zeros 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 \( 4 \).\[\begin{array}{rcl}
\dfrac{p}{q} & = & \dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}} \\[6pt]
& = & \dfrac{\text{factor of }-1}{\text{factor of } 4} \\[6pt]
\end{array} \nonumber\]The factors of \( -1 \) are \(\pm 1\) and the factors of \( 4 \) are \(\pm 1\),\(\pm 2\), and \(\pm 4\). The possible values for \(\frac{p}{q}\) are \(\pm 1\), \(\pm \frac{1}{2}\), and \(\pm \frac{1}{4}\). 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 \( 1 \).\[\begin{array}{r}
\underline{1} \mid & 4 & 0 & -3 & -1 \\
& & 4 & 4 & 1 \\
\hline & 4 & 4 & 1 & 0 \\
\end{array}\nonumber \]Dividing by \((x−1)\) gives a remainder of \( 0 \), so \( 1 \) is a zero of the function. The polynomial can be written as\[(x-1)\left(4 x^2+4 x+1\right). \nonumber \]The quadratic is a perfect square. Therefore, \(f(x)\) can be written as\[ (x-1)(2 x+1)^2 \nonumber \]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.\[ 2 x+1 =0 \implies x =-\dfrac{1}{2}. \nonumber \]The zeros of the function are \( 1 \) and \(-\frac{1}{2}\) with multiplicity \( 2 \).
Investigating Example \( \PageIndex{ 5 } \) a little more, we look 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 (\(2,4,6, \ldots \)) for the zero \( −0.5 \). At \(x=1\), the graph crosses the \( x \)-axis, indicating the odd multiplicity (\(1,3,5, \ldots \)) for the zero \(x=1\).
Figure \( \PageIndex{ 1 } \)
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 \(f(x)\).
If \( f(x) \) is a polynomial of degree \(n > 0\) with complex-numbered coefficients, then \(f(x)\) has at least one complex zero.
In truth, "a polynomial with complex-numbered coefficients" means that the coefficients can be any number we can imagine - from \( 2 \) to \( \sqrt{2} \) to \( \frac{1}{2} \) to \( 2 + 2i \). Therefore, most people write the Fundamental Theorem of Algebra as, "If \( f(x) \) is a polynomial of degree \(n > 0\), then \(f(x)\) has at least one complex zero."
Although the Fundamental Theorem of Algebra is a major theorem in Mathematics, it is the following corollary that makes the theorem useful.
If \(f(x)\) is a polynomial of degree \(n > 0\) with complex-numbered coefficients, then \(f(x)\) has exactly \(n\) linear factors over the complex number system. That is,\[f(x) = a \left(x-c_1\right)\left(x-c_2\right) \ldots \left(x-c_n\right), \nonumber \]where \(c_1, c_2, \ldots, c_n\) are complex numbers and \( a \in \mathbb{R} \).
Therefore, \(f(x)\) has \(n\) complex-numbered roots if we allow for multiplicities.
As with the Fundamental Theorem of Algebra, the Complete Factorization Theorem can be shortened to, "If \( f(x) \) is a polynomial of degree \(n > 0\), then \(f(x)\) has exactly \(n\) linear factors over the complex number system."
Does every polynomial have at least one imaginary zero?
No. A complex number is not necessarily imaginary. Real numbers are also complex numbers.
Find the zeros of\[f(x)=3 x^3+9 x^2+x+3. \nonumber \]
- Solution
-
The Rational Zeros 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 \( 3 \).\[\begin{array}{rcl}
\dfrac{p}{q} & = & \dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}} \\[6pt]
& = & \dfrac{\text{factor of } 3}{\text{factor of } 3} \\[6pt]
\end{array} \nonumber \]The factors of \( 3 \) are \(\pm 1\) and \(\pm 3\). The possible values for \(\frac{p}{q}\), and therefore the possible rational zeros for the function, are \(\pm 3\), \(\pm 1\), and\(\pm \frac{1}{3}\). 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.\[\begin{array}{r}
\underline{-3} \mid & 3 & 9 & 1 & 3 \\
& & -9 & 0 & -3 \\
\hline & 3 & 0 & 1 & 0 \\
\end{array}\nonumber \]Dividing by \((x+3)\) gives a remainder of \( 0 \), so \( -3 \) is a zero of the function. The polynomial can be written as\[(x+3)\left(3 x^2+1\right). \nonumber \]We can then set the quadratic equal to \( 0 \) and solve to find the other zeros of the function.\[\begin{array}{rrcl}
& 3 x^2+1 & = & 0 \\[6pt]
\implies & x^2 & = & -\dfrac{1}{3} \\[6pt]
\implies & x & = & \pm \sqrt{-\dfrac{1}{3}} \\[6pt]
& & = & \pm \dfrac{i \sqrt{3}}{3} \\[6pt]
\end{array} \nonumber \]The zeros of \(f(x)\) are \( -3 \) and \(\pm \frac{i \sqrt{3}}{3}\).
