Hi, I needed to factor x^4+5x^2+4. I understand the answer is (x-2)(x-1)(x+1)(x+2)

Can anyone tell me how to go about answering this though? How do I figure it out?

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- Jan 7th 2011, 08:39 PMfrankinaroundfactoring x^4+5x^2+4
Hi, I needed to factor x^4+5x^2+4. I understand the answer is (x-2)(x-1)(x+1)(x+2)

Can anyone tell me how to go about answering this though? How do I figure it out? - Jan 7th 2011, 08:43 PMdwsmith
let w^2=x^4

- Jan 7th 2011, 08:44 PMsnowtea
Are you sure it is not x^4 - 5x^2 + 4?

Substitute y = x^2 to get y^2 - 5y + 4.

First factor this, then resubstitue in x^2 and continue factoring. - Jan 7th 2011, 08:44 PMProve It
In other words, let $\displaystyle \displaystyle w = x^2$.

- Jan 7th 2011, 08:46 PMRandom Variable
If all else fails, search for a root and then use synthetic division.

- Jan 7th 2011, 08:49 PMProve It
- Jan 7th 2011, 09:01 PMTheCoffeeMachine
Or complete the square:

$\displaystyle x^4-5x^2+4 = \left(x^2+\frac{5}{2}\right)^2-\frac{9}{2}$.

Of course you know that $\displaystyle y^2-t = (y-\sqrt{t})(y+\sqrt{t})$. - Jan 7th 2011, 09:01 PMRandom Variable
Usually the functions aren't this nice.

In any case $\displaystyle x=i $ is clearly a root. And since the function has real coefficients, you know that $\displaystyle x=-i $ is also a root. Then a factor is $\displaystyle x^{2}+1 $. Now you can use synthetic division to find the other two.

But he might have meant $\displaystyle x^{4}-5x^{2}+4 $. - Jan 8th 2011, 05:22 AMArchie Meade
$\displaystyle (x-2)(x-1)(x+1)(x+2)$ is zero for x=2, 1, -1, -2.

$\displaystyle x^4+5x^2+4$ is not zero for any of these values of x, since the powers of x are all positive,

hence it is a different expression.

On the other hand... by factoring over 2 stages, the 2nd stage being "difference of squares"

$\displaystyle x^4-5x^2+4=\left(x^2-4\right)\left(x^2-1\right)=(x+2)(x-2)(x+1)(x-1)$ - Jan 10th 2011, 05:55 PMfrankinaround
thanks everyone. That substitution is great, but in the future how can I know to use that kind of substitution? meaning what if it was a little bit different right? then how would you solve it? or how did you know to substitute y=x^2 ? The quadratic equation didnt work for x^4 thats why im asking. So does that mean you just need to find any way to make it ax^2+bx+c?

- Jan 10th 2011, 07:02 PMmr fantastic