Question

How do you factor `f(x)=x^3+3x^2-12x-18`?

Answer 1

`f(x)=(x-3)(x+3+sqrt(3))(x+3-sqrt(3))`

Explanation:

To factorize a n-degree-polynomial function, one must discover its roots that will allow to rearrange the function in this way
`f(x)=(x-x_1)(x-x_2)[...]"(x-x_n)`

For a polynomial of the third degree, when all coefficients are real numbers, there's at least one real root. The other two roots are either real numbers or complex conjugate numbers.

For a general solution of roots of the third degree, requiring some ability to work with complex numbers, I recommend this source:
Roots of a cubic function

But we can try to resolve the problem in the easy way.

The coefficient d may help us to find at least one real root.
`d=18=2*3^2`
This is why we should try the numbers 1, 2, 3, 6, 9, 18 and 1/2, 1/3, 1/6, 1/9 and 1/18 to verify if at least one of them is a root of the function.

In the present case, we discover that 3 is a root (`x_1=3`) of the function:
`f(x=3)=27+3*9-12*3-18=27+27-36-18=0`

Then we should divide the polynomial by `(x-x_1)=(x-3)`:
`" "x^3+3x^2-12x-18" "`|`" "x-3`
`-x^3+3x^2" "`|____
___#" "x^2+6x+6#
`" "6x^2-12x`
`" "-6x^2+18x`
`" "` __
`" "6x-18`
`" "-6x+18`
`" "` _____
`" "0`

The second degree factor can still be further factorized:
`x^2+6x+6=0`
`Delta=36-24=12`
`x=(-6+-2*sqrt(3))/2` => `x_2=-3-sqrt(3)`; `x_3=-3+sqrt(3)`

Then the function factorized is
`f(x)=(x-x_1)(x-x_2)(x-x_3)=(x-3)(x+3+sqrt(3))(x_3+3-sqrt(3))`