<s>[INST] What is the value of $N$? $N$ is the smallest positive integer such that for any integer $n > N$, the quantity $n^3 – 7n^2 + 11n – 5$ is positive. What is the value of $N$?
Note: $N$ is a single digit number.
 [/INST] Start by trying to factorise the expression. One thing to note is that the sum of the coefficients is zero: $(+1) + (-7) + (+11) + (-5) = 0$, and so $n=1$ is a solution. We are able to factorise:
$$ n^3 - 7n^2 + 11n -5 \equiv (n-1)(\text{quadratic}) \, . $$
Assume the quadratic is of the form $a_2n^2+a_1n+a_0$, for some yet to be determined numbers $a_2$, $a_1$ and $a_0$. We may expand and compare coefficients. We see that $a_2 = 1,$ $a_1 - a_2 = -7$, $a_0-a_1 = 11$ and $-a_0 = -5.$ Solving these linear equations gives: $a_2 = 1,$ $a_1 = -6$ and $a_0 = 5$. Thus: $n^3-7n^2+11n + 5 \equiv (n-1)(n^2-6n+5).$ Again, we can factorise the quadratic term and we see that $n^2-6n+5 \equiv (n-1)(n-5)$. Thus:
$$n^3-7n^2+11n-5 \equiv (n-1)^2(n-5) \, . $$
The graph $y = x^3-7x^2+11x-5$ cuts the $x$-axis when $x=5$ and is tangent there when $x=1$. These are the critical values: $x=1$ and $x=5$. The question is: what happens when $x < 1$, when $1 < x < 5$ and when $x > 5.$ We can substitute values, say $x = 0$, $x = 2$ and $x = 6$.
$$
\begin{array}{ccc}
0^3 - 7\times 0^2 + 11\times 0 - 5 & = & -5 \, , \\
2^3 - 7\times 2^2 + 11\times 2 - 5 & = & -2 \, , \\
6^3 - 7\times 6^2 + 11\times 6 - 5 & = & 25 \, . 
\end{array}
$$
It follows that $x^3-7x^2+11x-5 \le 0$ for all $x \le 5$ while $x > 0$ for all $x > 5$. Thus, for all integer $n > N = 5$ we have $n^3 - 7n^2 + 11n -5 > 0$.
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