Given that the sum of two of the roots of #x^3-3x^2-4x+12=0# is zero, how do you factor #x^3-3x^2-4x+12# ?

1 Answer
Mar 12, 2017

Answer:

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

Explanation:

If the roots are #alpha#, #beta# and #gamma# then:

#x^3-3x^2-4x+12 = (x-alpha)(x-beta)(x-gamma)#

#color(white)(x^3-3x^2-4x+12) = x^3-(alpha+beta+gamma)x^2+(alphabeta+betagamma+gammaalpha)x-alphabetagamma#

So, equating the coefficients of #x^2#, we find:

#alpha+beta+gamma = 3#

Since the sum of two of the roots is #0#, the remaining one must be #3#.

So #(x-3)# is a factor.

In fact we find:

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

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

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

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