# Polynomial Theorems

### Remainder Theorem If the polynomial that defines a function $f(x)$ is divided by $x - c$, the remainder is $f(c)$.
A polynomial function is a function whose rule is a polynomial in one variable, which is a sum or difference of terms of the form $ax^n$, where $a$ is a real number and $n$ is a nonnegative integer. For a polynomial function $f(x)$, the remainder theorem states that when the polynomial is divided by an expression in the form $x - c$, the remainder is $f(c)$. The value of $f(c)$ is the result when the value $c$ is substituted for $x$ in the polynomial. The remainder theorem can be used when the remainder is needed, but the full quotient is not of interest. In that case, completing the full division is unnecessary.
Step-By-Step Example
Use the Remainder Theorem
Determine the remainder:
$(x^3-4x^2+x-3) \div (x-2)$
Step 1
Write the dividend as the rule in a polynomial function.
$f(x)=x^3-4x^2+x-3$
Step 2
Substitute 2 for $x$ in $f(x)$.
\begin{aligned}f(x)&=x^3-4x^2+x-3\\f(2)&=2^3-4(2)^2+2-3\end{aligned}
Solution
Simplify.
\begin{aligned}f(2)&=2^3-4(2)^2+2-3\\&=8-16+2-3\\&=-9\end{aligned}

### Factor Theorem According to the factor theorem, for a polynomial that defines a function $f(x)$, $x - c$ is a factor if and only if $f(c) = 0$.

The factor theorem states that if $x - c$ is a factor of a polynomial that defines a function $f(x)$, then $f(c) = 0$. It also states that if $f(c) = 0$, then $x - c$ is a factor of the polynomial that defines $f(x)$. The factor theorem is an application of the fact that when an expression is divided by one of its factors, the remainder is zero.

• If $f(c) = 0$, then the remainder is zero, and $x - c$ is a factor of the polynomial.
• If $f(c)\neq0$, then the remainder is not zero, and $x - c$ is not a factor of the polynomial.
Step-By-Step Example
Use the Factor Theorem
Determine if $x+4$ is a factor of the expression:
$3x^3+8x^2-6x+1$
Step 1
Write the dividend as the rule in a polynomial function.
$f(x)=3x^3+8x^2-6x+1$
Step 2
Substitute –4 for $x$ in $f(x)$, and simplify.
\begin{aligned}f(-4)&=3(-4)^3+8(-4)^2-6(-4)+1\\&=-192+128+24+1\\&=-39\end{aligned}
Solution
Since $f(-4)\neq 0$, $x+4$ is not a factor of $f(x)$.

### Rational Zeros Theorem If $\frac{p}{q}$ is a rational zero of a polynomial function $f(x)$, then $q$ is a factor of the leading coefficient of the polynomial, and $p$ is a factor of the constant term of the polynomial.
A zero of a function is any input value that makes the output of the function equal to zero. The rational zeros theorem states that if $\frac{p}{q}$ is a rational zero of a polynomial function $f(x)$, then $q$ is a factor of the leading coefficient of the polynomial, and $p$ is a factor of the constant term of the polynomial. This means that any rational zero is a fraction formed from the factors of the constant term and leading coefficient of the polynomial that defines the function. All possible rational zeros of a polynomial function can be found by creating fractions with all combinations of the factors of the constant term and leading coefficient. Once the possible zeros are known, they can be tested to determine if they are zeros.
Step-By-Step Example
Use the Rational Zeros Theorem
Determine the possible rational zeros of the function:
$f(x)=2x^3+3x^2-8x+3$
Step 1

Determine the factors of the constant.

The constant is 3. Its factors are $\pm 1$ and $\pm 3$.

Step 2

Identify the factors of the leading coefficient.

The leading coefficient is 2. Its factors are $\pm 1$ and $\pm 2$.