Figure \( \PageIndex{ 2 } \) shows the graph of the function \( f \) from Example \( \PageIndex{ 6 } \). 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 \(3^{\text{rd}}\) 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. 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.
Figure \( \PageIndex{ 2 } \)
Find the zeros of\[f(x)=2 x^3+5 x^2-11 x+4. \nonumber\]
Using the Complete 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 Complete 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 \(a+b i\), where \(b \neq 0\), is a zero of \(f(x)\). Then, by the Factor Theorem, \(x-(a+b i)\) 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.
If the polynomial function \(f\) has real coefficients and a complex zero of the form \(a+bi\), then the complex conjugate of the zero, \(a−bi\), is also a zero of \( f \).
Find a fourth degree polynomial with real coefficients that has zeros of \( -3 \), \( 2 \), and \(i\), 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.\[\begin{array}{rcl}
f(x) & = & a(x+3)(x-2)(x-i)(x+i) \\[6pt]
& = & a\left(x^2+x-6\right)\left(x^2+1\right) \\[6pt]
& = & a\left(x^4+x^3-5 x^2+x-6\right) \\[6pt]
\end{array} \nonumber \]We need to find \(a\) to ensure \(f(-2)=100\). Substitute \(x=-2\) and \(f(2)=100\) into \(f(x)\).\[\begin{array}{rrcl}
& 100 & = & a\left((-2)^4+(-2)^3-5(-2)^2+(-2)-6\right) \\[6pt]
\implies & 100 & = & a(-20) \\[6pt]
\implies & -5 & = & a \\[6pt]
\end{array} \nonumber \]So the polynomial function is\[f(x)=-5\left(x^4+x^3-5 x^2+x-6\right) \nonumber \]or\[f(x)=-5 x^4-5 x^3+25 x^2-5 x+30. \nonumber \]
In Example \( \PageIndex{ 7 } \), we 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 \(i\).
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.
Find a third degree polynomial with real coefficients that has zeros of \( 5 \) and \(−2i\) such that \(f(1)=10\).
It's probably good to summarize the theorems and corollaries we have encountered so far in this section, along with their requirements.
| Theorem or Corollary | Requirement | Result |
|---|---|---|
| Rational Zeros Theorem | Coefficients must be integers | Rational zeros of the polynomial are always ratios of factors of the constant term and the leading coefficient. |
| Fundamental Theorem of Algebra | Coefficients can be any number | The polynomial has at least one complex zero. |
| Complete Factorization Theorem | Coefficients can be any number | The polynomial can be factored into \( n \) linear factors over the complex number system. |
| Complex Conjugate Theorem | Coefficients must be real numbers | If \( a + bi \) is a zero, then \( a - bi \) is also a zero. |
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.
