Hi
Originally Posted by
magentarita Are you saying to replace n with the given exponents in the function and solve for x^(n - 1)?
No, that's not what Moo meant. The degree of $\displaystyle P(x)=x^3 + 5x^2 + 5x - 2$ is 3 which means that this polynomial has 3 roots. (they can be real or complex numbers) Let's denote these roots by $\displaystyle r_1,\,r_2$ and $\displaystyle r_3$ and factor the polynomial : $\displaystyle P(x)=(x-r_1)(x-r_2)(x-r_3)$. Expanding this expression, what do we get ? Let's see :
$\displaystyle \begin{aligned}
P(x)&=(x-r_1)(x-r_2)(x-r_3) \\
&=(x-r_1)(x^2-(r_2+r_3)x+r_2r_3)\\
&=x^3-(r_2+r_3)x^2+r_2r_3x-r_1x^2+r_1(r_2+r_3)x-r_1r_2r_3\\
&=x^3{\color{blue}-(r_1+r_2+r_3)}x^2+(r_1r_2+r_1r_3+r_2r_3)x-r_1r_2r_3\\
\end{aligned}$
What we learn from this is that the coefficient of $\displaystyle x^2$ is the opposite of the sum of the three roots. As this coefficient equals 5, ($\displaystyle P(x)=x^3 + {\color{blue}5}x^2 + 5x - 2$) we have $\displaystyle 5=-(r_1+r_2+r_3)$. Knowing that one of the roots is -2, you can solve this for the sum of the two remaining roots.
What Moo meant is that if the polynomial were $\displaystyle x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_0$ then the sum of the roots would have been equal to $\displaystyle -a_{n-1}$.
Is it clearer ?