I'm kind of struggling with factoring when encountering multiple exponents. Can someone please explain the steps to factor the following?
$\displaystyle 4x^4+12x^3-40x^2$
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:
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.
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: $\displaystyle x^4-61x^2+900 = (x-5)(x+5)(x-6)(x+6)$