Question

What are the roots of the equation `x^3 +4x^2-4x- 16=0`?

Answer 1

The roots are:

`x = 2`, `x = -2` and `x=-4`

Explanation:

The difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

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

Given:

`x^3+4x^2-4x-16 = 0`

Note that the ratio between the first and second terms is the same as the ratio between the third and fourth terms, so this cubic factors by grouping:

`0 = x^3+4x^2-4x-16`

`color(white)(0) = (x^3+4x^2)-(4x+16)`

`color(white)(0) = x^2(x+4)-4(x+4)`

`color(white)(0) = (x^2-4)(x+4)`

`color(white)(0) = (x^2-2^2)(x+4)`

`color(white)(0) = (x-2)(x+2)(x+4)`

Hence the roots are:

`x = 2`, `x = -2` and `x=-4`

Answer 2

`x in {2,-4,-2}`

Explanation:

`x^3+4x^2-4x-16=0`

`rArr x^3+4x^2=4x+16`

`rarr x * x^2 +4 * x^2 = x * 2^2 + 4 * 2^2`

`rarr`(at least one possible solution is) `x=2`
(i.e. `(x-2)` is a factor)

Dividing `x^3+4x^2-4x-16` by `(x-2)` (using either synthetic or long division) gives:
`color(white)("XXX")x^2+6x+8`
which factors using standard operations as:
`color(white)("XXX")(x+4)(x+2)`

Therefore
`x^3+4x^2-4x-16=0`
`rArr (x-2)(x+4)(x+2)=0`

`rArr x in {2,-4,-2}`