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