How do you factor 27x^3-(1/8)?

2 Answers
May 9, 2015

This is an example of the difference of cubes, in which (a^3-b^3)=(a-b)(a^2+ab+b^2) .

Factor 27x^3-1/8 .

a=3x
b=1/2

(3x-1/2)((3x)^2+(3x)(1/2)+(1/2)^2) =

(3x-1/2)(9x^2+(3x)/2+1/4)

May 9, 2015

Notice that 27, x^3 and 1/8 are all cubes. If you were allowed complex numbers as coefficients, then this could be completely factored as (3x - 1/2)(3 omega x - 1/2)(3 omega ^2x - 1/2), where omega is the complex cube root of 1.

As it is, you are only interested in real numbers, not complex ones. So the only linear factor is (3x - 1/2). The other factor is (9x^2+(3/2)x+1/4).

Here are a couple of ways of finding that quadratic factor:

(1) Using complex arithmetic:
(3 omega x - 1/2)(3 omega ^2x - 1/2) = 9 omega^3 x^2 - (3/2)(omega + omega^2)x + 1/4 = 9x^2 + (3/2)x + 1/4
(since omega^3 = 1 and omega + omega^2 = -1).

(2) Using real arithmetic, solve:
27x^3 - 1/8 = (3x - 1/2)(ax^2 + bx + c)
= 3ax^3 + (3b - (1/2)a)x^2 + (3c - (1/2)b)x - (1/2)c
Comparing coefficients, we get:
3a = 27
(3b - (1/2)a) = 0
(3c - (1/2)b) = 0
(1/2)c = 1/8
yielding a = 9, b = 3/2 and c = 1/4.