Confused, I’d appreciate help!

Question

asked Jun 23, 2020 60.2k views

1 Answer

Ababneh A · Answer 1 · 2020-06-29T13:37:39+0000

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.

$(4x^3+32x^2+79x+60)/(x+4)=4x^2+16x+15=4\left(x+\frac32\right)\left(x+\frac52\right)$

which further tells us that
$a=-\frac32$ and
$b=-\frac52$ .