A new bakery offers decorated, multi-tiered cakes for display and cutting at Quinceañera and wedding celebrations, as well as sheet cakes to serve most of the guests. The bakery wants the volume of a small sheet 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
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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 \(h=\frac{1}{3} w\). Let’s write the volume of the cake in terms of width of the cake.\[\begin{array}{rcl}
V & = & (w+4)(w)\left(\dfrac{1}{3} w\right) \\[6pt]
& = & \dfrac{1}{3} w^3+\dfrac{4}{3} w^2 \\[6pt]
\end{array} \nonumber \]Substitute the given volume into this equation.\[\begin{array}{rrclcl}
& 351 & = & \dfrac{1}{3} w^3+\dfrac{4}{3} w^2 & \quad & \left(\text {substituting} \right) \\[6pt]
\implies & 1053 & = & w^3+4 w^2 & \quad & \left( \text{multiplying both sides by } 3 \right) \\[6pt]
\implies & 0 & = & w^3+4 w^2-1053 & \quad & \left(\text{subtracting } 1053 \text{ from both sides} \right) \\[6pt]
\end{array} \nonumber \]The Rational Zeros Theorem tells us that the possible rational zeros are \(\pm 1\), \(\pm 3\), \(\pm 9\), \(\pm 13\), \(\pm 27\), \(\pm 39\), \(\pm 81\), \(\pm 117\), \(\pm 351\), and \(\pm 1053\). 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 \(x=1\).\[\begin{array}{r}
\underline{1} \mid & 1 & 4 & 0 & -1053 \\
& & 1 & 5 & 5 \\
\hline & 1 & 5 & 5 & -1048 \\
\end{array}\nonumber \]Since 1 is not a solution, we will check \(x=3\).\[\begin{array}{r}
\underline{3} \mid & 1 & 4 & 0 & -1053 \\
& & 3 & 21 & 63 \\
\hline & 1 & 7 & 21 & -990 \\
\end{array}\nonumber \]Since 3 is not a solution either, we will test \(x=9\).\[\begin{array}{r}
\underline{9} \mid & 1 & 4 & 0 & -1053 \\
& & 9 & 117 & 1053 \\
\hline & 1 & 13 & 117 & 0 \\
\end{array}\nonumber \]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.\[l=w+4=9+4=13 \text { and } h=\dfrac{1}{3} w = \dfrac{1}{3}(9)=3. \nonumber \]>The sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches.
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?
Access these online resources for additional instruction and practice with zeros of polynomial functions.


7.2: Muscles
Muscles make up the bulk of an animal’s body and account for about half its weight. The meat on the chop or roast is muscle and is composed mainly of protein. The cells that make up muscle tissue are elongated and able to contract to a half or even a third of their length when at rest. There are three different kinds of muscle based on appearance and function: smooth, cardiac and skeletal muscle.
Types of Muscle
- Smooth muscle : Smooth or Involuntary muscle carries out the unconscious routine tasks of the body such as moving food down the digestive system, keeping the eyes in focus and adjusting the diameter of blood vessels. The individual cells are spindle-shaped, being fatter in the middle and tapering off towards the ends with a nucleus in the center of the cell. They are usually found in sheets and are stimulated by the non-conscious or autonomic nervous system as well as by hormones (see Chapter 3).
- Cardiac muscle : Cardiac muscle is only found in the wall of the heart. It is composed of branching fibers that form a three-dimensional network. When examined under the microscope, a central nucleus and faint stripes or striations can be seen in the cells. Cardiac muscle cells contract spontaneously and rhythmically without outside stimulation, but the sinoatrial node (natural pacemaker) coordinates the heart beat. Nerves and hormones modify this rhythm (see Chapter 3).
- Skeletal muscle : Skeletal muscle is the muscle that is attached to and moves the skeleton, and is under voluntary control. It is composed of elongated cells or fibers lying parallel to each other. Each cell is unusual in that it has several nuclei and when examined under the microscope appears striped or striated. This appearance gives the muscle its names of striped or striated muscle. Each cell of striated muscle contains hundreds, or even thousands, of microscopic fibers each one with its own striped appearance. The stripes are formed by two different sorts of protein that slide over each other making the cell contract (see diagram 7.1).
Diagram 7.1 - A striped muscle cell
Muscle Contraction
Muscle contraction requires energy and muscle cells have numerous mitochondria. However, only about 15% of the energy released by the mitochondria is used to fuel muscle contraction. The rest is released as heat. This is why exercise increases body temperature and makes animals sweat or pant to rid themselves of this heat.
What we refer to as a muscle is made up of groups of muscle fibers surrounded by connective tissue. The connective tissue sheaths join together at the ends of the muscle to form tough white bands of fiber called tendons . These attach the muscles to the bones. Tendons are similar in structure to the ligaments that attach bones together across a joint (see diagrams 7.2a and b).
Diagram 7.2 a and b - The structure of a muscle
Tendons Tie muscles to bones and Ligaments Link bones at joints
Structure of a muscle
A single muscle is fat in the middle and tapers towards the ends. The middle part, which gets fatter when the muscle contracts, is called the belly of the muscle. If you contract your biceps muscle in your upper arm you may feel it getting fatter in the middle. You may also notice that the biceps is attached at its top end to bones in your shoulder while at the bottom it is attached to bones in your lower arm. Notice that the bones at only one end move when you contract the biceps. This end of the muscle is called the insertion . The other end of the muscle, the origin , is attached to the bone that moves the least (see diagram 7.3).
