# Thread: How to factor with multiple exponents?

1. ## 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?

$\displaystyle 4x^4+12x^3-40x^2$

2. 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:
$\displaystyle 4x^2(x+5)(x-2)$

3. I don't get it. Can you please break it down step by step as to how you got the 5 and -2?

4. 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 $\displaystyle 4x^2(x+5)(x-2)$

5. 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.

6. 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.

7. 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.

8. 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: $\displaystyle x^4-61x^2+900 = (x-5)(x+5)(x-6)(x+6)$

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