5.6: Zeros of Polynomial Functions

Section Preview

Note to the Instructor (click to expand)

This very dense section contains the Factor and Remainder theorems along with the Rational Roots Theorem (which I called the Rational Zero Theorem and was too lazy to change back to the Rational Roots Theorem). You also cover the Fundamental Theorem of Algebra, Complete Factorization Theorem, and the Conjugate Zeros Theorem. Like I said, this is a dense section. Expect to spend at least two class sessions here. This is necessary theory for Calculus.

Operational Prerequisites (i.e., "Completing the Homework"): The following prerequisite topics (which have not already been listed as prerequisites in previous sections of this text) are needed to complete the homework.

(click to expand)

• Arithmetic
  • Factors, Prime Factorizations, and Least Common Multiples

The following is a list of learning objectives for this section.

Learning Objectives (click to expand)

• Evaluate a polynomial using the Remainder Theorem.
• Use the Factor Theorem to solve a polynomial equation.
• Use the Rational Zeros Theorem to find rational zeros.
• Find zeros of a polynomial function.
• Use the Complete Factorization Theorem to find polynomials with given zeros.
• Solve real-world applications of polynomial equations.

A new bakery offers decorated, multi-tiered cakes for display and cutting at Quinceañera and wedding celebrations, as well as sheet cakes to serve most of the guests. The bakery wants the volume of a small sheet cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be?

This problem can be solved by writing a cubic function and solving a cubic equation for the volume of the cake. In this section, we will discuss a variety of tools for writing polynomial functions and solving polynomial equations.

Evaluating a Polynomial Using the Remainder Theorem

In the last section, we learned how to divide polynomials. We can now use polynomial division to evaluate polynomials using the Remainder Theorem.

Theorem: Remainder Theorem

If a polynomial \(f(x)\), where \(\text{deg}(f(x)) \geq 1\), is divided by \(x−c\), the remainder is \(f(c)\). That is,\[f(x) = (x - c) \cdot Q(x) + f(c)\nonumber\]for some polynomial \(Q(x)\) whose degree is one less than the degree of \(f(x)\).

Proof

Let \( f(x) \) be a polynomial of degree one or greater. The Division Algorithm states there exists a quotient, \( Q(x) \), and remainder, \( R(x) \), such that\[ \dfrac{f(x)}{x-c} = Q(x) + \dfrac{R(x)}{x - c}. \nonumber \]Multiplying both sides by \( x - c \), we get\[ f(x) = Q(x) (x - c) + R(x). \nonumber \]We now use this to evaluate \( f(c) \).\[ f(c) = Q(c)(c - c) + R(c) = Q(c) \cdot 0 + R(c) = R(c). \nonumber \]Hence, \( f(c) = R(c) \). However, the remainder must be of a degree strictly less than that of the divisor. Since the divisor, \( x - c \), is of degree one, the remainder must be of degree zero. That is, \( R(c) = r \) is constant. Thus,\[ f(c) = r = \text{the remainder when }f(x)\text{ is divided by }x - c. \nonumber \]

The Remainder Theorem gives us an alternative (and sometimes faster) way to evaluate polynomial functions. Let's see how this works.

Example \(\PageIndex{1}\): Using the Remainder Theorem to Evaluate a Polynomial

Use the Remainder Theorem to evaluate \[f(x)=6 x^4-x^3-15 x^2+2 x-7 \text{ at } x=2.\nonumber\]

Solution

To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by \(x-2\).\[\begin{array}{r}
\underline{2} \mid & 6 & -1 & -15 & 2 & -7 \\
& & 12 & 22 & 14 & 32 \\
\hline & 6 & 11 & 7 & 16 & 25 \\
\end{array}\nonumber\]The remainder is 25. Therefore, \(f(2)=25\).

