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# 2.6: Finding Real Zeros of Polynomial Functions

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Use polynomial long division and synthetic division to find roots
• Apply synthetic division to find the rational zeros of an unfactored polynomial (of degree >2)

## Synthetic Division

As you know from years of algebra, any linear equation $ax+b=c$, can be easily solved. You have also learned how to find a solution to a quadratic equation using the quadratic formula. In this section, we will learn how to solve polynomial equations of higher degrees. We begin with a process called the synthetic division. It is a special case of the polynomial long division method. To develop the method, we will present the following example in which the divisor has the form $x-c$. We will first use the method of long division and then redo the division by the synthetic division method.

Keep in mind that if we wish to divide

$\frac{f(x)}{D(x)}$

where $f(x)$ is the dividend and $D(x)$ is the divisor, then

$\frac{f(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}$

where $Q(x)$ is the quotient and $R(x)$ is the remainder. We can also rewrite this equation in the following form

$f(x)=D(x)Q(x)+R(x)$

as a solution.

Let's say we are interested in dividing $f(x)=x^{3}+x^{2}-10x+13$ by $D(x)=x-2$.

By long division, we get

$& \qquad \qquad \qquad \quad \ \ x^2 + 3x - 4 \qquad \qquad \quad \leftarrow\text{Quotient}\\& \text{Divisor}\rightarrow x-2 \ \big ) \overline{x^{3} +x^{2} -10x +13 } \qquad \leftarrow \text{Dividend}\\& \qquad \qquad \qquad \quad \ \ x^{3} -2x^{2}\\& \qquad \qquad \qquad \quad \ \ \searrow\\& \qquad \qquad \qquad \qquad \quad \ \ 3x^{2}-10x\\& \qquad \qquad \qquad \qquad \quad \ \ 3x^{2}-6x\\& \qquad \qquad \qquad \qquad \qquad \searrow\\& \qquad \qquad \qquad \qquad \qquad \quad \ -4x+13\\& \qquad \qquad \qquad \qquad \qquad \quad \ \ -4x+8\\& \qquad \qquad \qquad \qquad \qquad \qquad \ \searrow\\& \qquad \qquad \qquad \qquad \qquad \qquad \qquad +5 \qquad \ \leftarrow \text{Remainder}$

As you can see, all the work involved is to manipulate the coefficients of the variables and the constants. Thus we could just as easily complete the division by omitting the variables, as long as we write the coefficients in the proper places. So the division problem will look like this:

$& \qquad \quad \ \ x^2 + 3x - 4 \qquad \quad \leftarrow\text{Quotient}\\& 1-2 \ \big ) \overline{1 +1 -10 +13 } \qquad \leftarrow \text{Dividend}\\& \qquad \quad 1-2\\& \qquad \ \ \searrow\\& \qquad \qquad +3 -10\\& \qquad \qquad \quad \ 3 -6\\& \qquad \qquad \quad \ \searrow\\& \qquad \qquad \qquad -4+13\\& \qquad \qquad \qquad \ -4+8\\& \qquad \qquad \qquad \quad \searrow\\& \qquad \qquad \qquad \qquad \ +5 \qquad \leftarrow \text{Remainder}$

Since the underlined numerals are repetitions of those immediately above them, we can shorten the process by simply deleting them. Further, since these underlined numbers are products of the numbers in the quotient by the 1 in the divisor, we eliminate this 1. Thus we get the following

$& \qquad \quad \ \ 1 + 3 - 4 \qquad \quad \ \ \leftarrow\text{Quotient}\\& -2 \ \big ) \overline{1 +1 -10 +13 } \qquad \leftarrow \text{Dividend}\\& \qquad \quad -2\\& \qquad \ \searrow\\& \qquad \quad \ +3 -10\\& \qquad \qquad \quad \ -6\\& \qquad \qquad \ \searrow\\& \qquad \qquad \quad \ -4+13\\& \qquad \qquad \qquad \quad +8\\& \qquad \qquad \qquad \ \searrow\\& \qquad \qquad \qquad \qquad \ +5 \quad \leftarrow \text{Remainder}$