Step 3
Write all possible fractions using factors of the constant in the numerator and factors of the leading coefficient in the denominator.
$\pm\frac{1}{1},\;\pm\frac{1}{2},\;\pm\frac{3}{1},\;\pm\frac{3}{2}$
Solution
Simplify the fractions. The simplified fractions are the possible rational zeros.
$\pm1,\;\pm\frac{1}{2},\;\pm3,\;\pm\frac{3}{2}$

### Intermediate Value Theorem For a polynomial function $f(x)$, if $a < b$ and $f(a)$ and $f(b)$ have opposite signs, there is at least one real zero of $f(x)$ between $a$ and $b$.

The intermediate value theorem states that if there are two function values that have opposite signs, then there is at least one real zero between their corresponding $x$-values. This means that in order for a function to change from having positive outputs to having negative outputs (or from negative outputs to positive outputs), the graph of the function must cross the $x$-axis.

For example, examine the points of a function: $(-4, 30)$, $(0,-6)$, and $(2,60)$.

• The outputs ($y$-values) of the points $(-4, 30)$ and $(0,-6)$ have opposite signs. So, there must be at least one zero between the $x$-coordinates of the points. Likewise, the outputs of $(0,-6)$ and $(2,60)$ have opposite signs, so there must be at least one zero between the $x$-coordinates of the points.
• There must be at least one zero between –4 and zero, and at least one zero between zero and 2.
• When graphing the function, the graph will show that –3, –2, –1, and 1 are all $x$-intercepts. So, the function has three zeros between –4 and zero, and one zero between zero and 2. If the values of a polynomial change signs between two points on the graph, then the graph must cross the xxx-axis. This means that there must be at least one zero of the function between the two points.

### Fundamental Theorem of Algebra Every polynomial function $f(x)$ of degree $n>0$ has at least one complex zero.

The fundamental theorem of algebra states that if a polynomial function $f(x)$ has degree $n>0$ and the coefficients are complex, then the function has at least one complex zero. A complex zero is a zero in the form $a+bi$, where $a$ and $b$ are real numbers and $i=\sqrt{-1}$. Real zeros can be written as complex zeros of the form $a+bi$, where $b = 0$. If $a+bi$ is a zero of $f(x)$, then its complex conjugate $a-bi$ is also a zero.

The fundamental theorem of algebra leads to several important results. Let $f(x)$ be a polynomial function with degree $n>0$ and leading coefficient $a$ be a real number. Then:

• $f(x)$ has $n$ complex zeros, including multiplicities.
• There are $n$ complex numbers $c_1,c_2,...,c_n$ (that are not necessarily distinct) such that $f(x)=a(x-c_1)(x-c_2)\dots(x-c_n)$.
• The polynomial has $n$ linear factors of the form $(x-c)$, where $c$ is a complex number.
• If a linear factor is repeated, then the corresponding zero has a multiplicity greater than 1.

Complex zeros that are real zeros correspond to $x$-intercepts, but those that are not real do not represent $x$-intercepts.

### Complex Zeros

Polynomial Function with Four $x$-intercepts Polynomial Function with Two $x$-intercepts
The graph shows a polynomial function of a degree of four. Given $n=4$, the polynomial has four complex zeros and four linear factors.

In the function, each linear factor corresponds to an $x$-intercept, which is consistent with the four $x$-intercepts on the graph.
The graph shows a polynomial function of a degree of four. Given $n=4$, the polynomial has four complex zeros and four linear factors.

The graph shows only two $x$-intercepts, so each linear factor does not correspond to an $x$-intercept.

Although there are four complex zeros, there are only two real zeros. The zeros that are not real numbers are not represented by $x$-intercepts.

Polynomial theorems and features of polynomial functions can be used together to graph a polynomial function.

To make a sketch of a polynomial function:

1. Apply the rational zeros theorem to identify the possible rational zeros.

2. Test the possible zeros to determine which are actual zeros of the function.

3. Determine the number of complex zeros, and identify zeros with multiplicity greater than 1.

4. Find additional points on the graph, including the $y$-intercept. If the function is even or odd, use symmetry to locate additional points.

