Let a be largest real value of $x$ for which $x^3 - 8x^2 - 2x + 3 = 0.$ Let $a$ be largest real value of $x$ for which $x^3 - 8x^2 - 2x + 3 = 0$. Determine the integer closest to $a^2$.
How I tried to do this:
This is a third-degree polynomial, thus there are 3 positive/negative values of $x$. If I find the roots of this polynomial and take the largest one and square it, I should get $a^2$. I don't get the question's logic of "closest to $a^2$", why would I want to find that? How would I be able to find an integer "closest" to $a^2$? You either find it or you don't. I'd begin with testing small values, namely $0, -1, 1, 2, -2,$ etc. and then trying to find something that yields me an answer close to $0$. I'd then go on from there...
Thanks, would really appreciate help w/ this problem!

Let $f(x)=x^3-8x^2-2x+3$. Note that 
$$f(-0.75)\lt 0\lt f(-0.6)$$
$$f(0.6)\lt 0\lt f(0.5)$$
$$f(8.16)\lt 0\lt f(8.21)$$
So, since we have
$$8.16\lt a\lt 8.21\Rightarrow (66.5 \lt)\ 66.5856\lt a^2\lt 67.4041\ (\lt 67.5),$$
the integer closest to $a^2$ is $67$. 
