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

2 Answers
May 16, 2015

First, we must identify the common elements in this function. Let's just disaggregate them for a while:

#color(green)(2*x)*x*x+2*color(green)(2*x)*x-2*2*color(green)(2*x)#

Note that the three of them are multiplied by #2x#. Therefore, this element comes out:

#2x(x^2+2x-4)#

As inside the parenthesis we have a quadratic, we can solve it and find its roots, which are #x_1=-1-sqrt(5)# and #x_2=-1+sqrt(5)#.

We can rewrite these roots as

#x_1+1+sqrt(5)=0# and #x_2+1-sqrt(5)=0#

...and use these as our factors:

#2x(x+(1+sqrt(5)))(x+(1-sqrt(5)))#

May 16, 2015

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

#x^2+2x-4# is of the form #ax^2+bx+c# with #a=1#, #b=2# and #c=-4#.

The discriminant of this quadratic is

#Delta = b^2-4ac = 2^2-4xx1xx-4 = 4+16 = 20#

Since this is positive, the quadratic equation #x^2+2x-4 = 0# has 2 distinct real roots. Unfortunately, #20# is not a perfect square, so those roots are not rational.

The roots of #x^2+2x-4 = 0# are:

#x = (-b+-sqrt(Delta))/(2a)#

#=(-2+-sqrt(20))/2#

#=(-2+-2sqrt(5))/2#

#=-1+-sqrt(5)#

So #(x - (-1+sqrt(5))) = (x+1-sqrt(5))#

and #(x - (-1-sqrt(5))) = (x+1+sqrt(5))# are both factors.

#x^2+2x-4 = (x+1-sqrt(5))(x+1+sqrt(5))#

So if we are allowed irrational factors,

#2x^3+4x^2-8x = 2x(x+1-sqrt(5))(x+1+sqrt(5))#

Otherwise, we have to stop at

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