# Thread: help in factoring 6 roots

1. ## help in factoring 6 roots

i need help factoring this eqn.
x^6 + 9x^4 + 24x^2 + 16 = 0

how did it get
(x^2+1)(x^2+4)^2

i need help factoring this eqn.
x^6 + 9x^4 + 24x^2 + 16 = 0

how did it get
(x^2+1)(x^2+4)^2
So that the large powers don't get to you (and to make long division easier), let $y = x^2$, then we need to solve:

$y^3 + 9y^2 + 24y + 16 = 0$

we see that $y = -1$ is a root, by the remainder theorem, thus $y + 1$ is a factor by the factor theorem

doing $y^3 + 9y^2 + 24y + 16 \div y + 1$ by long or synthetic division, we find that

$y^3 + 9y^2 + 24y + 16 = (y + 1)(y^2 + 8y + 16) = (y + 1)(y + 4)^2$

now replace $y$ with $x^2$ and you're done

3. $y:=x^2$, then the polinom: $f(y)=y^3+9y^2+24y+16$
Now you should notice that if we substitute $-1$: $f(y)=-1+9-24+16=0$; and for $-4$: $f(y)=-64+144-96+16=0$. So $-1$ and $-4$ are roots. Therefore: $f(y)=(y+1)(y+4)^2$
And if you write back $x^2$ you get what you want.

Sorry Jhevon I didn't recognize your post..