We can check our answer by evaluating \(f(2)\).\[\begin{array}{rrcl}
& f(x) & = & 6 x^4-x^3-15 x^2+2 x-7 \\[6pt]
\implies & f(2) & = & 6(2)^4-(2)^3-15(2)^2+2(2)-7 \\[6pt]
& & = & 6 (16) - (8) - 15 (4) + 2(2) - 7 \\[6pt]
& & = & 96 - 8 - 60 + 4 - 7 \\[6pt]
& & = & 25 \\[6pt]
\end{array}\nonumber\]

If you pay attention to the amount of work required to evaluate \(f(2)\) in Example \(\PageIndex{1}\) using traditional methods, you should notice that it's actually longer than using the Remainder Theorem. Moreover, synthetic division only requires the operations of addition and multiplication, whereas our traditional function evaluation method requires addition, multiplication, and powers.

Checkpoint \(\PageIndex{1}\)

Use the Remainder Theorem to evaluate \[f(x)=2 x^5-3 x^4-9 x^3+8 x^2+2 \text{ at } x=-3.\nonumber\]

Using the Factor Theorem to Solve a Polynomial Equation

The Factor Theorem is another theorem that helps us analyze polynomial equations. It tells us how the zeros of a polynomial are related to the factors.

Theorem: Factor Theorem

Let \( f(x) \) be a nonzero polynomial function having degree greater than zero. \(c\) is a zero of \(f(x)\) if and only if \((x−c)\) is a factor of \(f(x)\).

Proof

Let \( f(x) \) be a polynomial function of degree greater than zero.
(\( \implies \)): Suppose \( c \) is a zero of \( f(x) \). Then \( f(c) = 0 \). By the Remainder Theorem, if \( f(x) \) is divided by \( x - c \), then\[ \dfrac{f(x)}{x - c} = Q(x) + \dfrac{R(x)}{x - c} = Q(x) + \dfrac{f(c)}{x - c} = Q(x) + \dfrac{0}{x - c} = Q(x). \nonumber \]Multiplying both sides by \( x - c \), we get\[ f(x) = q(x)(x - c). \nonumber \]Hence, \( x - c \) is a factor of \( f(x) \).
(\( \impliedby \)): Suppose \( x - c \) is a factor of \( f(x) \). Then\[ f(x) = (x - c)g(x), \nonumber \]for some polynomial \( g(x) \). Therefore,\[ f(c) = (c - c)g(c) = 0\cdot g(c) = 0. \nonumber \]Thus, \( f(c) = 0 \), which means \( c \) is a zero of \( f(x) \).

We have been "dancing around" the Factor Theorem for a couple of sections now; however, it wasn't until now that we formally stated the result. As we will soon see, a polynomial of degree \(n\) with complex-numbered coefficients will have \(n\) complex-numbered zeros. We can use the Factor Theorem to completely factor a polynomial into the product of \(n\) factors. Once the polynomial has been completely factored, we can easily determine the zeros of the polynomial.

Example \(\PageIndex{2}\): Using the Factor Theorem to Solve a Polynomial Equation

Show that \((x+2)\) is a factor of \(x^3-6 x^2-x+30\). Find the remaining factors. Use the factors to determine the zeros of the polynomial.

Solution

We can use synthetic division to show that \((x+2)\) is a factor of the polynomial.\[\begin{array}{r}
\underline{-2} \mid & 1 & -6 & -1 & 30 \\
& & -2 & 16 & -30 \\
\hline & 1 & -8 & 15 & 0 \\
\end{array}\nonumber\]The remainder is zero, so \((x+2)\) is a factor of the polynomial. We can use the Division Algorithm to write the polynomial as the product of the divisor and the quotient:\[(x+2)\left(x^2-8 x+15\right).\nonumber\]We can factor the quadratic factor to write the polynomial as\[(x+2)(x-3)(x-5).\nonumber\]By the Factor Theorem, the zeros of \(x^3-6 x^2-x+30\) are \(-2\), \(3\), and \(5\).

Checkpoint \(\PageIndex{2}\)

Use the Factor Theorem to find the zeros of \(f(x)=x^3+4 x^2-4 x-16\) given that \((x-2)\) is a factor of the polynomial.

Example \(\PageIndex{3}\): Writing a Formula for a Polynomial Function from the Graph

Write a formula for the polynomial function shown in Figure \(\PageIndex{1}\).

