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Section 5-2 : Zeroes/Roots of Polynomials

We’ll start off this section by defining just what a root or zero of a polynomial is. We say that \(x = r\) is a root or zero of a polynomial, \(P\left( x \right)\), if \(P\left( r \right) = 0\). In other words, \(x = r\) is a root or zero of a polynomial if it is a solution to the equation \(P\left( x \right) = 0\).

In the next couple of sections we will need to find all the zeroes for a given polynomial. So, before we get into that we need to get some ideas out of the way regarding zeroes of polynomials that will help us in that process.

The process of finding the zeros of \(P\left( x \right)\) really amount to nothing more than solving the equation \(P\left( x \right) = 0\) and we already know how to do that for second degree (quadratic) polynomials. So, to help illustrate some of the ideas were going to be looking at let’s get the zeroes of a couple of second degree polynomials.

Let’s first find the zeroes for \(P\left( x \right) = {x^2} + 2x - 15\). To do this we simply solve the following equation.

\[{x^2} + 2x - 15 = \left( {x + 5} \right)\left( {x - 3} \right) = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}x = - 5,\,\,x = 3\]

So, this second degree polynomial has two zeroes or roots.

Now, let’s find the zeroes for \(P\left( x \right) = {x^2} - 14x + 49\). That will mean solving,

\[{x^2} - 14x + 49 = {\left( {x - 7} \right)^2} = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,\,\,\,\,\,\,x = 7\]

So, this second degree polynomial has a single zero or root. Also, recall that when we first looked at these we called a root like this a double root.

We solved each of these by first factoring the polynomial and then using the zero factor property on the factored form. When we first looked at the zero factor property we saw that it said that if the product of two terms was zero then one of the terms had to be zero to start off with.

The zero factor property can be extended out to as many terms as we need. In other words, if we’ve got a product of \(n\) terms that is equal to zero, then at least one of them had to be zero to start off with. So, if we could factor higher degree polynomials we could then solve these as well.

Let’s take a look at a couple of these.

Example 1 Find the zeroes of each of the following polynomials.
  1. \(P\left( x \right) = 5{x^5} - 20{x^4} + 5{x^3} + 50{x^2} - 20x - 40 = 5{\left( {x + 1} \right)^2}{\left( {x - 2} \right)^3}\)
  2. \(Q\left( x \right) = {x^8} - 4{x^7} - 18{x^6} + 108{x^5} - 135{x^4} = {x^4}{\left( {x - 3} \right)^3}\left( {x + 5} \right)\)
  3. \(R\left( x \right) = {x^7} + 10{x^6} + 27{x^5} - 57{x^3} - 30{x^2} + 29x + 20 = {\left( {x + 1} \right)^3}{\left( {x - 1} \right)^2}\left( {x + 5} \right)\left( {x + 4} \right)\)
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In each of these the factoring has been done for us. Do not worry about factoring anything like this. You won’t be asked to do any factoring of this kind anywhere in this material. There are only here to make the point that the zero factor property works here as well. We will also use these in a later example.


a \(P\left( x \right) = 5{x^5} - 20{x^4} + 5{x^3} + 50{x^2} - 20x - 40 = 5{\left( {x + 1} \right)^2}{\left( {x - 2} \right)^3}\) Show Solution

Okay, in this case we do have a product of 3 terms however the first is a constant and will not make the polynomial zero. So, from the final two terms it looks like the polynomial will be zero for \(x = - 1\) and \(x = 2\). Therefore, the zeroes of this polynomial are,

\[x = - 1\hspace{0.25in}{\mbox{and}}\,\,x = 2\]

b \(Q\left( x \right) = {x^8} - 4{x^7} - 18{x^6} + 108{x^5} - 135{x^4} = {x^4}{\left( {x - 3} \right)^3}\left( {x + 5} \right)\) Show Solution

We’ve also got a product of three terms in this polynomial. However, since the first is now an \(x\) this will introduce a third zero. The zeroes for this polynomial are,

\[x = - 5,\,\,\,x = 0,\,\,\,{\mbox{and}}\,\,\,x = 3\]

because each of these will make one of the terms, and hence the whole polynomial, zero.


c \(R\left( x \right) = {x^7} + 10{x^6} + 27{x^5} - 57{x^3} - 30{x^2} + 29x + 20 = {\left( {x + 1} \right)^3}{\left( {x - 1} \right)^2}\left( {x + 5} \right)\left( {x + 4} \right)\) Show Solution

With this polynomial we have four terms and the zeroes here are,

\[x = - 5,\,\,\,x = - 1,\,\,\,x = 1,\,\,\,{\mbox{and}}\,\,\,x = - 4\]

Now, we’ve got some terminology to get out of the way. If \(r\) is a zero of a polynomial and the exponent on the term that produced the root is \(k\) then we say that \(r\) has multiplicity \(k\). Zeroes with a multiplicity of 1 are often called simple zeroes.

