Factor: 100x to the power of 4 - 81
A. x^2 (10x+9) (10x -9)
B. (25x^2+9)(25x^2-9)
C. (10x^2+9)(10x^2-9)
D. (4x^2+9)(5x+3)(5x-3)

Question

anonymous · Answer

$$100x^4-81$$ start by factoring it as the difference of two squares $$a^2-b^2=(a+b)(a-b)$$ with $$a=10x^2,b=9$$

anonymous · Answer

then?

anonymous · Answer

@AccessDenied

accessdenied · Answer

What do you have so far after doing that factor?

anonymous · Answer

i don't understand how to factor

accessdenied · Answer

This is a general rule of factoring, called "difference of squares"
   $ a^2 - b^2 = (a + b)(a - b) $
Have you seen this particular form used before?

anonymous · Answer

nope

accessdenied · Answer

To show that it is true, you could work backwards:
  $ \begin{align} ( a + b ) ( a - b) &= a (a - b) + b (a - b) \quad &  \text{distribute} \
 & =a^2 - ab + ab - b^2 & \text{distribute again}\ 
 & = a^2 - b^2  & \text{cancel like terms}\end{align} $
I am not certain how to proceed if we don't have this form clear.
  $ a^2 - b^2 = (a + b)(a - b) $

accessdenied · Answer

Some common examples:
    $ \color{red}{x^2 - 1} = \color{gray}{(x)^2 - (1)^2} =\color{maroon}{ (x + 1)(x - 1)} $
    $ x^2 - 100 = \color{gray}{(x)^2 - (10)^2} = (x + 10)(x - 10) $
    $ 4x^2 - 9 = \color{gray}{(2x)^2 - (3)^2} =  (2x + 3)(2x - 3) $
    $ x^4 - 1 = \color{gray}{(x^2)^2 - 1^2} = (x^2 + 1)\color{red}{(x^2 - 1)} = (x^2 + 1)\color{maroon}{(x + 1)(x - 1)} $

Back to the original question:
Because with your original expression, you can rewrite as a difference of squares:
     $ 100x^4 - 81 $
  =   $ (\underbrace{10 x^2}_{a} )^2 - (\underbrace{9}_{b})^2 $
This form is met:  $a^2 - b^2 = (a + b)(a - b) $,  where $a = 10x^2$  and $b = 9$:
   =   $ (\underbrace{10x^2}_{a} + \underbrace{9}_{b})(\underbrace{10x^2}_{a} - \underbrace{9}_b) $

anonymous · Answer

mhm

accessdenied · Answer

It really just comes down to memorizing that $a^2 - b^2 = (a + b)(a - b) $.  Most often, a binomial with perfect squares as terms and subtraction between the two numbers can be dealt with in this way...

Sorry, I can't think of a better explanation at the moment if you've never used it.  :(

anonymous · Answer

can you help me answer it

anonymous · Answer

@AccessDenied