SEARCH HOME
 Math Central Quandaries & Queries
 Question from Indrajit, a student: 8x^3 + 4x -3

Hi Indrajit.

Factoring cubics is not an easy thing to do generally, but you may get lucky.

What I like to do is to remember that if f(x) can be factored, then there is some f(a) = 0. If I can find this value a, then I know that (x-a) is a factor of f(x).

f(x) = 8x3 + 4x - 3.

This is going to take some guesswork.
f(-1) = -8 - 4 - 3 = -15.
f(1) = 8 + 4 - 3 = 9.
f(0) = -3.
f(1/2) = 1 + 2 - 3 = 0 !!
Aha!

So I know that (x - 1/2) is a factor of 8x3 + 4x - 3.

I can use long division to find its quotient, which would have to be a quadratic.

x into 8x3 goes 8x2 times. NOTE THIS.
8x2 times (x - 1/2) is 8x3 - 4x2.
8x3 + 4x - 3 - (8x3 - 4x2) is 4x2 + 4x - 3.

Do it again:
x into 4x2 is 4x. NOTE THIS.
4x times (x - 1/2) is 4x2 - 2x.
4x2 + 4x - 3 - (4x2 - 2x) is 6x - 3.

Do it one more time:
x into 6x - 3 is 6. NOTE THIS.
6 times (x - 1/2) is 6x - 3.
6x - 3 - (6x - 3) is zero; there is no remainder, so indeed, (x-1/2)
is a factor of f(x).

The quotient is the sum of the terms we "noted": 8x2 + 4x + 6.

So 8x3 + 4x - 3 = (x - 1/2)(8x2 + 4x + 6).

Now you just have to factor down the quadratic.

Cheers,
Stephen La Rocque.

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.