# How to factor with multiple exponents?

• Sep 17th 2010, 05:59 AM
softwareguy
How to factor with multiple exponents?
I'm kind of struggling with factoring when encountering multiple exponents. Can someone please explain the steps to factor the following?

$\displaystyle 4x^4+12x^3-40x^2$
• Sep 17th 2010, 06:04 AM
e^(i*pi)
You need to find a common factor - when there are multiple exponents on the same base then the lowest exponent is a factor of all of them - in this case $\displaystyle x^2$ is a factor (and so is 4). In this case you will get a quadratic once you take out the factors - you should check if it factorises to (which it does in this case)

My answer is in a spoiler below

Spoiler:
$\displaystyle 4x^2(x+5)(x-2)$
• Sep 17th 2010, 07:13 AM
softwareguy
I don't get it. Can you please break it down step by step as to how you got the 5 and -2?
• Sep 17th 2010, 07:25 AM
e^(i*pi)
Do you understand that you can factor $\displaystyle 4x^2$ initally?

This gives: $\displaystyle 4x^2(x^2+3x-10)$

However, the $\displaystyle x^2+3x+10$ can also be factorised to give the answer of $\displaystyle 4x^2(x+5)(x-2)$
• Sep 17th 2010, 07:52 AM
softwareguy
Ah, ok, I got it. But what about when there is no exponent or x on the last number, such as:

$\displaystyle x^4-20x^2+64$

I know it is:

(x+2)(x-2)(x+4)(x-4)

But I want to know a simple way of break it down step by step to arrive at that. That is what I'm struggling with. Thanks.
• Sep 17th 2010, 07:57 AM
masters
Quote:

Originally Posted by softwareguy
Ah, ok, I got it. But what about when there is no exponent or x on the last number, such as:

$\displaystyle x^4-20x^2+64$

I know it is:

(x+2)(x-2)(x+4)(x-4)

But I want to know a simple way of break it down step by step to arrive at that. That is what I'm struggling with. Thanks.

Hi softwareguy,

$\displaystyle x^4-20x^2+64=(x^2-16)(x^2-4)$

Now, all you have to do is factor the two "difference of squares" to reach your desired result.

• Sep 17th 2010, 08:50 AM
softwareguy
Thanks, that helps.

One other thing. Is there any "trick" to more quickly figuring this out? For example, lets say you have a larger last number, such as:

$\displaystyle x^4-61x^2+900$

I'm struggling how to come up with the answer for this one.
• Sep 17th 2010, 09:08 AM
e^(i*pi)
That is a tough one, $\displaystyle (-30)^2=900$ but that can't be right because $\displaystyle -30 -30 \neq -61$ mind you it will be close so I will try 25 which is also a factor of 900:

$\displaystyle 900 = -25 \cdot -36$. Now to check if $\displaystyle -25-36=-61$ which works.

Hence $\displaystyle x^4-61x^2+900=(x^2-25)(x^2-36)$

Yet these are both the difference of two squares: $\displaystyle x^4-61x^2+900 = (x-5)(x+5)(x-6)(x+6)$