Figure \(\PageIndex{1}\)

Solution

This graph has three \(x\)-intercepts: \(x=-3\), \(2\), and \(5\). The \(y\)-intercept is located at \((0,-2)\). At \(x=-3\) and \(x=5\), the graph passes through the axis linearly, suggesting the corresponding factors of the polynomial will be linear. At \(x=2\), the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be second degree (quadratic). Together, this gives us\[f(x) = a(x+3) (x-2)^2 (x-5).\nonumber\]To determine the stretch factor, we utilize another point on the graph. We will use the \(y\)-intercept \((0,-2)\), to solve for \(a\).\[\begin{array}{rrclcl} & f(0) & = & a(0 + 3)(0 - 2)^2 (0 - 5) & \quad & \left( \text{substituting} \right) \\[6pt] \implies & -2 & = & a(0 + 3)(0 - 2)^2 (0 - 5) & \quad & \left( \text{substituting} \right) \\[6pt] \implies & -2 & = & -60a & \quad & \left( \text{simplifying} \right) \\[6pt] \implies & \dfrac{1}{30} & = & a & \quad & \left( \text{dividing both sides by }-60 \right) \\[6pt] \end{array}\nonumber\]The graphed polynomial appears to represent the function \(f(x)= \frac{1}{30} (x+3) (x-2)^2 (x-5)\).

Interactive Element: The Stretch Factor and the Fixed Zeros

The following Interactive Element might help you visualize how the stretch factor \(a\) reshapes a polynomial's graph without moving its zeros.

Interact: Drag the red point along the \(y\)-axis to change the stretch factor \(a\).

Observation: Stretching the graph vertically leaves the \(x\)-intercepts at \(x=-3,\ 2,\ 5\) fixed; only one value of \(a\) sends the graph through a given point off the axis.

Checkpoint \(\PageIndex{3}\)

Given the graph shown in Figure \(\PageIndex{2}\), write a formula for the function shown.

Figure \(\PageIndex{2}\)

Using the Rational Zeros Theorem to Find Rational Zeros

Another use for the Remainder Theorem is to test whether a rational number is a zero for a given polynomial. But first we need a pool of rational numbers to test. The Rational Zeros Theorem (also called the Rational Roots Theorem) helps us to narrow down the number of possible rational zeros using the ratio of the factors of the constant term and factors of the leading coefficient of the polynomial. Before stating the Rational Zeros Theorem, let's motivate why it works with a simple example.

Consider a quadratic function with two zeros, \(x=\frac{2}{5}\) and \(x=\frac{3}{4}\). By the Factor Theorem, these zeros have factors associated with them. Let us set each factor equal to \(0\) and then construct the original quadratic function absent its stretching factor.\[\begin{array}{rccccl}
& x-\dfrac{2}{5} = 0 & \text{or} & x - \dfrac{3}{4}=0 & \quad & \left(\text{setting each factor equal to } 0\right) \\[6pt]
\implies & 5 x - 2 = 0 & \text{or} & 4 x-3=0 & \quad & \left(\text{multiplying both sides of the equation to eliminate fractions}\right) \\[6pt]
\end{array}\nonumber\]\[\begin{array}{rrclcl} \implies & f(x) & = & (5 x-2)(4 x-3) & \quad & \left(\text{creating the quadratic function, multiplying the factors}\right) \\[6pt]
\implies & f(x) & = & 20 x^2-23 x+6 & \quad & \left(\text{expanding the polynomial}\right) \\[6pt]
\implies & f(x) & = & (5 \cdot 4) x^2-23 x+(2 \cdot 3) & & \\[6pt]
\end{array}\nonumber\]Notice that two of the factors of the constant term, \(6\), are the two numerators from the original rational roots: \(2\) and \(3\). Similarly, two of the factors from the leading coefficient, \(20\), are the two denominators from the original rational roots: \(5\) and \(4\).

If we used this single example to infer that the numerators of the rational roots will always be factors of the constant term and the denominators will be factors of the leading coefficient, we would be (somewhat) citing the Rational Zeros Theorem. However, as we know, examples are not proofs!

