Algebra Know-It-ALL

416 Cubic Equations in Real Numbers

Binomial-factor form

Here’s the general form of a cubic broken down into three binomial factors, assuming that the

equation can be expressed that way with real coefficients and real constants:

(a 1 x+b 1 )(a 2 x+b 2 )(a 3 x+b 3 )= 0

The three coefficients are a 1 ,a 2 , and a 3. They must all be nonzero, so the multiplied-out equa-

tion contains a nonzero multiple of x^3. The three stand-alone constants are b 1 ,b 2 , and b 3. Here

are some examples of binomial-factor cubics:

(x− 1)(x+ 2)(x− 3) = 0

(3x+ 2)(5x+ 6)(− 7 x− 1) = 0

x(x+ 6)(x+ 8) = 0

x^2 (− 4 x− 1) = 0

The third and fourth of these equations have one and two stand-alone constants equal to 0,

respectively. In “unabridged” binomial-factor form, these equations are

(x+ 0)(x+ 6)(x+ 8) = 0

and

(x+ 0)(x+ 0)(− 4 x− 1) = 0

Multiplying out

We’re fortunate if we can reduce a cubic to three binomial factors. Consider the equation

x^3 − 2 x^2 − 5 x+ 6 = 0

This doesn’t advertise that it can be factored into a product of binomials! But try multiplying

this out:

(x− 1)(x+ 2)(x− 3) = 0

Let’s work an example backward. Start with this:

x(x+ 6)(x+ 8) = 0

If we multiply the first factor by the second, we obtain

(x^2 + 6 x)(x+ 8) = 0

Multiplying these two factors out gives us

x^3 + 14 x^2 + 48 x= 0