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

1 Answer
Dec 31, 2015

Answer:

#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))#