How do you factor #4x^5-16x^3+12x# ?

1 Answer
Nov 26, 2017

Answer:

#4x^5-16x^3+12x = 4x(x-1)(x+1)(x-sqrt(3))(x+sqrt(3))#

Explanation:

The difference of squares identity can be written:

#A^2-B^2 = (A-B)(A+B)#

We will use this a couple of times, but first separate out the common factor #4x#...

#4x^5-16x^3+12x = 4x(x^4-4x^2+3)#

#color(white)(4x^5-16x^3+12x) = 4x((x^2)^2-4(x^2)+3)#

#color(white)(4x^5-16x^3+12x) = 4x(x^2-1)(x^2-3)#

#color(white)(4x^5-16x^3+12x) = 4x(x^2-1^2)(x^2-(sqrt(3))^2)#

#color(white)(4x^5-16x^3+12x) = 4x(x-1)(x+1)(x-sqrt(3))(x+sqrt(3))#