How do you factor completely #12x^3 - 3xy^2#?

2 Answers
Feb 19, 2017

Answer:

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

Explanation:

#12x^3-3xy^2#

take out all common factors

#3x(4x^2-y^2)#

use difference of squares

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

Feb 19, 2017

To factor, we have to divide out common factors in each term. Let's first write out the factors:

#12x^3# = #4*3*x*x*x#

#-3xy^2# = #3*-1*x*y*y#

Now look for the factors that each term share. We see that there is a #3# and an #x# that is found in each list. Now we have to take this out of each list.

#(3*x)(4*x*x)#

#(3*x)(-1*y*y)#

Remember that these were being subtracted, so rewrite:

#(3*x)[(4*x*x) -(1*y*y)]#

Simplify.

#(3x)(4x^2-y^2)#

Use difference of squares to simplify further.

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