(− 7 x− 12)^7 = 0
(− 118 x+ 59)^13 = 0
(− 3 x)^17 = 0
In the last case, the stand-alone constant is 0.
Multiplying out
You can multiply out an equation in binomial to the nth form to get a polynomial equation,
but the result usually looks more complicated. In some instances, the multiplied-out equation
becomes messy indeed, as would happen in the fifth example above. But the reverse process,
if you can carry it out, is useful. Once in awhile, a formidable polynomial equation can be
reduced to binomial to the nth form. Then it’s easy to solve!
What’s the real root?
Whenever we see an equation in binomial to the nth form, we can find the real root by
considering the binomial as a first-degree equation. When we do that with the above higher-
degree equations, we get
2 x− 3 = 0
6 x+ 1 = 0
23 x+ 77 = 0
− 7 x− 12 = 0
− 118 x+ 59 = 0
− 3 x= 0
These all resolve easily. The original equations, which are of degree 4, 5, 6, 7, 13, and 17, have
one real root apiece, of multiplicity 4, 5, 6, 7, 13, and 17 respectively.
Are you confused?
Perhaps you still wonder, “What’s all this fuss about root multiplicity? If an equation has one real root, isn’t
that all there is to be said about it?” The answer is, “Not exactly.” Let’s look at three equations that have
identical solution sets. The first-degree equation
2 x− 3 = 0has one real solution, x= 3/2. The fourth-degree equation
(2x− 3)^4 = 0has a single real root, x= 3/2. But there’s something about the fourth-degree equation that makes it con-
ceptually different than the first-degree equation. We can rewrite the fourth-degree equation as
(2x− 3)(2x− 3)(2x− 3)(2x− 3) = 0Binomial to the nth Power 433