Algebra Know-It-ALL

We have a root in every single case where one of these factors becomes equal to 0. There are not four
such instances here, but 10! We “hit the target” once when x=−1, twice when x= 2, three times
whenx=−3, and four times when x= 4. This is true even though the real solution set has only four
elements:

X= {−1, 2, −3, 4}

Incidentally, the degree of the original equation is equal to the sum of the exponents attached to the bino-
mial factors. That’s 1 + 2 + 3 + 4 = 10. On that basis, we can immediately see that

(x+ 1)(x− 2)^2 (x+ 3)^3 (x− 4)^4 = 0

is a 10th-degree equation in the variable x.

Here’s a challenge!

State the real roots of the following equation. Also state the real solution set X and the multiplicity of each
root. What is the degree of the equation?

(x+ 6)(2x− 5)^2 (7x)(− 3 x)= 0

Solution

We take each binomial individually, set it equal to 0, and then solve the resulting first-degree
equations:

x+ 6 = 0

2 x− 5 = 0

7 x= 0

− 3 x= 0

The real roots are

x=−6 or x= 5/2 or x= 0

and the real solution set is X= {−6, 5/2, 0}. The root −6 has multiplicity 1. The root 5/2 has multiplicity 2.
The root 0 has multiplicity 2. These facts can be clarified by stating the original equation as

(x+ 6)(2x− 5)(2x− 5)(7x)(− 3 x)= 0

The degree of the original equation is the sum of the exponents attached to the factors. In this example,
whether we write the equation in the original form or in the fully expanded binomial factor form, that sum
is 5, indicating that it’s a fifth-degree equation.

Binomial Factors 437