Algebra Know-It-ALL

696 Worked-Out Solutions to Exercises: Chapters 21 to 29

and

2 x− 3 = 0

Those roots are x=−3/2 or x= 3/2. Because each binomial is cubed, each of these

roots has multiplicity 3.

5. Here’s the original binomial factor equation, which we have been told to solve and
   scrutinize:

(x− 3/2)^2 (2x− 7)^2 (7x)^3 (− 3 x+ 5)^5 = 0

Let’s set each binomial equal to 0, and then solve the resulting first-degree equations.

Those equations are

x− 3/2 = 0

2 x− 7 = 0

7 x = 0

− 3 x+ 5 = 0

The solutions to these first-degree equations, and therefore the real roots of the higher-

degree equation, are

x= 3/2 or x= 7/2 or x= 0 or x= 5/3

The solution set is X= {3/2, 7/2, 0, 5/3}. The multiplicity of each root is the same as the

power to which its binomial is raised in the original equation. Therefore, the root x= 3/2

has multiplicity 2, the root x= 7/2 has multiplicity 2, the root x= 0 has multiplicity 3,

and the root x= 5/3 has multiplicity 5. The degree of the original equation is the sum of

the exponents attached to the factors, which is 2 + 2 + 3 + 5 = 12.

6. Here’s the original binomial factor equation once again, for reference:

(x+ 4)(2x− 8)^2 (x/3+ 12)^3 = 0

Let’s set each binomial equal to 0, and then solve the resulting first-degree equations.

Those equations are

x+ 4 = 0

2 x− 8 = 0

x/3+ 12 = 0

The solution to the first of these is x=−4. The solution to the second is x= 4. To solve

the third equation, we can subtract 12 from each side and then multiply through by 3,

obtainingx=−36. The roots of the original equation are therefore

x=−4 or x= 4 or x=− 36