2.4 Finding Zeros of Higher-Degree Polynomials

To find the zeros of higher-degree polynomials, we can use ordinary division (often called long division) and synthetic division. Before we dive into those methods, it is important to learn the underlying theorem that makes them work, which is called the rational zero theorem.

The Rational Zero Theorem

The rational zero theorem states that if $f(x)$ is a polynomial with integer coefficients, and $\frac{p}{q}$  is a zero of $f(x)$, then $p$ is a factor of the constant term of $f(x)$, and $q$ is a factor of the leading coefficient (the coefficient of the highest power of $x$) of $f(x)$. In other words, if

$f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+a_{n-2}{x}^{n-2}+...+a_{1}x+a_0$

with integer coefficients, and $\frac{p}{q}$  is a rational zero of $f(x)$, then $p$ is a factor of the constant term $a_0$, and $q$ is a factor of the leading coefficient $a_n$. This theorem will help us know which numbers are potential zeros of the polynomial function.

If $a_n\ne 0$, then $f$ is said to be degree $n$, denoted by $\text{deg}(f)=n$, and $a_n$ is called the leading coefficient of $f$.

Suppose we have the following polynomial equation:

$f(x)=2x^3-5x^2-4x+3$

If the rational number $\frac{p}{q}$  is a zero of $f(x)$, $p$ is a factor of 3, and $q$ is a factor of 2. The factors of 3 are $\pm 1$ and $\pm 3$. The factors of 2 are $\pm 1$ and $\pm 2$. Therefore, the potential zeros of this polynomial are found by dividing each factor of 3 by each factor of 2, as shown below:

$\pm 1,\pm 3,\pm \frac{1}{2},\pm \frac{3}{2}$

These are potential zeros, meaning they are not all necessarily zeros. For instance, $x=1$ is not a zero:

$f(x=1)=2(1{)}^3-5(1{)}^2-4(1)+3\ne 0$

On the other hand, $x=3$ is a zero, since

$f(x=3)=2(3{)}^3-5(3{)}^2-4(3)+3=0$

Let’s look at another example. Take the following equation:

$f(x)=2x^3-3x^2-11x+6$

We can follow the same process again. If the rational number $\frac{p}{q}$  is a zero of $f(x)$, $p$ is a factor of 6, and $q$ is a factor of 2. The factors of 6 are $\pm 1$, $\pm 2$, $\pm 3$, and $\pm 6$. The factors of 2 are $\pm 1$ and $\pm 2$. Therefore, the potential zeros of this polynomial are found by dividing each factor of 6 by each factor of 2, as follows:

$\pm 1,\pm 2,\pm 3,\pm \frac{1}{2},\pm \frac{3}{2}$

We can check which of these possibilities are the zeros by plugging each value into the polynomial function to see if it equals 0.

Ordinary Division

You may be wondering why we need to learn how to perform division with polynomials. Higher-degree polynomials are not easy to factor. Also, if the coefficients of the highest power of $x$ and the constant are large, there will be many potential zeros to test out, which would take a long time. If we can find just one of these zeros, we can factor the polynomials and end up with a polynomial that is one degree less and thus more manageable.

Let $f(x)$ and $g(x)$ be polynomials and $g(x)$ not the zero-degree polynomial. Then there exist unique polynomials $q(x)$ and $r(x)$ such that

$f(x)=g(x)q(x)+r(x)$

and either $r(x)$ is the zero-degree polynomial or $\text{deg}(r(x))<\text{deg}(g(x))$.

Let’s start with the first example from the previous section:

$f(x)=2x^3-5x^2-4x+3$

Since we know that one of the zeros is $x=3$, we can divide the polynomial function by $x-3$. First, there are two $x^3$ terms in the function, so we can multiply $x-3$ by $2x^2$ to get $2x^3-6x^2$. Then we subtract it from the original function, which returns $x^2-4x+3$.

