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?
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 $.
$\displaystyle (x-2)(x-1)(x+1)(x+2)$ 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"
$\displaystyle x^4-5x^2+4=\left(x^2-4\right)\left(x^2-1\right)=(x+2)(x-2)(x+1)(x-1)$
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?