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.
 A: 
$\text{We have}\quad f(n)=n^3-7n^2+11n-5\tag{1}.$ 

$\textrm{We want to solve for}\;\; n\;\textrm{ when}\;\;f(n) = n^3 - 7n^2 + 11n - 5 = 0\tag{2}$  
Note that replacing $n$ with $1$ in equation $(2)$, and then summing, gives us: 
$$1(1^3) - 7(1^2 + 11(1) - 5 =  1 - 7 + 11 - 5 = 0,$$ and so $n = 1$ is a solution to $(2)$, and so $(n - 1)$ is at least one factor of the polynomial $(1)$. 
Dividing $\;f(n) = n^3-7n^2+11n-5\;$ by $\;(n - 1)$ gives us $f(n)/(n-1) = (n^2 - 6n + 5).$ 
We are then able to factor $(1)$:
$$f(n)=(n-1)(n^2-6n+5)=(n-1)[(n-1)(n-5)]=(n-1)^2(n-5).\tag{3}$$
Now we look at the roots of $f(n)$; that is, when is it true that $f(n) = (n-1)^2(n-5) = 0\;$?. 
 - We already know that when $n = 1$, $f(n) = 0$. 
 - It is also the case that when $n = 5$, $f(n) = 0$. 
So what must the value of $n$ be if we must have $f(n) > 0$?. 


*

*This is true when $n < 1$, but then, since $n$ is an integer, $n = 0$ is the greatest integer less than $1$, and we know that $n > N$, meaning $N \le -1$. But we need a positive $N$. 

*When $1 < n < 5$, $f(n) < 0.$ No good. 

*When $n > 5$, $f(n) > 0$. So we must have $n = 6$. This is the smallest positive integer for which $f(n)>0$. 


Key Note: Now, knowing $n$ is NOT ENOUGH! This $n$ is NOT the answer to your question.


*

*We need to use $n$ to determine the integer $N$. We want the smallest positive integer $N$ such that for any $n > N$, $f(n)$ is positive.

*When $N = 5$, we have that for all $n > 5$, $f(n) > 0$: this is your answer to this question.

A: Let $f(x)=x^3-7x^2+11x-5$. $f(1)=f(5)=0$ and so $$f(x)=(x-1)(x^2-6x+5)=(x-1)(x-1)(x-5)=(x-1)^2(x-5)$$
Then $$f(x)>0\iff ... $$
A: The polynomial was chosen so that it would factor simply. So probably a solution through factoring is expected. But let's take a look at the polynomial $n^3-7n^2+11n-5$ as if factoring might be hard.   
If $n\ge 7$, then $n^3-7n^2\ge 0$. And $11n-5\gt 0$ for any positive integer $n$. So $n^3-7n^2+11n-5\gt 0$ for any $n\ge 7$.
Now ask whether our polynomial is $\gt 0$ for $n\ge 6$. So plug in $n=6$. We get a positive result. 
Is our polynomial $\gt 0$ for $n\ge 5$? Plug in $5$. We get $0$. This is not positive. So the required number $N$ is $5$.
A: 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 $n^3 - 7n^2 + 11n -5 > 0$.
A: $n^3-7n^2+11n-5=(n-1)(n^2-6n+5)=(n-1)^2(n-5)$ 
Now, for real $n,(n-1)^2 >0$ if $n \ne 1$ 
So, the sign of $(n-5)$ will determine the sign of the given polynomial if $n \ne 1$ 
