# factoring x^4+5x^2+4

• Jan 7th 2011, 09:39 PM
frankinaround
factoring 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, 09:43 PM
dwsmith
let w^2=x^4
• Jan 7th 2011, 09:44 PM
snowtea
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, 09:44 PM
Prove It
In other words, let $\displaystyle w = x^2$.
• Jan 7th 2011, 09:46 PM
Random Variable
If all else fails, search for a root and then use synthetic division.
• Jan 7th 2011, 09:49 PM
Prove It
Quote:

Originally Posted by Random Variable
If all else fails, search for a root and then use synthetic division.

No need, as a quadratic it factorises nicely as $\displaystyle 4 \cdot 1 = 4$ and $\displaystyle 4 + 1 = 5$.

Also, there aren't any real roots anyway, so it would be pretty hard to find one and then divide...
• Jan 7th 2011, 10:01 PM
TheCoffeeMachine
Or complete the square:

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

Of course you know that $y^2-t = (y-\sqrt{t})(y+\sqrt{t})$.
• Jan 7th 2011, 10:01 PM
Random Variable
Quote:

Originally Posted by Prove It
No need, as a quadratic it factorises nicely as $\displaystyle 4 \cdot 1 = 4$ and $\displaystyle 4 + 1 = 5$.

Also, there aren't any real roots anyway, so it would be pretty hard to find one and then divide...

Usually the functions aren't this nice.

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

But he might have meant $x^{4}-5x^{2}+4$.
• Jan 8th 2011, 06:22 AM
Quote:

Originally Posted by frankinaround
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?

$(x-2)(x-1)(x+1)(x+2)$ is zero for x=2, 1, -1, -2.

$x^4+5x^2+4$ 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"

$x^4-5x^2+4=\left(x^2-4\right)\left(x^2-1\right)=(x+2)(x-2)(x+1)(x-1)$
• Jan 10th 2011, 06:55 PM
frankinaround
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, 08:02 PM
mr fantastic
Quote:

Originally Posted by frankinaround
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?

Experience teaches you to recognise the underlying quadratic form.