It is also unnecessary to bring down the -10 and 13:

$& \qquad \quad \ \ 1 + 3 - 4 \qquad \quad \ \ \leftarrow\text{Quotient}\\& -2 \ \big ) \overline{1 +1 -10 +13 } \qquad \leftarrow \text{Dividend}\\& \qquad \quad -2\\& \qquad \ \searrow\\& \qquad \quad \ +3\\& \qquad \qquad \quad \ -6\\& \qquad \qquad \ \searrow\\& \qquad \qquad \quad \ -4\\& \qquad \qquad \qquad \quad +8\\& \qquad \qquad \qquad \ \searrow\\& \qquad \qquad \qquad \qquad \ +5 \quad \leftarrow \text{Remainder}$

Moving the numerals upward, we get

$& \qquad \quad \ \ 1 + 3 - 4 \qquad \quad \ \ \leftarrow\text{Quotient}\\& -2 \ \big ) \overline{1 +1 -10 +13 } \qquad \leftarrow \text{Dividend}\\& \qquad \quad \ \underline{ -2\ -6 \ +8\;\;}\\& \qquad \quad \ +3-4 \ +5 \qquad \ \leftarrow \text{Remainder}$

When the numeral 1 (the first number in the quotient) is brought down to the last line, it will contain the remainder and the quotient,

$& \qquad \quad -2 \ \big ) \overline{1 +1 -10 +13 } \qquad \leftarrow \text{Dividend}\\& \qquad \qquad \qquad\ \underline{ -2\ -6 \ +8\;\;}\\& \text{Quotient} \rightarrow 1 + 3 \ -4 \ +5 \qquad \ \leftarrow \text{Remainder}=+5$

Therefore, the quotient is $Q(x)=x^{2}+3x-4$ and the remainder is $R(x)=5$. This is the method of the synthetic division.

Example 1

Use synthetic division to find the quotient and the remainder of

$\frac{3x^{3}-8x+1}{x+2}$

Solution

First we write the divisor $x+2$ in the form $x-c$, as $x-(-2)$. Then use -2 as a “divisor” in the synthetic division as follows:

$& \-2 \ \big ) \overline{3 \ \ \ 0 \ -8 \ \ \ \ 1}\\& \quad \ \underline{\downarrow -6 \ \ \ 12 -8}\\& \quad \ 3 -6 \ \ \ 4 \ -7$

Notice that 0 is used as the coefficient of the “missing” $x^{2}$ term. Also, we wrote the coefficients of the dividend in descending order. The process of synthetic division is as follows: Bring down the first coefficient 3 and multiply by -2 (the divisor) to get -6, and then add 0 to -6 to get -6. Next, multiply -6 by -2 (the divisor) to get 12, and then add -8 to 12 to get 4. Finally, multiply 4 by -2 to get -8, and then add 1 to -8 to get the remainder, -7. As a result of this process, the quotient is

$Q(x)=3x^{2}-6x+4$

and the remainder is

$R(x)=-7$

In other words, since

$f(x) & = D(x)Q(x)+R(x)\\& = (x+2)(3x^2-6x+4)+(-7)$

Remember, this method will only work when the divisor is in the form $x-c$, that is, when the coefficient of $x$ in the divisor is 1.

## The Remainder Theorem and The Factor Theorem

This example leads us to the Remainder Theorem.

