How do you factor # x^4 + 7x^3 - 8x - 56 = 0 #?

1 Answer
Sep 15, 2016

Answer:

#x^4+7x^3-8x-56 = (x-2)(x^2+2x+4)(x+7)#

#color(white)(x^4+7x^3-8x-56) = (x-2)(x+1-sqrt(3)i)(x+1+sqrt(3)i)(x+7)#

with corresponding zeros #x=2#, #x=-7# and #x = -1+-sqrt(3)i#

Explanation:

The difference of cubes identity can be written:

#a^3-b^3=(a-b)(a^2+ab+b^2)#

We will use this with #a=x# and #b=2# later.

#color(white)()#
Note that the ratio between the first and second terms is the same as that between the third and fourth terms. So this quartic will factor by grouping:

#x^4+7x^3-8x-56 = (x^4+7x^3)-(8x+56)#

#color(white)(x^4+7x^3-8x-56) = x^3(x+7)-8(x+7)#

#color(white)(x^4+7x^3-8x-56) = (x^3-8)(x+7)#

#color(white)(x^4+7x^3-8x-56) = (x^3-2^3)(x+7)#

#color(white)(x^4+7x^3-8x-56) = (x-2)(x^2+2x+2^2)(x+7)#

#color(white)(x^4+7x^3-8x-56) = (x-2)(x^2+2x+4)(x+7)#

The Real zeros corresponding to the linear factors we have found are #x=2# and #x=-7#

The remaining quadratic factor has no Real zeros, so will only factor using Complex coefficients.

#x^2+2x+4 = x^2+2x+1+3#

#color(white)(x^2+2x+4) = (x+1)^2-(sqrt(3)i)^2#

#color(white)(x^2+2x+4) = ((x+1)-sqrt(3)i)((x+1)+sqrt(3)i)#

#color(white)(x^2+2x+4) = (x+1-sqrt(3)i)(x+1+sqrt(3)i)#