and the solution set X is {0, −6,−8}. A new root shows up this time: x= 0! The fact that one of the roots
is 0 caused us to inadvertently divide by 0 when we divided the equation through by x. This “blinded” us
to the existence of that root.
Binomial Times Trinomial
When a cubic can be expressed as a binomial multiplied by a trinomial, the equation is in
binomial-trinomial form. (Actually, I’ve never seen that expression used in other texts. But it’s
easy to remember, don’t you think?) A cubic in this form is not particularly difficult to solve
for real roots. The technique shown in this section will also reveal the complex-number roots
of a cubic equation, if any such roots exist.
Binomial-trinomial form
Suppose that a 1 and a 2 are nonzero real numbers. Also suppose that b 1 ,b 2 , and c are real num-
bers, any or all of which can equal 0. The binomial-trinomial form of a cubic equation in the
variable x can be written as follows:
(a 1 x+b 1 )(a 2 x^2 +b 2 x+c)= 0
Here are some examples of cubics in the binomial-trinomial form:
(− 4 x− 3)(− 7 x^2 + 6 x− 13) = 0
(3x+ 5)(16x^2 − 56 x+ 49) = 0
(3x)(4x^2 − 7 x− 10) = 0
(− 21 x+ 2)(3x^2 − 14) = 0
In the third case above, the stand-alone constant is 0 in the binomial. In the fourth case, the
coefficient of x is 0 in the trinomial.
Multiplying out
Let’s take a specific example of a cubic in binomial-trinomial form and multiply it out. Here’s
a good one, with plenty of sign changes to make it interesting:
(− 4 x− 3)(− 7 x^2 + 6 x− 13) = 0
Using the expanded product of sums rule, we obtain
28 x^3 − 24 x^2 + 52 x+ 21 x^2 − 18 x+ 39 = 0
Consolidating the terms for x^2 and x, we get
28 x^3 − 3 x^2 + 34 x+ 39 = 0
Binomial Times Trinomial 419