For example, the polynomial \(P\left( x \right) = {x^2} - 10x + 25 = {\left( {x - 5} \right)^2}\) will have one zero, \(x = 5\), and its multiplicity is 2. In some way we can think of this zero as occurring twice in the list of all zeroes since we could write the polynomial as,

\[P\left( x \right) = {x^2} - 10x + 25 = \left( {x - 5} \right)\left( {x - 5} \right)\]

Written this way the term \(x - 5\) shows up twice and each term gives the same zero, \(x = 5\). Saying that the multiplicity of a zero is \(k\) is just a shorthand to acknowledge that the zero will occur \(k\) times in the list of all zeroes.

Example 2 List the multiplicities of the zeroes of each of the following polynomials.
  1. \(P\left( x \right) = {x^2} + 2x - 15\)
  2. \(P\left( x \right) = {x^2} - 14x + 49\)
  3. \(P\left( x \right) = 5{x^5} - 20{x^4} + 5{x^3} + 50{x^2} - 20x - 40 = 5{\left( {x + 1} \right)^2}{\left( {x - 2} \right)^3}\)
  4. \(Q\left( x \right) = {x^8} - 4{x^7} - 18{x^6} + 108{x^5} - 135{x^4} = {x^4}{\left( {x - 3} \right)^3}\left( {x + 5} \right)\)
  5. \(R\left( x \right) = {x^7} + 10{x^6} + 27{x^5} - 57{x^3} - 30{x^2} + 29x + 20 = {\left( {x + 1} \right)^3}{\left( {x - 1} \right)^2}\left( {x + 5} \right)\left( {x + 4} \right)\)
Show Solution

We’ve already determined the zeroes of each of these in previous work or examples in this section so we won’t redo that work. In each case we will simply write down the previously found zeroes and then go back to the factored form of the polynomial, look at the exponent on each term and give the multiplicity.

a In this case we’ve got two simple zeroes : \(x = - 5,\,\,x = 3\).

b Here \(x = 7\) is a zero of multiplicity 2.

c There are two zeroes for this polynomial : \(x = - 1\) with multiplicity 2 and \(x = 2\) with multiplicity 3.

d We have three zeroes in this case. : \(x = - 5\) which is simple, \(x = 0\) with multiplicity of 4 and \(x = 3\) with multiplicity 3.

e In the final case we’ve got four zeroes. \(x = - 5\) which is simple, \(x = - 1\) with multiplicity of 3, \(x = 1\) with multiplicity 2 and \(x = - 4\) which is simple.

This example leads us to several nice facts about polynomials. Here is the first and probably the most important.

Fundamental Theorem of Algebra

If \(P\left( x \right)\) is a polynomial of degree n then \(P\left( x \right)\) will have exactly \(n\) zeroes, some of which may repeat.

This fact says that if you list out all the zeroes and listing each one \(k\) times where \(k\) is its multiplicity you will have exactly \(n\) numbers in the list. Another way to say this fact is that the multiplicity of all the zeroes must add to the degree of the polynomial.

We can go back to the previous example and verify that this fact is true for the polynomials listed there.

This will be a nice fact in a couple of sections when we go into detail about finding all the zeroes of a polynomial. If we know an upper bound for the number of zeroes for a polynomial then we will know when we’ve found all of them and so we can stop looking.

Note as well that some of the zeroes may be complex. In this section we have worked with polynomials that only have real zeroes but do not let that lead you to the idea that this theorem will only apply to real zeroes. It is completely possible that complex zeroes will show up in the list of zeroes.

The next fact is also very useful at times.

The Factor Theorem

For the polynomial \(P\left( x \right)\),

  1. If \(r\) is a zero of \(P\left( x \right)\) then \(x - r\) will be a factor of \(P\left( x \right)\).

  2. If \(x - r\) is a factor of \(P\left( x \right)\) then \(r\) will be a zero of \(P\left( x \right)\).

Again, if we go back to the previous example we can see that this is verified with the polynomials listed there.

The factor theorem leads to the following fact.

Fact 1

If \(P\left( x \right)\) is a polynomial of degree \(n\) and \(r\) is a zero of \(P\left( x \right)\) then \(P\left( x \right)\) can be written in the following form.

\[P\left( x \right) = \left( {x - r} \right)Q\left( x \right)\]

where \(Q\left( x \right)\) is a polynomial with degree \(n - 1\). \(Q\left( x \right)\) can be found by dividing \(P\left( x \right)\) by \(x - r\).

