How do I factor this?

Question

12x^6 - 27y^4

Answer 1

12x^6 - 27y^4
first, factor out a 3.
= 3(4x^6 - 9y^4)
the term in paranthesis is a difference between squares:
= 3[ (2x^3)^2 - (3y^2)^2 ]
so:
= 3(2x^3 + 3y^2)(2x^3 - 3y^2)

Answer 2

You find what each term has in common first. Both have a 3 in common, so 12x^6-27y^4=3(4x^6-9y^4). Now the number inside the parentheses is a difference of two perfect squares, so 3(4x^6-9y^4)=3(2x^3-3y^2)(2x^3+3y^2).

3(2x^3-3y^2)(2x^3+3y^2) is as far as it goes.

Answer 3

Your first step is to pull out the biggest common factor. In this case, the greater common factor is 3.

3(4x^6 - 9y^4)

Now, we note carefully that 4x^6 - 9y^4 is a difference of squares. 4 and 9 are square numbers, and x and y are to even powers (i.e. x^6 = x^3 * x^3 and y^4 = y^2 * y^2). There's a method of factoring difference of squares.

3(2x^3 - 3y^2) (2x^3 + 3y^2)

(Generally, whenever we have a difference of squares a^2 - b^2, it factors into (a-b)(a+b))

Answer 4

12x^6-27y^4

the 2 terms have a common factor,namely 3.

3(4x^6-9y^4)
=3(2x^3+3y^2)(2x^3-3y^2)

Answer 5

3*(2x^3-3y^2)*(2x^3 + 3y^2)

did i win?

... nope