The Remainder Theorem

If a polynomial $f(x)$ of degree $n>0$ is divided by $x-c$, then the remainder $R$ is a constant and it is equal to the value of the polynomial when $c$ is substituted for $x$. That is

$f(c)=R$

Proof

The division algorithm states

$f(x)=(x-c)Q(x)+R(x)$

since the degree of the remainder must be less than the degree of the division, $x-c$, then $R(x)$ must be a constant. We will denote it by $R$. In addition, if $x=c$, we will find,

$f(c) & = (c-c) Q(x)+R\\& = 0 \cdot Q(x)+R\\& = R$

The Factor Theorem

If $f(x)$ is a polynomial of degree $n>0$ and $f(c)=0$, then $x-c$ is a factor of the polynomial $f(x)$. Further, if $x-c$ is a factor, then $c$ is a zero of $f$.

Proof

If $f(c)=0$, then by the Remainder Theorem,

$f(c)=R=0$

and the division algorithm becomes,

$f(x) & = (x-c)Q(x)+R(x)\\& = (x-c)Q(x)$

This shows that $(x-c)$ is a factor of $f(x)$.

In addition, since $x-c$ is a factor of $f(x)$, then

$f(x) & = (x-c)Q(x)\\f(c) & = (c-c)Q(c)\\& = 0$

and $c$ is thus a zero of $f$.

Example 2

Use synthetic division and the remainder and factor theorems to find the quotient $Q(x)$ and the remainder $R$ if $f(x)=2x^{3}-3x^{2}+6$ is divided by $x-5$.

Solution

$& \ 5 \ \big ) \overline{2 -3 \ \ \ 0 \ \ \ \ \ 6}\\& \quad \ \ \underline{\downarrow \ 10 \ \ 35 \ \ 175}\\& \quad \ \ 2 \ \ \ 7 \ \ 35 \ \ 181$

Hence

$2x^{3}-3x^{2}+6=(2x^{2}+7x+35)(x-5)+181$

Notice that the remainder is 181 and it can also be obtained if we simply substituted $x=5$ into $f(x)$,

$f(5) & = 2(5)^3 - 3(5)^2+6\\& = 250-75+6\\& = 181$

Example 3

If $h(x)=x^{3}-2x^{2}+5x-3$, evaluate

• $h(1)$
• $h(-2)$
• $h\left ( \frac{1}{2} \right )$

Solution

We can simply use the synthetic division to evaluate $h(x)$ at the given values. By the remainder theorem, the remainder is equal to $h(c)$.

• Using synthetic division,

$& \ 1 \ \big ) \overline{1 -2 \ \ \ 5 \ -3}\\& \quad \ \ \underline{\downarrow \ \ 1 -1 \ \ \ \ 4}\\& \quad \ \ 1 -1 \ \ \ 4 \ \ \ \ 1$

Hence, $h(1)=1$.

• By synthetic division,

$& \ -2 \ \big ) \overline{1 -2 \ \ \ 5 \ -3\;}\\& \qquad \quad \underline{\downarrow \ \ 2 \ \ \ 8 \ \ \ \ 26\;}\\& \qquad \quad 1 -4 \ 13 -29$

Hence, $h(-2)=-29$.

• By synthetic division,

$& \ \frac{1}{2} \ \big ) \overline{1 -2 \ \ \ 5 \ -3\;}\\& \qquad \underline{\downarrow \ \ \frac{1}{2} \ \frac{-3}{4} \ \ \frac{17}{8}\;}\\& \qquad 1 \ \ \frac{-3}{2} \ \frac{17}{4} \ \frac{-7}{8}$

Hence, $h\left ( \frac{1}{2} \right )=\frac{-7}{8}$.

Example 4

Show that $x+3$ is a factor of $g(x)=x^{4}+2x^{3}-3x^{2}+4x+12$. Find the quotient $Q(x)$ and express $f(x)$ is factored form.