Diagram 7.3 - Antagonistic muscles, flexion and extension
Antagonistic muscles
Skeletal muscles usually work in pairs. When one contracts the other relaxes and vice versa. Pairs of muscles that work like this are called antagonistic muscles . For example the muscles in the upper forearm are the biceps and triceps (see diagram 7.3). Together they bend the elbow. When the biceps contracts (and the triceps relaxes) the lower forearm is raised and the angle of the joint is reduced. This kind of movement is called flexion . When the triceps is contracted (and the biceps relaxes), the angle of the elbow increases. The term for this movement is extension .
When you or animals contract skeletal muscle it is a voluntary action. For example, you make a conscious decision to walk across the room, raise the spoon to your mouth, or smile. There is however, another way in which contraction of muscles attached to the skeleton happens that is not under voluntary control. This is during a reflex action , such as jerking your hand away from the hot stove you have touched by accident. This is called a reflex arc and will be described in detail in chapter 14-15.


7.2D: Vitamin K
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Vitamin K (Figure 1) is a group of structurally similar, fat-soluble vitamins the human body requires for complete synthesis of certain proteins that are needed for blood coagulation ( blood clotting ) and which the body also needs for controlling the binding of calcium in bones and other tissues. The vitamin K-related modification of the proteins allows them to bind calcium ions, which they cannot do otherwise.
In menaquinone, the side chain is composed of a varying number of isoprenoid residues. The most common number of these residues is four, since animal enzymes normally produce menaquinone-4 from plant phylloquinone. The three synthetic forms of vitamin K are vitamins K 3 (menadione), K 4 , and K 5 , which are used in many areas, including the pet food industry (vitamin K 3 ) and to inhibit fungal growth (vitamin K 5 ).
Figure 1: (left) Vitamin K 1 (phylloquinone) – both forms of the vitamin contain a functional naphthoquinone ring and an aliphatic side chain. Phylloquinone has a phytyl side chain. (right) Vitamin K 2 (menaquinone).
The MK-4 form of vitamin K 2 is produced by conversion of vitamin K 1 in the testes, pancreas, and arterial walls. While major questions still surround the biochemical pathway for this transformation, the conversion is not dependent on gut bacteria, as it occurs in germ-free rats and in parenterally-administered K 1 in rats. In fact, tissues that accumulate high amounts of MK-4 have a remarkable capacity to convert up to 90% of the available K 1 into MK-4. There is evidence that the conversion proceeds by removal of the phytyl tail of K 1 to produce menadione as an intermediate, which is then condensed with an activated geranylgeranyl moiety (see also prenylation) to produce vitamin K 2 in the MK-4 (menatetrione) form.
Health effects of Vitamin K
Without vitamin K, blood coagulation is seriously impaired, and uncontrolled bleeding occurs. Preliminary clinical research indicates that deficiency of vitamin K may weaken bones, potentially leading to osteoporosis, and may promote calcification of arteries and other soft tissues. Our main sources of vitamin K are from leafy green vegetables and from the production by bacteria in our large intestine.
- Osteoporosis : A review of 2014 concluded that there is positive evidence that monotherapy using MK-4, one of the forms of Vitamin K 2 , reduces fracture incidence in post-menopausal women with osteoporosis, and suggested further research on the combined use of MK-4 with bisphosphonates. In contrast, an earlier review article of 2013 concluded that there is no good evidence that vitamin K supplementation helps prevent osteoporosis or fractures in postmenopausal women. A Cochrane systematic review of 2006 suggested that supplementation with Vitamin K 1 and with MK4 reduces bone loss; in particular, a strong effect of MK-4 on incident fractures among Japanese patients was emphasized. A review article of 2016 suggested to consider, as one of several measures for bone health, increasing the intake of foods rich in vitamins K 1 and K 2 .
- Cardiovascular health: Adequate intake of vitamin K is associated with the inhibition of arterial calcification and stiffening, but there have been few interventional studies and no good evidence that vitamin K supplementation is of any benefit in the primary prevention of cardiovascular disease. One 10-year population study, the Rotterdam Study, did show a clear and significant inverse relationship between the highest intake levels of menaquinone (mainly MK-4 from eggs and meat, and MK-8 and MK-9 from cheese) and cardiovascular disease and all-cause mortality in older men and women.
