factoring x^4+5x^2+4

**frankinaround** · January 7th 2011, 08:39 PM

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?

**dwsmith** · January 7th 2011, 08:43 PM

let w^2=x^4

**snowtea** · January 7th 2011, 08:44 PM

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.

**Prove It** · January 7th 2011, 08:44 PM

In other words, let $\displaystyle w = x^2$ .

**Random Variable** · January 7th 2011, 08:46 PM

If all else fails, search for a root and then use synthetic division.

**Prove It** · January 7th 2011, 08:49 PM

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...

**TheCoffeeMachine** · January 7th 2011, 09:01 PM

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 $y^2-t = (y-\sqrt{t})(y+\sqrt{t})$ .

**Random Variable** · January 7th 2011, 09:01 PM

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 $x^{4}-5x^{2}+4$ .

**Archie Meade** · January 8th 2011, 05:22 AM

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"

$x^4-5x^2+4=\left(x^2-4\right)\left(x^2-1\right)=(x+2)(x-2)(x+1)(x-1)$

**frankinaround** · January 10th 2011, 05:55 PM

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?

**mr fantastic** · January 10th 2011, 07:02 PM

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.

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