Solution

By the factor theorem, if $f(c)=0$, then $x-c$ is a factor of the polynomial. In other words, if the synthetic division produces a remainder equal to zero, then $c$ is a factor of the polynomial. Using the synthetic division with $c=-3$,

$& \ -3 \ \big ) \overline{1 \ \ \ 2 -3 \ \ 4 \ \ \ 12\;\;}\\& \qquad \quad \underline{\downarrow -3 \ \ \ 3 \ \ 0 -12}\\& \qquad \quad 1 -1 \ \ \ 0 \ \ 4 \ \ \ \ 0$

Hence, $g(-3)=0$, and the quotient is

$Q(x)=x^{3}-x^{2}+4$

so that $g(x)$ can be written as

$g(x) & = (x-(-3))(x^3-x^2+4)\\g(x) & = (x+3)(x^3-x^2+4)$

## Rational Zero Theorem

Recall that the zeros of a polynomial represent the $x-$intercepts on the graph. It is easy to find the zeros of first or second degree polynomials but as the degree increases, the problem of determining the zeros becomes more difficult. There is a general method however for finding the zeros of polynomials (if they exist) for degrees $n>2$. Before we state the method it is important to recall one of the fundamental properties of integers: Every positive integer can be expressed as a product of prime numbers. For example, $25=5^{2}, 22=2\cdot 11$, and so on. Here 5, 2, and 11 are prime numbers.

The Rational Zero Theorem

Given the polynomial

$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$

$a_{n}\ne 0$ and $n$ is a positive integer. If the coefficients are integers and $\frac{p}{q}$ is a rational zero in lowest terms, then $p$ is a divisor of $a_{0}$ and $q$ is a divisor of $a_{n}$.

Example 5

Use the theorem above and synthetic division to find all the possible rational zeros of the polynomial

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

Solution

From the rational zero theorem, $\frac{p}{q}$ is a rational zero of the polynomial $f$. So $p$ is a divisor of 2 and $q$ is a divisor of 1. Hence, $p$ can take the following values: -1, 1, -2, 2 and $q$ can be either -1 or 1. Therefore, the possible values of $\frac{p}{q}$ are

$\frac{p}{q}:-1,1,-2,2$

So there are four possible zeros. Of these four, not more than three can be zeros of $f$ because $f$ is a polynomial with degree 3. To test which of the four possible candidates are zeros of $f$, we use the synthetic division. Recall from the remainder theorem if $f(c)=0$, then $c$ is a zero of $f$. We have

$& \ 2 \ \big ) \overline{1 \ -2 \ -1 \ \ \ \ 2\;}\\& \quad \ \ \underline{\downarrow \ \ \ 2 \ \ \ \ \ 0 \ -2}\\& \quad \ \ 1 \ \ \ \ 0 \ -1 \ \ \ \ 0$

Hence, 2 is a zero of $f$. Further, by the division algorithm,

$f(x) & = (x-c)Q(x)+R(x)\\& = (x-2)(x^2-1)+0$

The remaining zeros of $f$ are simply the zeros of $Q(x)=x^{2}-1$ which is easier to manipulate,

$Q(x) & = x^2-1\\& = (x-1)(x+1)$

and thus the remaining zeros are -1 and 1. Thus the rational zeros of $f$ are -1, 1, and 2.

Example 6

Use the theorem above and synthetic division to find all the possible rational zeros of the polynomial

$f(x)=x^{3}-2x^{2}-5x+6$

Solution

Assume $\frac{p}{q}$ is a rational zero of $f$. By the rational zero theorem, $p$ is a divisor of 6 and $q$ is a divisor of 1. Thus $p$ and $q$ can assume the following respective values

$p:1,-1,2,-2,3,-3,6,-6$

and

$q:-1,1$

Therefore, the possible rational zeros will be

$\frac{p}{q}:-1,1,-2,2,-3,3,-6,6$

Notice that with these choices for $p$ and $q$ there could be $8 \cdot 2=16$ rational zeros. But, eight of them are duplicates. For example $\frac{1}{-1}=\frac{-1}{1}=-1$. The next step is to test all these values by the synthetic division (we'll let you do this on your own for practice) and we finally find that

$1,-2, \ \text{and} \ 3$

are zeros of $f$. That is

$f(x) & = x^3 - 2x^2 - 5x +6\\& = (x-1)(x+2)(x-3)$

Keep in mind that the rational zero theorem has only a limited ability in finding the zeros of a polynomial equation. For example, the equation $x^{2}-3x+1=0$ can not be solved by using the rational zero theorem because the roots of this equation are irrational numbers.