- Cancer : Vitamin K has been promoted in supplement form with claims it can slow tumor growth; there is however no good medical evidence that supports such claims.
- Coumarin poisoning: Vitamin K is part of the suggested treatment regime for poisoning by rodenticide (coumarin poisoning).
Although allergic reaction from supplementation is possible, no known toxicity is associated with high doses of the phylloquinone (vitamin K 1 ) or menaquinone (vitamin K 2 ) forms of vitamin K, so no tolerable upper intake level (UL) has been set. Blood clotting (coagulation) studies in humans using 45 mg per day of vitamin K2 (as MK-4)[12] and even up to 135 mg per day (45 mg three times daily) of K2 (as MK-4),[13] showed no increase in blood clot risk. Even doses in rats as high as 250 mg/kg, body weight did not alter the tendency for blood-clot formation to occur. Unlike the safe natural forms of vitamin K 1 and vitamin K 2 and their various isomers, a synthetic form of vitamin K, vitamin K3 (menadione), is demonstrably toxic at high levels. The U.S. FDA has banned this form from over-the-counter sale in the United States because large doses have been shown to cause allergic reactions, hemolytic anemia, and cytotoxicity in liver cells.
Phylloquinone (K 1 ) or menaquinone (K 2 ) are capable of reversing the anticoagulant activity of the anticoagulant warfarin (tradename Coumadin). Warfarin works by blocking recycling of vitamin K, so that the body and tissues have lower levels of active vitamin K, and thus a deficiency of vitamin K. Supplemental vitamin K (for which oral dosing is often more active than injectable dosing in human adults) reverses the vitamin K deficiency caused by warfarin, and therefore reduces the intended anticoagulant action of warfarin and related drugs. Sometimes small amounts of vitamin K are given orally to patients taking warfarin so that the action of the drug is more predictable. The proper anticoagulant action of the drug is a function of vitamin K intake and drug dose, and due to differing absorption must be individualized for each patient. The action of warfarin and vitamin K both require two to five days after dosing to have maximum effect, and neither warfarin or vitamin K shows much effect in the first 24 hours after they are given.
Absorption and Dietary Need
Previous theory held that dietary deficiency is extremely rare unless the small intestine was heavily damaged, resulting in malabsorption of the molecule. Another at-risk group for deficiency were those subject to decreased production of K 2 by normal intestinal microbiota, as seen in broad spectrum antibiotic use. Taking broad-spectrum antibiotics can reduce vitamin K production in the gut by nearly 74% in people compared with those not taking these antibiotics . Diets low in vitamin K also decrease the body's vitamin K concentration. Those with chronic kidney disease are at risk for vitamin K deficiency, as well as vitamin D deficiency, and particularly those with the apoE4 genotype. Additionally, in the elderly there is a reduction in vitamin K 2 production. Like other lipid-soluble vitamins (A, D and E), vitamin K is stored in the fatty tissue of the human body.
The National Academy of Medicine (NAM) updated an estimate of what constitutes an Adequate Intake (AI) for vitamin K in 2001. The NAM does not distinguish between K 1 and K 2 – both are counted as vitamin K. At that time there was not sufficient evidence to set the more rigorous Estimated Average Requirement (EAR) or recommended dietary allowance (RDA) given for most of the essential vitamins and minerals. The current daily AIs for vitamin K for adult women and men are 90 μg and 120 μg respectively. The AI for pregnancy and lactation is 90 μg. For infants up to 12 months the AI is 2–2.5 μg, and for children aged 1 to 18 years the AI increases with age from 30 to 75 μg. As for safety, the FNB also sets tolerable upper intake levels (known as ULs) for vitamins and minerals when evidence is sufficient. In the case of vitamin K no UL is set, as evidence for adverse effects is not sufficient. Collectively EARs, RDAs, AIs and ULs are referred to as dietary reference intakes.
For U.S. food and dietary supplement labeling purposes, the amount in a serving is expressed as a percentage of daily value (%DV). For vitamin K labeling purposes the daily value was 80 μg, but as of May 2016 it has been revised upwards to 120 μg. A table of the pre-change adult daily values is provided at Reference Daily Intake (Table \(\PageIndex{1}\)). Food and supplement companies have until 28 July 2018 to comply with the change.