Theorem: Rational Zeros Theorem

Suppose\[f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+a_{0}\nonumber \]is a polynomial of degree \(n\) with \(n \geq 1\), and \(a_{0}\), \(a_{1}\), \(\ldots\), \(a_{n}\) are integers. If \(r\) is a rational zero of \(f\), then \(r\) must be of the form \(\pm \frac{p}{q}\), where \(p\) is a factor of the constant term \(a_{0}\), and \(q\) is a factor of the leading coefficient \(a_{n}\).

Proof

Suppose \(c\) is a zero of \(f\) and \(c=\frac{p}{q}\) in lowest terms. This means \(p\) and \(q\) have no common factors. Since \(f(c) = 0\), we have\[ f(c) = f\left( \dfrac{p}{q} \right) = a_{n}\left(\dfrac{p}{q}\right)^{n}+a_{n-1}\left(\dfrac{p}{q}\right)^{n-1}+\ldots+a_{1}\left(\dfrac{p}{q}\right)+a_{0}=0.\nonumber \]That is,\[ a_{n}\left(\dfrac{p}{q}\right)^{n}+a_{n-1}\left(\dfrac{p}{q}\right)^{n-1}+\ldots+a_{1}\left(\dfrac{p}{q}\right)+a_{0}=0.\nonumber \]Multiplying both sides of this equation by \(q^{n}\), we clear the denominators to get\[a_{n} p^{n}+a_{n-1} p^{n-1} q+\ldots+a_{1} p q^{n-1}+a_{0} q^{n}=0.\nonumber \]Keeping the term \( a_n p^n \) on the left side and moving everything else to the right side, we get\[a_{n} p^{n}=-a_{n-1} p^{n-1} q-\ldots-a_{1} p q^{n-1}-a_{0} q^{n}.\nonumber \]However, each term on the right side has a factor of \( q \) in common. Therefore,\[a_{n} p^{n}=\left(-a_{n-1} p^{n-1} -\ldots-a_{1} p q^{n-2}-a_{0} q^{n - 1}\right)q.\nonumber \]Recall that \( p \), \( q \), and \( a_k \) are integers (for all \( k \)). Moreover, \( n \geq 1 \). This shows that \( -a_{n-1} p^{n-1} -\ldots-a_{1} p q^{n-2}-a_{0} q^{n - 1} \) must be an integer (can you see why?). Hence, \( a_n p^n \) is an integer multiple of \( q \). Moreover, since \( a_n p^n = \left( a_n p^{n -1} \right)p \), \( a_n p^n \) is also an integer multiple of \( p \). However, \(p\) and \(q\) have no common factors. Thus, \( a_n \) must be the integer multiple of \( q \). This shows that \( q \) is a factor of the leading coefficient \( a_n \).

If we rearrange the equation\[a_{n} p^{n}+a_{n-1} p^{n-1} q+\ldots+a_{1} p q^{n-1}+a_{0} q^{n}=0\nonumber \]as\[a_{0} q^{n}=-a_{n} p^{n}-a_{n-1} p^{n-1} q-\ldots-a_{1} p q^{n-1},\nonumber \]we can play the same game and conclude \(a_{0}\) is a multiple of \(p\), and we have the result we desire.

When the leading coefficient is 1, the Rational Zeros Theorem simplifies to stating that the possible rational zeros are the factors of the constant term.

Caution: The Rational Zeros Theorem Requires Integer Coefficients

An often overlooked requirement to using the Rational Zeros Theorem is that the coefficients of the polynomial must be integers. Do not make the mistake of trying to use the Rational Zeros Theorem on polynomials with coefficients of the form \(\frac{3}{7}\), \(\sqrt{2}\), or \(3 + 2i\).

Example \(\PageIndex{4}\): Listing All Possible Rational Zeros

List all possible rational zeros of \(f(x)=2 x^4-5 x^3+x^2-4\)

Solution

The only possible rational zeros of \(f(x)\) are the quotients of the factors of the last term, \(-4\), and the factors of the leading coefficient, \(2\).

The factors of \(-4\) are \(p= \pm 1, \pm 2, \pm 4\).

The factors of \(2\) are \(q= \pm 1, \pm 2\).