$\begin{array}{r}2x^2\phantom{0000000)}\\ x-3\overline{)2x^3-5x^2-4x+3}\\ \underset{\_}{2x^3-6x^2\phantom{0000000)}}\\ x^2-4x+3\end{array}$

To eliminate $x^2$ from the remainder, we multiply $x-3$ by $x$, and then subtract the result, $x^2-3x$, from $x^2-4x+3$. This yields the remainder of $-x+3$.

$\begin{array}{r}2x^2+x\phantom{0000)}\\ x-3\overline{)2x^3-5x^2-4x+3}\\ \underset{\_}{2x^3-6x^2\phantom{0000000)}}\\ x^2-4x+3\\ \underset{\_}{x^2-3x\phantom{(\mathrm{00 }}}\\ -x+3\end{array}$

Then we can multiply $x-3$, and subtract it from the remainder to get 0.

$\begin{array}{r}2x^2+\phantom{0}x-1\\ x-3\overline{)2x^3-5x^2-4x+3}\\ \underset{\_}{2x^3-6x^2\phantom{0000000)}}\\ x^2-4x+3\\ \underset{\_}{x^2-3x\phantom{\mathrm{000 }}}\\ -x+3\\ \underset{\_}{-x+3}\\ 0\end{array}$

So, we can factor the third-degree polynomial by $x-3$, then factor $2x^2+x-1$ as follows:

$\begin{array}{rl}f(x)& =2x^3-5x^2-4x+3\\ & =(x-3)(2x^2+x-1)\\ & =(x-3)(2x-1)(x+1)\end{array}$

Therefore, the zeros are $x=-1,\frac{1}{2},3$. When $x=0$, we have a y-intercept of (0, 3). When $x$ is less than -1, $f(x)$ is negative (you can add some number less than -1 to check). When $x$ is between -1 and $\frac{1}{2}$, $f(x)$ is positive. If $x$ is between $\frac{1}{2}$  and 3, $f(x)$ is negative, and if $x$ is greater than 3, $f(x)$ is positive.

Given all the information available, the shape of this graph looks like this:

Note that $f(-1)=f\left(\frac{1}{2}\right)=f(3)=0$, so $x=-1,\frac{1}{2} ,\mathrm{ 3}$ are roots of $f$. In general, if $x-a$ is a factor of a polynomial $f(x)$, then $a$ must be a root of $f$. The converse is also true by the following theorem:

Let $f(x)$ be a polynomial and $a$ be a real number. Then $a$ is a (real) root of $f(x)$ if and only if $x-a$ is a factor of $f(x)$, that is, $f(x)$ can be written as $f(x)=(x-a)q(x)$ for some polynomial $q(x)$.

Synthetic Division

There is a faster and easier way to divide polynomials than long division: synthetic division. We set up synthetic division in such a way that a zero of the polynomial is on the top left, and then in the rest of the top row will be the coefficients of the terms of the polynomial in order from the highest degree to the lowest degree (the constant). When the coefficient of a term is 0 (e.g., $3x^3-2x+1=3x^3+0x^2-2x+1$ ), then we write 0. An example will show this process more clearly than explaining how to do it in words.

Let’s use the following polynomial as an example:

$f(x)=2x^3-5x^2-4x+3$

We know one of the zeros is -1. We can set up synthetic division as follows:

1. Place a zero of the polynomial in the top left corner.

2. In the same row, write down the coefficients of each $x$ term in order from the highest power of $x$ to the lowest.

3. If any $x$ terms are missing, do not skip it but write a 0 in its place.

$\begin{array}{crrrr}-1& 2& -5& -4& 3\\ & & & & & \\ \end{array}$

First, we drop the coefficient of the highest $x$ term under the horizontal line.

$\begin{array}{crrrr}-1& 2& -5& -4& 3\\ & & & & & \\ & 2& & & \end{array}$

Then, we multiply this number by the zero we wrote in the top left corner, and the product is placed under the second coefficient.

$\begin{array}{crrrr}-1& 2& -5& -4& 3\\ & & -2& & & \\ & 2& & \end{array}$

Add the two numbers in the third column and write the result under the horizontal line.