Deficiency
Average diets are usually not lacking in vitamin K, and primary deficiency is rare in healthy adults. Newborn infants are at an increased risk of deficiency. Other populations with an increased prevalence of vitamin K deficiency include those who suffer from liver damage or disease (e.g. alcoholics), cystic fibrosis, or inflammatory bowel diseases, or have recently had abdominal surgeries. Secondary vitamin K deficiency can occur in people with bulimia, those on stringent diets, and those taking anticoagulants. Other drugs associated with vitamin K deficiency include salicylates, barbiturates, and cefamandole, although the mechanisms are still unknown. Vitamin K 1 deficiency can result in coagulopathy, a bleeding disorder. Symptoms of K 1 deficiency include anemia, bruising, nosebleeds and bleeding of the gums in both sexes, and heavy menstrual bleeding in women.
| Food | Serving size |
Vitamin
K 1 (μg) |
Food | Serving size |
Vitamin
K 1 (μg) |
|
|---|---|---|---|---|---|---|
| Kale, cooked | 1 ⁄ 2 cup | 531 | Parsley, raw | 1 ⁄ 4 cup | 246 | |
| Spinach, cooked | 1 ⁄ 2 cup | 444 | Spinach, raw | 1 cup | 145 | |
| Collards, cooked | 1 ⁄ 2 cup | 418 | Collards, raw | 1 cup | 184 | |
| Swiss chard, cooked | 1 ⁄ 2 cup | 287 | Swiss chard, raw | 1 cup | 299 | |
| Mustard greens, cooked | 1 ⁄ 2 cup | 210 | Mustard greens, raw | 1 cup | 279 | |
| Turnip greens, cooked | 1 ⁄ 2 cup | 265 | Turnip greens, raw | 1 cup | 138 | |
| Broccoli, cooked | 1 cup | 220 | Broccoli, raw | 1 cup | 89 | |
| Brussels sprouts, cooked | 1 cup | 219 | Endive, raw | 1 cup | 116 | |
| Cabbage, cooked | 1 ⁄ 2 cup | 82 | Green leaf lettuce | 1 cup | 71 | |
| Asparagus | 4 spears | 48 | Romaine lettuce, raw | 1 cup | 57 | |
| Table from "Important information to know when you are taking: Warfarin (Coumadin) and Vitamin K", Clinical Center, National Institutes of Health Drug Nutrient Interaction Task Force. |
Contributors and Attributions
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3.2: The Meaning of Lines
Remember to read blueprints and create effective drawings; you must understand the use of lines. The alphabet of lines is the common language of the technician and the engineer. In drawing an object, the different views are arranged in a certain way, and various lines convey the information. Here is another example of the use of standard lines in a simple drawing is shown in Figure \(\PageIndex{1}\). Line characteristics, such as width, breaks in the line, and zigzags, are shown below in Figure \(\PageIndex{2}\).
Line Standards
Table \(\PageIndex{1}\): Line characteristics and conventions for drawings.
| Name | Convention | Description and Application | Example |
| Center Lines |
Thin lines made up of long and short dashes alternatively spaces and consistent in length. Used to indicate symmetry about an axis and location of centers. |
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| Visible Lines |
Heavy unbroken lines. Used to indicate visible edges of an object. |
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| Hidden Lines |
Medium lines with short evenly spaced dashes. Used to indicate concealed edges. |
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| Extension Lines |
Thin unbroken lines. Used to indicate extent of dimensions. |
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| Dimension Lines |
Thin lines terminated with arrow heads at each end. Used to indicate distance measured. |
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| Leader |
Thin line terminated with arrowhead or dot at one end. Used to indicate a part, dimension or other reference. |
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| Break (Long) |
Thin, solid ruled lines with freehand zigzags. Used to reduce size of drawing required to delineate object and reduce detail. |
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| Break (Short) |
Thick, solid free hand lines. Used to indicate a short break. |
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| Phantom or Datum Line |
Medium series of one long dash and two short dashes evenly spaced ending with long dash. Used to indicate alternate position of parts, repeated detail or to indicate a datum plane. |
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| Stitch Line |
Medium line of short dashes evenly spaced and labeled. Used to indicate stitching or sewing. |
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| Cutting or Viewing Plan - Viewing Plane Optional | Thick solid lines with arrowhead to indicate direction in which section or plane is viewed or taken. | ||
| Cutting Plane for Complex or Offset Views |
Thick short dashed. Used to show offset with arrowheads to show direction viewed. |