# 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 $\displaystyle y = x^2$, then we need to solve:

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

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

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

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

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

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

Sorry Jhevon I didn't recognize your post..