## Graphing Polynomials Using Rational Zero Theorem

Example 7

Graph the polynomial function $h(x)=2x^{3}-9x^{2}+12x-5$.

Solution

Notice that the leading term is $2x^{3}$, where $n=3$ odd and $a_{n}=2>0$. This tells us that the end behavior will take the shape of a power function with an odd exponent.

Here, as you can see, there is no straight-forward way to find the zeros of $h(x)$. However, with the use of the factor theorem and the synthetic division, we can find the rational roots of $h(x)$. First, we use the rational zero theorem and find that the possible rational zeros are

$\frac{p}{q}:-1,1,-2,2,-5,5,-\frac{5}{2},\frac{5}{2}$

testing all these numbers by the synthetic division,

$& \ -1 \ \big ) \overline{2 \ -9 \ \ 12 \ -5\;}\\& \qquad \ \ \underline{\downarrow \ -2 \ \ \ 11 \ -23}\\& \qquad \ \ 2 \ -11 \ 23 \ -28$

-1 is not a root. Now let's test $x=1$.

$& \ 1 \ \big ) \overline{2 \ -9 \ \ \ 12 \ -5\;}\\& \quad \ \ \underline{\downarrow \ -2 \ -7 \ \ \ \ \ 5\;}\\& \quad \ \ 2 \ -7 \ \ \ 5 \ -28$

we find that 1 is a zero of $h$ and so we can re-write $h(x)$,

$h(x)=(2x^{2}-7x+5)(x-1)$

$2x^{2}-7x+5=(2x-5)(x-1)$

and so

$h(x)=(2x-5)(x-1)^{2}$

Thus 1 and $\frac{5}{2}$ are the $x-$intercepts of $h(x)$. The $y-$intercept is

$h(0)=-5$

Further, the synthetic division can be also used to form a table of values for the graph of $h(x)$:

$& x && -1 && \ \ \ 0 && 1 && \ \ \ 2 && \frac{5}{2} && 3\\& h(x) && -28 && -5 && 0 && -1 && 0 && 4$

We choose test points from each interval and find $g(x)$.

Interval Test Value $x$ $h(x)$ Sign of $h(x)$ Location of points on the graph
$(-\infty,1)$ -1 -28 - below the $x-$axis
$\left ( 1,\frac{5}{2} \right )$ $\frac{3}{2}$ $\frac{-11}{4}$ - below the $x-$axis
$\left ( \frac{5}{2},\infty \right )$ 3 4 + above the $x-$axis

From this information, the graph of $h(x)$ is shown in the two graphs below. Notice that the second graph is a magnification of $h(x)$ in the vicinity of the $x-$axis.

## Descartes’ Rule of Signs

One challenge for finding the zeros of polynomials is that you may not know how many roots a polynomial actually has. Descartes Rule of Signs helps with this problem. Descartes' Rule of Signs is given as a procedure for finding the number of possible positive and negative roots of a polynomial.

Descartes Rule of Signs

Given any polynomial, $p(x)$,

1. Write it with the terms in descending order, i.e. from the highest degree term to the lowest degree term.
2. Count the number of sign changes of the terms in $p(x)$. Call the number of sign changes $n$.
3. Then the number of positive roots of $p(x)$ is less than or equal to $n$.
4. Further, the possible number of positive roots is $n, n-2, n-4, \ldots$
5. To find the number of negative roots of $p(x)$, write $p(-x)$ in descending order as above (i.e. change the sign of all terms in $p(x)$ with odd powers), and repeat the process above. Then the maximum number of negative roots is $n$.

