Algebra Know-It-ALL

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

(a) The real root is found by solving

x+ 3 = 0

That root is x=−3. Because the binomial is raised to the fourth power, this single real

root has multiplicity 4.

(b) The real root is found by solving

x− 2 = 0

That root is x= 2. Because the binomial is raised to the sixth power, this single real

root has multiplicity 6.

(c) The real root is found by solving

4 x− 3 = 0

We can add 3 to each side and then divide through by 4, obtaining the root x= 3/4.

Because the binomial is raised to the eighth power, this single real root has multiplicity 8.

3. In each of these situations, the trinomial can be factored into the product of two
   different binomials. Then that product is raised to the indicated power.
   (a) Here is the original equation:

(x^2 − 3 x+ 2)^2 = 0

In the trinomial, the coefficient of x^2 is 1, the coefficient of x is −3, and the stand-

alone constant is 2. This trinomial factors into the product of (x− 1) and (x− 2).

Therefore, the original equation can be rewritten as

[(x− 1)(x− 2)]^2 = 0

which can be further broken down to

(x− 1)^2 (x− 2)^2 = 0

(b) Here is the original equation:

(− 3 x^2 − 5 x+ 2)^5 = 0

In the trinomial, the coefficient of x^2 is −3, the coefficient of x is −5, and the constant

is 4. This trinomial factors into the product of (x+ 2) and (− 3 x+ 1). Therefore, we

can rewrite the original equation as

[(x+ 2)(− 3 x+ 1)]^5 = 0