5. Determine the end behavior of the function.

A graphing utility, such as a graphing calculator, is a useful tool for graphing complicated polynomial functions.

Step-By-Step Example
Apply Polynomial Theorems to Graph a Function
Sketch the graph of the function:
$f(x)=-x^4+9x^2+4x-12$
Step 1
Determine whether the function is even, odd, or neither.
$f(x)=-x^4+9x^2+4x-12$
The function has some terms with an even degree and some terms with an odd degree. So, the function is neither even nor odd.
Step 2

Use the rational zeros theorem to determine the possible rational zeros.

The constant term is –12. Factors of the constant term are $\pm1$, $\pm2$, $\pm3$, $\pm4$, $\pm6$, $\pm12$.

The leading coefficient is –1. Factors of the leading coefficient are $\pm1$.

The possible zeros are: $\pm1$, $\pm2$, $\pm3$, $\pm4$, $\pm6$, $\pm12$.

Step 3

Test the possible zeros to determine which are zeros of the function.

Substitute each possible zero in the function.
\begin{aligned}f(x)&=-x^{4}+9x^{2}+4x-12\\f({\color{#c42126}{1}})&=-(1)^{4}+9(1)^{2}+4(1)-12={\color{#c42126}{0}}\\f(-1)&=-(-1)^4+9(-1)^2+4(-1)-12=-8\\f(2)&=-(2)^4+9(2)^2+4(2)-12=16\\f({\color{#c42126}{-2}})&=-(-2)^{4}+9(-2)^2+4(-2)-12={\color{#c42126}{0}}\\f({\color{#c42126}{3}})&=-(3)^{4}+9(3)^{2}+4(3)-12={\color{#c42126}{0}}\\f(-3)&=-(-3)^{4}+9(-3)^{2}+4(-3)-12=-24\\f(4)&=-(4)^4+9(4)^2+4(4)-12=-108\\f(-4)&=-(-4)^{4}+9(-4)^{2}+4(-4)-12=-140\\f(6)&=-(6)^{4}+9(6)^{2}+4(6)-12=-960\\f(-6)&=-(-6)^{4}+9(-6)^2+4(-6)-12=-1\text{,}008\\f(12)&=-(12)^{4}+9(12)^{2}+4(12)-12=-19\text{,}404\\f(-12)&=-(-12)^{4}+9(-12)^{2}+4(-12)-12=-19\text{,}500\end{aligned}
The rational zeros are –2, 1, and 3.

The graph has $x$-intercepts at $x=-2$, $x=1$, and $x=3$.

Step 4

Determine the number of complex zeros.

The degree of the polynomial is four, so there are four complex zeros including multiplicities. Three zeros have been identified.

The linear factors that represent the three zeros in Step 3 are:
$\begin{gathered}(x+2)\\\\(x-1)\\\\(x-3)\end{gathered}$
If there are nonreal complex zeros, then they must occur in conjugate pairs. Since three real zeros are identified, the fourth zero must be real.
Step 5
Identify the fourth zero by first identifying the product of the three linear factors:
$x^3-2x^2-5x+6$
Then, divide the polynomial by the product of the known linear factors to determine the fourth factor.
$\frac{-x^4+9x^2+4x-12}{x^3-2x^2-5x+6}=-x-2$
The fourth factor is:
$-x-2$
It can be written as:
$-1(x+2)$
So, the fourth zero is –2.

Since –2 was already identified as a zero, it has multiplicity 2.

Step 6

Identify the additional points on the graph, including the $y$-intercept. Use points that result from testing the possible rational zeros in Step 2.

$x$ $y$
–2 0
–1 –8
0 –12
1 0
2 16
3 0
Step 7

Determine the end behavior of the function.

The degree is four, an even number. So, the graph has the same end behavior as $x$ approaches both positive and negative infinity.