There is one more fact that we need to get out of the way.

Fact 2

If \(P\left( x \right) = \left( {x - r} \right)Q\left( x \right)\) and \(x = t\) is a zero of \(Q\left( x \right)\) then \(x = t\) will also be a zero of \(P\left( x \right)\).

This fact is easy enough to verify directly. First, if \(x = t\) is a zero of \(Q\left( x \right)\) then we know that,

\[Q\left( t \right) = 0\]

since that is what it means to be a zero. So, if \(x = t\) is to be a zero of \(P\left( x \right)\) then all we need to do is show that \(P\left( t \right) = 0\) and that’s actually quite simple. Here it is,

\[P\left( t \right) = \left( {t - r} \right)Q\left( t \right) = \left( {t - r} \right)\left( 0 \right) = 0\]

and so \(x = t\) is a zero of \(P\left( x \right)\).

Let’s work an example to see how these last few facts can be of use to us.

Example 3 Given that \(x = 2\) is a zero of \(P\left( x \right) = {x^3} + 2{x^2} - 5x - 6\) find the other two zeroes.
Show Solution

First, notice that we really can say the other two since we know that this is a third degree polynomial and so by The Fundamental Theorem of Algebra we will have exactly 3 zeroes, with some repeats possible.

So, since we know that \(x = 2\) is a zero of \(P\left( x \right) = {x^3} + 2{x^2} - 5x - 6\) the Fact 1 tells us that we can write \(P\left( x \right)\) as,

\[P\left( x \right) = \left( {x - 2} \right)Q\left( x \right)\]

and \(Q\left( x \right)\) will be a quadratic polynomial. Then we can find the zeroes of \(Q\left( x \right)\) by any of the methods that we’ve looked at to this point and by Fact 2 we know that the two zeroes we get from \(Q\left( x \right)\) will also by zeroes of \(P\left( x \right)\). At this point we’ll have 3 zeroes and so we will be done.

So, let’s find \(Q\left( x \right)\). To do this all we need to do is a quick synthetic division as follows.

\[\begin{align*}\left. {\underline {\,2 \,}}\! \right| & \,\,\,\begin{array}{*{20}{l}}1&{\,\,2}&{ - 5}&{ - 6}\end{array}\\ & \,\,\,\,\,\,\underline {\,\,\begin{array}{*{20}{l}}{}&2&{\,\,\,\,8}&{\,\,\,\,\,6}\end{array}} \\ & \,\,\,\,\,\begin{array}{*{20}{l}}{\,1}&{4}&{\,\,3}&{\,\,\,\,\,0}\end{array}\end{align*}\]

Before writing down \(Q\left( x \right)\) recall that the final number in the third row is the remainder and that we know that \(P\left( 2 \right)\) must be equal to this number. So, in this case we have that \(P\left( 2 \right) = 0\). If you think about it, we should already know this to be true. We were given in the problem statement the fact that \(x = 2\) is a zero of \(P\left( x \right)\) and that means that we must have \(P\left( 2 \right) = 0\).

So, why go on about this? This is a great check of our synthetic division. Since we know that \(x = 2\) is a zero of \(P\left( x \right)\) and we get any other number than zero in that last entry we will know that we’ve done something wrong and we can go back and find the mistake.

Now, let’s get back to the problem. From the synthetic division we have,

\[P\left( x \right) = \left( {x - 2} \right)\left( {{x^2} + 4x + 3} \right)\]

So, this means that,

\[Q\left( x \right) = {x^2} + 4x + 3\]

and we can find the zeroes of this. Here they are,

\[Q\left( x \right) = {x^2} + 4x + 3 = \left( {x + 3} \right)\left( {x + 1} \right)\hspace{0.25in}\,\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,x = - 3,\,\,x = - 1\]

So, the three zeroes of \(P\left( x \right)\) are \(x = - 3\), \(x = - 1\) and \(x = 2\).

As an aside to the previous example notice that we can also now completely factor the polynomial \(P\left( x \right) = {x^3} + 2{x^2} - 5x - 6\). Substituting the factored form of \(Q\left( x \right)\) into \(P\left( x \right)\) we get,

\[P\left( x \right) = \left( {x - 2} \right)\left( {x + 3} \right)\left( {x + 1} \right)\]

This is how the polynomials in the first set of examples were factored by the way. Those require a little more work than this, but they can be done in the same manner.