Algebra Know-It-ALL

434 Polynomial Equations in Real Numbers

If we substitute 3/2 for x here, we get

(2× 3/2 − 3)(2 × 3/2 − 3)(2 × 3/2 − 3)(2 × 3/2 − 3) = 0

which reduces to

(3− 3)(3 − 3)(3 − 3)(3 − 3) = 0

and further to

0 × 0 × 0 × 0 = 0

To carry out the substitution and simplification process completely, we must repeat it for each of the four

binomials. The root x= 3/2 exists, in effect, “four times over.” Now look at this:

(2x− 3)^345 = 0

Here, the single real root x= 3/2 exists “345 times over.”

The solution sets are identical for the first-degree, the fourth-degree, and the 345th-degree equations in

this example. But the equations themselves are vastly different!

Here’s a challenge!

Consider the following equation, which is in the form of a trinomial squared. Find all the real roots, and

state the multiplicity of each.

(x^2 + 2 x+ 1)^2 = 0

Solution

Look closely at the trinomial. Suppose we set it equal to 0, so it becomes the quadratic

x^2 + 2 x+ 1 = 0

We can factor this to get

(x+ 1)^2 = 0

If we substitute (x+ 1)^2 for the trinomial in the original equation, we have

[(x+ 1)^2 ]^2 = 0

The product of powers rule from Chap. 9 tells us that this is the same as

(x+ 1)(2×2)= 0

which can be simplified to

(x+ 1)^4 = 0