Solving a polynomial with grouping

Question

I am a student learning competition math, and in a past test I found, you need to find the sum of the reciprocals of the roots of the polynomial $x^4-7x^3+4x^2+7x-4 = 0$. I have watched some videos on how to solve higher order polynomials, and I found that I like grouping. However, when I try to solve this with grouping, I end up with $x^4+4(x^2-1)-7x(x^2-1)$ which can be simplified to $x^4 + (x^2-1)(4-7x)$. I am not sure what to do now, and am having trouble figuring it out from here. Any help would be appreciated!

Answer 1

If $r$ is a root of the polynomial $p(x) = x^4-7x^3+4x^2+7x-4 $, then $1/r$ is a root of $p(1/x) = x^{-4}-7x^{-3}+4x^{-2}+7x^{-1}-4$, and vice-versa, because $x = 0$ is not a root of either one.

That means that $1/r$ is also a root of $x^4 p(1/x)$, which is a polynomial, namely, $$ x^4 p(1/x) = 1-7x+4x^2+7x^3-4x^4 = -4(-\frac{1}{4}+\frac{7}{4}x-x^2-\frac{7}{4}x^3+x^4). $$

Now if you recall, when you have a monic polynomial (coefficient of highest power is $1$), the coefficient of the NEXT highest power is the negative of the sum of the roots. So $\frac{7}{4}$ is the sum of the roots of the polynomial in parens, which has the same roots as $x^4 p(1/x)$, i.e., the number you're looking for.

Notice that I never had to actually find any of the roots to know what their sum was (or the sum of their reciprocals).