$\begin{array}{crrrr}-1& 2& -5& -4& 3\\ & & -2& & & \\ & 2& -7& \end{array}$

Repeat the same process for the rest of the columns.

$\begin{array}{crrrr}-1& 2& -5& -4& 3\\ & & -2& 7& -3& \\ & 2& -7& 3& 0\end{array}$

The numbers we wrote under the horizontal line are the coefficients of a polynomial after factoring by $x+1$. In this example, the original function is a third-degree polynomial, so we have a quadratic function after factoring. Note that the last number is the remainder of the division (if it is 0, there is no remainder). Based on this result, we can rewrite the original polynomial function as follows:

$\begin{array}{rl}f(x)& =2x^3-5x^2-4x+3\\ & =(x+1)(2x^2-7x+3)\end{array}$

We can factor the quadratic part, $2x^2-7x+3$, to get:

$\begin{array}{rl}f(x)& =2x^3-5x^2-4x+3\\ & =(x-3)(2x^2+x-1)\\ & =(x+1)(x-3)(2x-1)\end{array}$

which provides the same result as the long division example.

Let’s try synthetic division with the second equation from the previous section:

$f(x)=2x^3-3x^2-11x+6$

Do you remember how to set it up? From the previous section, we know that one of the zeros is -2.

$\begin{array}{crrrr}-2& 2& -3& -11& 6\\ & & & & & \\ & & & & & \end{array}$

Follow the steps for synthetic division to find all the coefficients of the quadratic part of the function after factoring by $x+2$.

$\begin{array}{crrrr}-2& 2& -3& -11& 6\\ & & -4& 14& -6\\ & 2& -7& 3& 0& \end{array}$

After factoring by $x+2$, we have $2x^2-7x+3$, so we can factor further to get

$\begin{array}{rl}f(x)& =2x^3-3x^2-11x+6\\ & =(x+2)(2x^2-7x+3)\\ & =(x+2)(2x-1)(x-3)\end{array}$

Therefore, the zeros of this polynomial are $x=-2,\frac{1}{2} , 3$.

Example

Let’s use the synthetic division method to find the zeros of a polynomial function, and then graph the function as practice. Suppose that there is a fourth-degree polynomial function such that

$f(x)=x^3-10x-3$

Graph the function, then label all of the x-intercepts and the y-intercept.

If the rational number $\frac{p}{q}$  is a zero of $f(x)$, $p$ is a factor of -3, and $q$ is a factor of 1. The factors of -3 are $\pm 1$ and $\pm 3$. The factors of 1 are $\pm 1$. Therefore, the potential zeros of this polynomial are, by dividing each factor of -3 by each factor of 1,

$\pm 1,\pm 3$

To find one of the zeros, we can pick one to plug into the function. Let’s start with $x=-3$:

$f(x=-3)=(-3{)}^3-10(-3)-3=0$

We can factor by $x-1$. Then, we set up synthetic division as follows:

$\begin{array}{crrrr}-3& 1& 0& -10& -3\\ & & & & & \\ & & & & & \end{array}$

so that we can calculate the coefficient of the polynomial left after factoring by $x+3$.

$\begin{array}{crrrr}-3& 1& 0& -10& -3\\ & & -3& 9& 3\\ & 1& -3& -1& 0\end{array}$

Therefore, we have

$\begin{array}{rl}f(x)& =x^3-10x-3\\ & =(x+3)(x^2-3x-1)\end{array}$

Now, let’s focus on the second-degree polynomial part. Since it is not easy to factor, we can either use the quadratic formula or complete the square. We’ll use the quadratic formula here:

$x=\frac{3\pm \sqrt{(-3{)}^2-4(1)(-1)}}{2(1)}=\frac{3\pm \sqrt{13}}{2}$

The y-intercept is (0, -3). Finally, let’s check the signs of the graph between the zeros.

The graph of the polynomial function looks like this: