Question

Confused, I’d appreciate help!

Answer 1

The fundamental theorem of algebra says this cubic has three roots, so we could write it as

$4x^3+32x^2+79x+60=4(x-a)(x-b)(x-c)$

We're told one of the roots is equal to the sum of the other two, so we could take $c=a+b$:

$4x^3+32x^2+79x+60=4(x-a)(x-b)(x-a-b)$

If we expand the right side, we get

$4\bigg(x^3-2(a+b)x^2+(a^2+3ab+b^2)x-(a^2b+ab^2)\bigg)$

For two polynomial to be equal, the coefficients of terms of the same degree must match, so that

$\begin{cases}4=4&(x^3)\\32=-8(a+b)&(x^2)\\79=4(a^2+3ab+b^2)&(x)\\60=-4(a^2b+ab^2)&\text{(constant)}\end{cases}$

Now we can find $a,b,c$.

$32=-8(a+b)=-8c\implies c=-4$

This tells us that $x+4$ is a factor, so dividing the original cubic by this returns a remainder of 0.

$\dfrac{4x^3+32x^2+79x+60}{x+4}=4x^2+16x+15=4\left(x+\dfrac32\right)\left(x+\dfrac52\right)$

which further tells us that $a=-\dfrac32$ and $b=-\dfrac52$.