Since the coefficient of the first term, $-x^{4}$, of the polynomial is –1, there are two things to consider:

• As $x$ approaches $-\infty$, $f(x)$ approaches $-\infty$.
• As $x$ approaches $\infty$, $f(x)$approaches $-\infty$.
Solution
Use all of the information from previous steps to make a sketch of the polynomial function.
Step-By-Step Example
Apply Polynomial Theorems to Graph an Even Function
Sketch the graph of the function:
$f(x)=-x^4+3x^2+4$
Step 1

Determine whether the function is even, odd, or neither.

The function has only terms with an even degree. So, it is even. This means that the graph is symmetric about the $y$-axis.

Step 2

Use the rational zeros theorem to determine the possible rational zeros.

The constant is 4. Its factors are $\pm 1$, $\pm 2$, and $\pm 4$.

The leading coefficient is –1. Its factors are $\pm1$.

Write all possible fractions using factors of the constant in the numerator and factors of the leading coefficient in the denominator.
$\frac{\pm1}{\pm1}, \frac{\pm2}{\pm1}, \frac{\pm4}{\pm1}$
Simplify the fractions to get the possible rational zeros. The possible rational zeros are $\pm1$, $\pm2$, $\pm4$.
Step 3

Test the possible zeros to determine which are zeros of the function. If the function equals zero when the value is substituted for $x$, the value is a zero of the function.

Since the function is even, it is symmetric about the $y$-axis. So, $f(-x)=f(x)$. Test only the positive possible rational zeros.
\begin{aligned}f(x)&=-x^4+3x^2+4\\f(1)&=-(1)^4+3(1)^2+4=6\\f({\color{#c42126}{2}})&=-(2)^4+3(2)^2+4={\color{#c42126}{0}}\\f(4)&=-(4)^4+3(4)^2+4=-204\end{aligned}
The rational zeros are –2 and 2.

The graph has $x$-intercepts at $x=-2$ and $x=2$.

Step 4

Determine the number of complex zeros.

The degree of the polynomial is four, so there are four complex zeros, including multiplicities. Two zeros have been identified.

The linear factors that represent the two zeros from Step 2 are:
$\begin{gathered}(x+2)\\\\(x-2)\\\end{gathered}$
The other two zeros could be irrational real numbers or nonreal complex numbers. It is also possible that one or both of the identified zeros could have a multiplicity greater than 1.
Step 5
To determine the other two zeros, first determine the product of the two linear factors:
$x^2-4$
Then, divide the polynomial by the product of the known linear factors.
$\frac{-x^4+3x^2+4}{x^2-4}=-x^2-1$
The quotient is not a linear factor. Set the quotient equal to zero, and solve.
\begin{aligned}-x^2-1&=0\\-1&=x^2\\\pm\sqrt{-1}&=\sqrt{x^2}\\\pm i&=x\end{aligned}
So, the other two zeros are $x=i$ and $x=-i$. Since they are not real numbers, the zeros are not represented on the graph by $x$-intercepts.
Step 6

Identify additional points on the graph, including the $y$-intercept. Use points that result from testing the possible rational zeros in Step 2.

 $x$ $y$ 0 4 1 6 2 0 3 –50

Since the function is even and is symmetric about the $y$-axis, the points $(-1, 6)$, $(-2,0)$, and $(-3, -50)$ are also on the graph.

The value –50 may not show on the graph, depending on the range chosen for the $y$-axis, but it helps to indicate the shape of the graph.

Step 7

Determine the end behavior of the function.

The degree is 4, an even number. So, the graph has the same end behavior as $x$ approaches both positive and negative infinity.

Since the coefficient of the first term, $-x^{4}$, is –1, there are two things to consider:

• As $x$ approaches $-\infty$, $f(x)$ approaches $-\infty$.
• As $x$ approaches $\infty$, $f(x)$ approaches $-\infty$.
Solution
Use all the information to make a sketch of the polynomial function.