If any of the four real zeros are rational zeros, then they will be of one of the following factors of \(-4\) divided by one of the factors of \(2\).\[\dfrac{p}{q}= \pm \dfrac{1}{1}, \pm \dfrac{1}{2} \quad \dfrac{p}{q}= \pm \dfrac{2}{1}, \pm \dfrac{2}{2} \quad \dfrac{p}{q}= \pm \dfrac{4}{1}, \pm \dfrac{4}{2}.\nonumber\]Note that \(\frac{2}{2}=1\) and \(\frac{4}{2}=2\), which have already been listed. So we can shorten our list.\[\dfrac{p}{q} = \dfrac{\text{Factors of the last}}{\text{Factors of the first}}= \pm 1, \pm 2, \pm 4, \pm \dfrac{1}{2}.\nonumber\]

Example \(\PageIndex{5}\): Using the Rational Zeros Theorem to Find Rational Zeros

Use the Rational Zeros Theorem to find the rational zeros of \(f(x)=2 x^3+x^2-4 x+1\).

Solution

The Rational Zeros Theorem tells us that if \(\frac{p}{q}\) is a zero of \(f(x)\), then \(p\) is a factor of \(1\) and \(q\) is a factor of \(2\).\[\begin{array}{rcl}
\dfrac{p}{q} & = & \dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}} \\[6pt]
& = & \dfrac{\text{factor of } 1}{\text{factor of } 2} \\[6pt]
\end{array}\nonumber\]The factors of \(1\) are \(\pm 1\) and the factors of \(2\) are \(\pm 1\) and \(\pm 2\). The possible values for \(\frac{p}{q}\) are \(\pm 1\) and \(\pm \frac{1}{2}\). These are the possible rational zeros for the function. We can determine which of the possible zeros are actual zeros by substituting these values for \(x\) in \(f(x)\).\[\begin{array}{rcl}
f(-1) & = & 2(-1)^3+(-1)^2-4(-1)+1=4 \\[6pt]
f(1) & = & 2(1)^3+(1)^2-4(1)+1=0 \\[6pt]
f\left(-\dfrac{1}{2}\right) & = & 2\left(-\dfrac{1}{2}\right)^3+\left(-\dfrac{1}{2}\right)^2-4\left(-\dfrac{1}{2}\right)+1=3 \\[6pt]
f\left(\dfrac{1}{2}\right) & = & 2\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^2-4\left(\dfrac{1}{2}\right)+1=-\dfrac{1}{2} \\[6pt]
\end{array}\nonumber\]Of those, \(-1\), \(-\frac{1}{2}\), and \(\frac{1}{2}\) are not zeros of \(f(x)\). \(1\) is the only rational zero of \(f(x)\).

Checkpoint \(\PageIndex{5}\)

Use the Rational Zeros Theorem to find the rational zeros of\[f(x)=x^3-5 x^2+2 x+1.\nonumber\]

Finding the Zeros of Polynomial Functions

The Rational Zeros Theorem helps us to narrow down the list of possible rational zeros for a polynomial function. Once we have done this, we can use synthetic division repeatedly to determine all of the zeros of a polynomial function.

Example \(\PageIndex{6}\): Finding the Zeros of a Polynomial Function with Repeated Real Zeros

Find the zeros of \(f(x)=4 x^3-3 x-1\).

