Question

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

Answer 1

`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)`