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