Solution

First, note that the coefficients of the given polynomial function are all integers—remember, we cannot use the Rational Zeros Theorem if this is not the case. The Rational Zeros Theorem tells us that if \(\frac{p}{q}\) is a zero of \(f(x)\), then \(p\) is a factor of \(-1\) and \(q\) is a factor of \(4\).\[\begin{array}{rcl}
\dfrac{p}{q} & = & \dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}} \\[6pt]
& = & \dfrac{\text{factor of }-1}{\text{factor of } 4} \\[6pt]
\end{array}\nonumber\]The factors of \(-1\) are \(\pm 1\) and the factors of \(4\) are \(\pm 1\),\(\pm 2\), and \(\pm 4\). The possible values for \(\frac{p}{q}\) are \(\pm 1\), \(\pm \frac{1}{2}\), and \(\pm \frac{1}{4}\). These are the possible rational zeros for the function. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of \(0\). Let's begin with \(1\).\[\begin{array}{r}
\underline{1} \mid & 4 & 0 & -3 & -1 \\
& & 4 & 4 & 1 \\
\hline & 4 & 4 & 1 & 0 \\
\end{array}\nonumber\]Dividing by \((x-1)\) gives a remainder of \(0\), so \(1\) is a zero of the function. The polynomial can be written as\[(x-1)\left(4 x^2+4 x+1\right).\nonumber\]The quadratic is a perfect square. Therefore, \(f(x)\) can be written as\[(x-1)(2 x+1)^2.\nonumber\]We already know that \(1\) is a zero. The other zero will have a multiplicity of \(2\) because the factor is squared. To find the other zero, we can set the factor equal to 0.\[2 x+1 =0 \implies x =-\dfrac{1}{2}.\nonumber\]The zeros of the function are \(1\) and \(-\frac{1}{2}\) with multiplicity \(2\).

Investigating Example \(\PageIndex{6}\) a little more, we look at the graph of the function \(f\) in Figure \(\PageIndex{3}\). Notice, at \(x=-0.5\), the graph bounces off the \(x\)-axis, indicating the even multiplicity (\(2,4,6, \ldots\)) for the zero \(-0.5\). At \(x=1\), the graph crosses the \(x\)-axis, indicating the odd multiplicity (\(1,3,5, \ldots\)) for the zero \(x=1\).

Figure \(\PageIndex{3}\)

Interactive Element: Multiplicity and the Shape of a Graph at a Zero

The following Interactive Element might help you visualize how the multiplicity of a zero controls the behavior of a polynomial's graph at that zero.

Interact: Drag the red point up and down to change the multiplicity \(m\) of the zero at \(x=0\).

Observation: When \(m\) is odd the graph crosses the \(x\)-axis; when \(m\) is even it bounces off it. The larger \(m\) becomes, the flatter the graph lies near the zero.

Using the Fundamental Theorem of Algebra

Now that we can find rational zeros for a polynomial function, we will look at a theorem that discusses the number of complex zeros of a polynomial function. The Fundamental Theorem of Algebra tells us that every polynomial function has at least one complex zero. This theorem forms the foundation for solving polynomial equations.

Theorem: Fundamental Theorem of Algebra

Every nonconstant polynomial with complex coefficients has at least one complex zero.

Proof

The proof of this theorem is beyond the scope of lower-division mathematics.

The Fundamental Theorem of Algebra is an example of an "existence theorem" in Mathematics. It guarantees the existence of at least one complex-valued zero but gives us no algorithm to use in finding it. It took mathematicians hundreds of years to prove the theorem in its full generality, and some of that history is recorded here. Note that the Fundamental Theorem of Algebra applies to not only polynomial functions with real coefficients but to those with complex number coefficients as well.

Suppose \(f\) is a polynomial function of degree four, and \(f(x)=0\). The Fundamental Theorem of Algebra states that there is at least one complex-valued solution, call it \(z_1\). By the Factor Theorem, we can write \(f(x)\) as a product of \(x-z_1\) and a polynomial quotient. Since \(x-z_1\) is linear, the polynomial quotient will be of degree three. Now we apply the Fundamental Theorem of Algebra to the third-degree polynomial quotient. It will have at least one complex zero, call it \(z_2\). So we can write the polynomial quotient as a product of \(x-z_2\) and a new polynomial quotient of degree two. Continue to apply the Fundamental Theorem of Algebra until all of the zeros are found. There will be four of them, and each one will yield a factor of \(f(x)\).

In truth, "a polynomial with complex-valued coefficients" means that the coefficients can be any number we can imagine—from \(2\) to \(\sqrt{2}\) to \(\frac{1}{2}\) to \(2 + 2i\). Therefore, most people write the Fundamental Theorem of Algebra as, "If \(f(x)\) is a polynomial of degree \(n \geq 1\), then \(f(x)\) has at least one complex zero," thereby omitting the needless "with complex-valued coefficients."

