## Multiplicity of Roots

On this page you’ll learn about multiplicity of roots, or zeros, or solutions. One of the main take-aways from the Fundamental Theorem of Algebra is that a polynomial function of degree *n* will have *n* solutions.

So, if we have a function of degree 8 called *f*(*x*), then the equation *f*(*x*) = 0, there will be *n* solutions.

The solutions can be Real or Imaginary, or even repeated. The frequency of a repeated root is called its multiplicity. Learn about how this concept works and see some examples by navigating the tabs below.

The *x* – intercepts are often called zeros or roots. The function graphed to the left has a degree of 5, meaning it could have 1 to roots. This function only has 3 Real roots.

In mathematics, the words “Real,” and “Imaginary,” are misnomers. A misnomer is a misleading name, a name that can give a false impression to the uninformed. A classic example is a driveway. A driveway is not where someone drives, but in fact where they do not drive, but just park their car! This is familiar enough to not cause confusion. However, real and imaginary numbers are likely not familiar and can be confusing.

Before we get too far into what real and imaginary numbers are, here’s what they have to do with the roots of polynomial functions. The total number of roots, real and imaginary combined, equals the degree, always! A polynomial of degree 5 will always have 5 roots. The example we used previous has 3 real roots, which means that there are two imaginary roots.

So, if we have a polynomial function, say *f*(*x*), of degree *n*, then *f*(*x*) = 0 will have *n* solutions total.

**Fact: The number of solutions to a polynomial function of degree n, f(x) = 0 will be n.**

In this section we will explore those imaginary solutions. If you’ve already learned about complex numbers, then this will not be new information to you.

Let’s start with an example. Say, *f*(*x*) = 4(*x* + 4)^{2} + 9. This is quadratic and has a vertex (the single turning point) at (-4, 9). That point is either the minimum or the maximum of the function. The vertex is in quadrant number two. The leading coefficient is positive, so this graph goes up. The vertex is the minimum, and it is above the *x* – axis. This function does not have any *x* – intercepts. It has zero **real solutions**.

But, there are two solutions to the equation 0 = 4(*x* + 4)^{2} + 9. Since there are not any real *x* – intercepts, there will be two imaginary solutions. Let’s use inverse operations to find them.

At this point we see that there are not any real *x* – intercepts. The square root of a negative number is NOT on the real number line. Here’s why.

Square roots are numbers with a quirky property. If you square a square root number, multiply it by itself, the product is the radicand. For example, if you multiply the square root of nine times itself, you get the whole number nine as the product:

The reason we can say that the square root of nine is three (or even -3 for that matter), is because three and the square root of 9 have the same property. If you square either, you get the same product.

Sometimes square roots are not rational, like There is not an integer (or rational) number squared that equals two. The number squared that is 2 is irrational, which means we can only approximate it with our decimal system (it is a non-terminating, non-repeating decimal). So, the best we can do is write But still, if you multiply these irrational radical numbers by themselves, you square them, you still get the radical.

In a way, when you’re simplifying a square root, what you’re asking is if there’s another number squared that equals the radicand? For the square root of nine, we are asking, is there another number squared that is nine? If so, we can write that number in its place, it is simpler.

Here’s where the problem comes in. Is there a real number squared that is equal to -1? Well, 1 × 1 = 1, and -1 × -1 = 1, because a negative times a negative is positive. And 1 × -1 is not a number squared because -1 and 1 are different.

There is a number squared that is equal to -1, it’s called “*i”. *Here’s the deal:

Let’s look at our original quadratic equation’s solution.

This is the same as:

We can now simplify this because we know what the square root of -1 equals: *i*.

On the coordinate plane we only have room for real numbers. So, imaginary solutions do not show up!

Let’s see another example. Once we find all of the solutions (roots), let’s check each one to show that they’re true. Remember, a solution is a value that makes an equation true.

Example: Find all *x* – intercepts for the function below.

This function is degree 4. It will have three roots because the degree is three. If only one real root exists, the other two are imaginary.

Let’s check each root to make sure they satisfy the equation *x*^{2}(*x*^{2} – 2*x* + 17) = 0.

The first is simple because zero times anything is zero. So (0)^{2}(0^{2} – 2(0) + 17) = 0 is obviously true. The other two solutions are not so obvious. Let’s work through them carefully.

Here we need to follow the order of operations, so we’ll square (1 + 4*i*) first as scratch work.

Combine like terms and square *i*.

Remember, *i* is defined as the number squared that equals -1. So, *i* × *i* = -1.

Anything times zero is zero.

Our example function has a degree of four. That means that the number of roots will be four, combining both real and imaginary. We found three *unique* roots, 0, 1 + 4*i*, and 1 – 4*i*. However, there are four solutions. The solution of zero has what is called a ** multiplicity **of two. It came from

*x*

^{2}= 0. This is

*x*×

*x*= 0. If we take each factor and set it equal to zero, we get

*x*= 0 and

*x*= 0. That’s two solutions of zero.

## Multiplicity of Roots

Consider the function *f*(*x*) = (*x*^{2} + 1)(*x* + 4)^{2}. This function has a degree of four. On its graph (to the left), you can see it has exactly one *x* – intercept, at (0, -4). Yet, we have learned that because the degree is four, the function will have four solutions to *f*(*x*) = 0.

There are two imaginary solutions that come from the factor (*x*^{2} + 1). Let’s set that factor equal to zero and solve it.

The real solution(s) come from the other factors.

This is called a multiplicity of two. The roots to this function are *i, *–*i¸ *–4 (multiplicity of two). The solution *i* has a multiplicity of one as does the solution – *i*.

**Summary:**

Given a polynomial function, *f*(*x*), of degree *n*, there will be *n* solutions to the equation *f*(*x*) = 0.

Multiplicity of One: The solution is unique.

Multiplicity: How many times a solution is found within a function.

The number *i *is defined as the number squared that is -1. So, *i*^{2} = –1, and