We now have a binomial to the nth equation with n= 4. We can find the solitary real root by setting the
binomial equal to 0:
x+ 1 = 0
This first-degree equation resolves to x=−1. The real solution set of the original trinomial-squared equa-
tion is X= {−1}. The single root has multiplicity 4.
Binomial Factors
A higher-degree equation can appear as a product of binomials. Such an equation has one
real root for each factor. You can find each root by setting every binomial equal to 0 and then
creating first-degree equations from them. Each root has multiplicity equal to the number of
times its binomial appears in the product. If a binomial is raised to a power, then its root has
multiplicity equal to that power.
The general form
Letx be a real-number variable. Suppose that a 1 ,a 2 ,a 3 , ... an are nonzero real coefficients of x,
andb 1 ,b 2 ,b 3 , ... bn are real stand-alone constants. Consider the equation
(a 1 x+b 1 )(a 2 x+b 2 )(a 3 x+b 3 ) ··· (anx+bn)= 0
This is the binomial factor form for an equation of degree n. Here are some examples of equa-
tions in this form where n > 3:
(x+ 1)(x− 2)(x+ 3)(x− 4) = 0
(x+ 1)(x− 2)^2 (x+ 3)^3 (x− 4)^4 = 0
(3x+ 7)^6 (4x− 5)^2 = 0
(−x+ 1)(− 7 x+2)^7 = 0
(− 3 x+ 21)^2 (− 8 x+ 5)^3 = 0
(x+ 6)(2x− 5)^2 (7x)(− 3 x)= 0
In all but the first of these equations, some of the binomials are raised to powers. That’s the
equivalent of repeating those binomials in the products by the number of times the power
indicates. In the last equation, two of the stand-alone constants are equal to 0.
Multiplying out
You can multiply out an equation that appears in the binomial factor form to get a polyno-
mial equation. If you want to do this for each of the six equations in the previous paragraph,
go ahead! You’ll end up with equations that are difficult to solve for anyone who comes across
them for the first time.
Binomial Factors 435