Although the Fundamental Theorem of Algebra is a major theorem in Mathematics, it is the following corollary that makes the theorem useful.

Corollary: Complete Factorization Theorem

Let \(f(x)\) be a polynomial of degree \(n \geq 1\) with complex coefficients and leading coefficient \(a\). Then \(f(x)\) has exactly \(n\) linear factors over the complex number system. That is, there exists complex numbers \(z_1, z_2, \ldots, z_n\) such that\[f(x) = a \left(x-z_1\right)\left(x-z_2\right) \ldots \left(x-z_n\right). \nonumber \]

Proof

The proof of this theorem is done using mathematical induction and the Fundamental Theorem of Algebra. Since this topic is usually covered prior to learning proofs by mathematical induction, the proof is omitted.

As with the Fundamental Theorem of Algebra, the Complete Factorization Theorem can be shortened to, "If \(f(x)\) is a polynomial of degree \(n \geq 1\), then \(f(x)\) has exactly \(n\) linear factors over the complex number system."

Q & A

Does every polynomial have at least one imaginary zero?

No. A complex number is not necessarily imaginary. Real numbers are also complex numbers.

Example \(\PageIndex{7}\): Finding the Zeros of a Polynomial Function with Complex Zeros

Find the zeros of\[f(x)=3 x^3+9 x^2+x+3.\nonumber\]

Solution

The Rational Zeros Theorem tells us that if \(\frac{p}{q}\) is a zero of \(f(x)\), then \(p\) is a factor of \(3\) and \(q\) is a factor of \(3\).\[\begin{array}{rcl}
\dfrac{p}{q} & = & \dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}} \\[6pt]
& = & \dfrac{\text{factor of } 3}{\text{factor of } 3} \\[6pt]
\end{array}\nonumber\]The factors of \(3\) are \(\pm 1\) and \(\pm 3\). The possible values for \(\frac{p}{q}\), and therefore the possible rational zeros for the function, are \(\pm 3\), \(\pm 1\), and \(\pm \frac{1}{3}\). We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let's begin with \(-3\).\[\begin{array}{r}
\underline{-3} \mid & 3 & 9 & 1 & 3 \\
& & -9 & 0 & -3 \\
\hline & 3 & 0 & 1 & 0 \\
\end{array}\nonumber\]Dividing by \((x+3)\) gives a remainder of \(0\), so \(-3\) is a zero of the function. The polynomial can be written as\[(x+3)\left(3 x^2+1\right).\nonumber\]We can then set the quadratic equal to \(0\) and solve to find the other zeros of the function.\[\begin{array}{rrcl}
& 3 x^2+1 & = & 0 \\[6pt]
\implies & x^2 & = & -\dfrac{1}{3} \\[6pt]
\implies & x & = & \pm \sqrt{-\dfrac{1}{3}} \\[6pt]
& & = & \pm \dfrac{i \sqrt{3}}{3} \\[6pt]
\end{array}\nonumber\]The zeros of \(f(x)\) are \(-3\) and \(\pm \frac{i \sqrt{3}}{3}\).

Figure \(\PageIndex{4}\) shows the graph of the function \(f\) from Example \(\PageIndex{7}\). Notice that, at \(x=-3\), the graph crosses the \(x\)-axis, indicating an odd multiplicity (1) for the zero \(x=-3\). Also note the presence of the two turning points. This means that, since there is a \(3^{\text{rd}}\) degree polynomial, we are looking at the maximum number of turning points. So, the end behavior of increasing without bound to the right and decreasing without bound to the left will continue. Thus, all the \(x\)-intercepts for the function are shown. Either the multiplicity of \(x=-3\) is \(1\) and there are two complex solutions, which is what we found, or the multiplicity at \(x=-3\) is three. Either way, our result is correct.

Figure \(\PageIndex{4}\)

Checkpoint \(\PageIndex{7}\)

Find the zeros of\[f(x)=2 x^3+5 x^2-11 x+4.\nonumber\]

Using the Complete Factorization Theorem to Find Polynomials with Given Zeros

A vital implication of the Fundamental Theorem of Algebra, as we stated above, is that a polynomial function of degree \(n\) will have \(n\) zeros in the set of complex numbers, if we allow for multiplicities. This means that we can factor the polynomial function into \(n\) factors. The Complete Factorization Theorem tells us that a polynomial function will have the same number of factors as its degree, and that each factor will be in the form \((x-c)\), where \(c\) is a complex number.

