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

##### 1 Answer

#### Answer:

with corresponding zeros

#### 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

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

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)#