Question

Factor 16s^4-81

Answer 1

This is a difference of two squares, so you follow \[a^{2}-b^{2}=\left( a-b \right)\left( a+b \right)\]

Answer 2

I do, but I how do I break it down?

Answer 3

First, you'll have to find out what the square root of each term is. sqrt81 is pretty straightforward, it's 9. To find the sqrt of 16s^4, you follow\[\sqrt{ab}=\sqrt{a}\sqrt{b}\]. To take a sqrt of a number, it can also be said that you are dividing the power of that number by 2, such that\[\sqrt{a^{x}}=(a^{x})^{1/2}=a ^{x/2}\]

Answer 4

(4s^2-9)(4s^2+9)
Can you do this one problem so I can see how you do it? Step by step? I know how to do it, but I don't remember because it's been about a month since I've factored.

Answer 5

@Kuromeru

Answer 6

You've factored correctly so far, but now you can do the difference of two squares again on your left term "(4s^2-9)"

Answer 7

That's where I get stuck @Kuromeru

Answer 8

Ok, now consider 4s^2-9 as a separate case and disregard 4s^2+9 for now. You can factor 4s^2-9 in the same way that you factored 16s^4-81, so that you get an expression in the form (a-b)(a+b), but then you will multiply that term by the 4s^2+9 that you left out earlier.
Are you having trouble determining what the actual factors of the term "(4s^2-9)" are?

Answer 9

If so, refer to my second post on this page, which is still applicable to this problem.