and break it down to
(x+ 2)^5 (− 3 x+ 1)^5 = 0
(c) Here is the original equation:
(4x^2 − 9)^3 = 0
Here, the coefficient of x^2 is 4, the coefficient of x is 0, and the constant is −9. This
trinomial factors into the product of (2x+ 3) and (2x− 3). Therefore, we can rewrite
the original equation as
[(2x+ 3)(2x− 3)]^3 = 0
and break it down to
(2x+ 3)^3 (2x− 3)^3 = 0
- In each case, we can remove the exponents from the binomials, and then consider each
binomial separately as a first-degree equation when it is set equal to 0. The real roots of
the higher-degree equation are equal to the solutions of the first-degree equations. The
multiplicity of each root is the power to which its associated binomial is raised.
(a) The real roots are found by solving
x− 1 = 0
and
x− 2 = 0
Those roots are x= 1 or x= 2. Because each binomial is squared, each of these roots
has multiplicity 2.
(b) The real roots are found by solving
x+ 2 = 0
and
− 3 x+ 1 = 0
Those roots are x=−2 or x= 1/3. Because each binomial is raised to the fifth power,
each of these roots has multiplicity 5.
(c) The real roots are found by solving
2 x+ 3 = 0
Chapter 26 695
