(a^{2}+1)^{2}-7(a^{2}+1)+10

ans = Its wrong I tried a^{4}-1+7a^{2}+16

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- Jun 6th 2012, 06:53 PMzbest1966factor radicals
(a

^{2}+1)^{2}-7(a^{2}+1)+10

ans = Its wrong I tried a^{4}-1+7a^{2}+16 - Jun 6th 2012, 06:58 PMReckonerRe: factor radicals
There are no radicals here. Did you make a mistake when typing the problem?

Anyway, this is a quadratic in form. Let $\displaystyle u=a^2+1$. Then

$\displaystyle \left(a^2+1\right)^2-7\left(a^2+1\right)+10 = u^2-7u+10$.

Now factor that trinomial on the right, and then substitute for $\displaystyle u$ to get everything back in terms of $\displaystyle a$. - Jun 6th 2012, 07:34 PMzbest1966Re: factor radicals
a^4 + 1-7a^2-7+10

a^4-6-7a^2+10

a^4-7a^2+4

Im stuck - Jun 6th 2012, 07:42 PMReckonerRe: factor radicals
Did you not read what I wrote? We have a quadratic in $\displaystyle a^2+1$. You don't need to multiply everything out. And if you were to multiply everything out, you're doing it wrong: $\displaystyle \left(a^2+1\right)^2\neq a^4+1$.

I've shown you how to reduce the problem to factoring $\displaystyle u^2-7u+10$. Can you factor this trinomial?