How do you factor # 3x^4 - 6x^3 - 4x^2 + 8x#?

1 Answer
Feb 27, 2017

Answer:

#3x^4-6x^3-4x^2+8x = x(3x^2-4)(x-2)#

#color(white)(3x^4-6x^3-4x^2+8x) = 3x(x-(2sqrt(3))/3)(x+(2sqrt(3))/3)(x-2)#

Explanation:

We can separate out the common factor #x# then factor by grouping as follows:

#3x^4-6x^3-4x^2+8x = x(3x^3-6x^2-4x+8)#

#color(white)(3x^4-6x^3-4x^2+8x) = x((3x^3-6x^2)-(4x-8))#

#color(white)(3x^4-6x^3-4x^2+8x) = x(3x^2(x-2)-4(x-2))#

#color(white)(3x^4-6x^3-4x^2+8x) = x(3x^2-4)(x-2)#

#color(white)(3x^4-6x^3-4x^2+8x) = 3x(x^2-4/3)(x-2)#

#color(white)(3x^4-6x^3-4x^2+8x) = 3x(x^2-((2sqrt(3))/3)^2)(x-2)#

#color(white)(3x^4-6x^3-4x^2+8x) = 3x(x-(2sqrt(3))/3)(x+(2sqrt(3))/3)(x-2)#