Example 8

Use Descartes Rule of Signs to identify the possible number of positive and negative roots of $f(x)=-2x^{3}+x^{2}-3x^{5}+5x-1$.

Solution

First, re-write $f(x)$ in descending order

$f(x)=-3x^{5}-2x^{3}+x^{2}+5x-1.$

The number of sign changes of $f(x)$ is 2, so the number of positive roots is either 2 or 0.

For the negative roots, write

$f(-x)=-3x^{5}+2x^{3}+x^{2}-5x-1$

The number of sign changes of $f(-x)$ is 2, so the maximum number of negative roots is 2.

The graph of $f(x)$ below shows that there is one negative root and two positive roots.

## Exercises

1. Use (i) long division and (ii) synthetic division to perform the divisions. Express each result in the form $f(x)=D(x)\cdot Q(x)+R$.
1. $5x^{5}-3x^{4}+2x^{3}+x^{2}-7x+3$ by $x-2$
2. $-4x^{6}-5x^{3}+3x^{2}+x+7$ by $x-1$
3. $2x^{3}-5x^{2}+5x+11$ by $x-\frac{1}{2}$
2. Use synthetic division to find $Q(x)$ and $f(c)$ so that $f(x)=(x-c)Q(x)+f(c)$ if $f(x)=-3x^{4}-3x^{3}+3x^{2}+2x-4$ and $c=-2$
3. If $f(x)=x^{3}+2x^{2}-10x+10$, use synthetic division to determine the followings:
1. $f(-1)$
2. $f(-3)$
3. $f(0)$
4. $f(4)$
5. What are the factors of $f(x)$?
4. Find $k$ so that $x-2$ is a factor of $f(x)=3x^{3}+4x^{2}+kx-20$
5. Use synthetic division to determine all the zeros of the polynomials.
1. $f(x)=3x^{3}-7x^{2}+8x-2$
2. $g(x)=4x^{4}-4x^{3}-7x^{2}+4x+3$
6. Graph the polynomial function $f(x)=x^{3}-2x^{2}-5x+6$ by using synthetic division to find the $x-$intercepts and locate the $y-$intercepts.
7. Graph the polynomial function $h(x)=x^{3}-3x^{2}+4$ by using synthetic division to find the $x-$intercepts and locate the $y-$intercepts.
8. Let $f(x)=2x^{3}-5x^{2}-4x+3$. Find the solution of the following equations:
1. $f(x)=0$
2. $f(2x)=0$
9. Refer back to Problem 6 above. Use the graph of $f(x)$ to find the solution set of the inequality $x^{3}-2x^{2}-5x+6\le0$.
10. Use the graph of $f(x)=x(x-1)(x+2)$ to find the solution set of the inequality $x(x-1)(x+2)>0$.

1. $f(x)=(x-2)(5x^{4}+7x^{3}+16x^{2}+33x+59)+121$
2. $f(x)=(x-1)(-4x^{5}-4x^{4}-4x^{3}-9x^{2}-6x-5)+2$
3. $f(x)=(x-\frac{1}{2})(2x^{2}-4x+3)+\frac{25}{2}$
2. $Q(x)=-3x^{3}+3x^{2}-3x+8, f(-2)=-20$
1. 21
2. 31
3. 10
4. 66
3. $k=-10$
1. $\frac{1}{3}$
2. $-1,-\frac{1}{2},1,\frac{3}{2}$
5. $x-$intercepts: -2, 1, 3; $y-$intercepts: 6
6. $x-$intercepts: -1, 2; $y-$intercepts: 4
1. $-1,\frac{1}{2},3$
2. $-\frac{1}{2},\frac{1}{4},\frac{3}{2}$
8. $x\in(-\infty,-2]\cup[1,3]$
9. $x\in(-2,0)\cup(1,+\infty)$

Feb 23, 2012

Dec 29, 2014