Question

factor completely 25c^2-81

Answer 1

Do you know how to approach these problems? I suggest you look at the example problems on your textbook. This is a very simple one...

I always start out with ( )( )

Then I find two numbers that when you multiply together will give you a 25. Ask yourself, what 2 numbers would that be? In this case, you only have 2 options: either 1 x 25 or 5 x 5... I know it from experience, 5 x 5 is going to be it... and there's a reason for this and you'll see why.

So now I have (5c )(5c )

Then I think of two numbers that multiply together would gives me -81 AND the sum of the multiplication to the each of the two numbers that I picked(5 and 5), it would be 0. It has to be 0 because your equation doesn't have any "c" term (You only have a c^2 and a constant).

So now ask yourself, what two # would fit that criteria? It's very obvious... the choice is 9 and -9 because 9*-9 = 81, and 9*5c=45c, -9*5c=-45c, and the sum of 45c + (-45c) = 0. So 9 and -9 works.

and so now you get (5c+9)(5c-9)=0

You are done here, but if you want to go a step further, go ahead and solve for c by setting up the equation:

(5c+9) = 0 and c = -9/5
(5c-9) = 0 and c = 9/5

so c = +/- 9/5.

How you get it... I know it's confusing, I can't think of other ways to explain it. Practice more problems and you'll find it very easy to find the fitting numbers.

Answer 2

Recognize that both 25c² and 81 are perfect squares, you can use the standard identity:
a² - b² = (a+b)(a-b)
by substituting
a=√(25c²)=5c, and
b=&radic(81)=9