Theorem: Conjugate Zeros Theorem

Let \(f(x)\) be a polynomial with real coefficients. If the complex number \(z = a + bi\) is a zero of \(f(x)\), then \(\overline{z} = a - bi\) is also a zero of \(f(x)\), and \(z\) and \(\overline{z}\) have the same multiplicity.

Proof

The proof requires several theorems from complex numbers. These are mentioned within the proof. If \( z = a+bi \) is a zero of a polynomial function \( f(x) = a_n x^n + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0 \), then\[ \begin{array}{rclcl}
f\left( \overline{z} \right) & = & a_n \left( \overline{z} \right)^n + a_{n - 1} \left( \overline{z} \right)^{n - 1} + \cdots + a_1 \overline{z} + a_0 & \quad & \left( \text{substituting} \right) \\[6pt]
& = & a_n \overline{z^n} + a_{n - 1} \overline{z^{n - 1}} + \cdots + a_1 \overline{z} + a_0 & \quad & \left( \overline{z^n} = \left( \overline{z} \right)^n \right) \\[6pt]
& = & \overline{a_n} \overline{z^n} + \overline{a_{n - 1}} \overline{z^{n - 1}} + \cdots + \overline{a_1} \overline{z} + \overline{a_0} & \quad & \left( \text{if }a_i \in \mathbb{R}, \text{ then }a_i = \overline{a_i} \right) \\[6pt]
& = & \overline{a_n z^n} + \overline{a_{n - 1} z^{n - 1}} + \cdots + \overline{a_1 z} + \overline{a_0} & \quad & \left( \overline{z_1} \cdot \overline{z_2} = \overline{z_1 \cdot z_2} \right) \\[6pt]
& = & \overline{a_n z^n + a_{n - 1} z^{n - 1} + \cdots + a_1 z + a_0} & \quad & \left( \overline{z_1} + \overline{z_2} + \cdots + \overline{z_k} = \overline{z_1 + z_2 + \cdots + z_k} \right) \\[6pt]
& = & \overline{f(z)} & & \\[6pt]
& = & \overline{0} & \quad & \left( z \text{ is a zero of }f(x) \right) \\[6pt]
& = & 0 & \quad & \left( \overline{k} = k \text{ for real numbers }k \right) \\[6pt]
\end{array} \nonumber \]Hence, \( \overline{z} \) (the conjugate) is also a zero of \( f(x) \).

To prove that multiplicities must be the same, one can argue via contradiction. This part of the proof is left for the reader.

Example \(\PageIndex{8}\): Finding a Polynomial with Given Zeros

Find a fourth degree polynomial with real coefficients that has zeros of \(-3\), \(2\), and \(i\), such that \(f(-2)=100\).

Solution

Because \(x=i\) is a zero, by the Conjugate Zeros Theorem \(x=-i\) is also a zero. The polynomial must have factors of \((x+3)\), \((x-2)\), \((x-i)\), and \((x+i)\). Since we are looking for a degree 4 polynomial, and now have four zeros, we have all four factors. Let's begin by multiplying these factors.\[\begin{array}{rcl}
f(x) & = & a(x+3)(x-2)(x-i)(x+i) \\[6pt]
& = & a\left(x^2+x-6\right)\left(x^2+1\right) \\[6pt]
& = & a\left(x^4+x^3-5 x^2+x-6\right) \\[6pt]
\end{array}\nonumber\]We need to find \(a\) to ensure \(f(-2)=100\). Substitute \(x=-2\) and \(f(-2)=100\) into \(f(x)\).\[\begin{array}{rrcl}
& 100 & = & a\left((-2)^4+(-2)^3-5(-2)^2+(-2)-6\right) \\[6pt]
\implies & 100 & = & a(-20) \\[6pt]
\implies & -5 & = & a \\[6pt]
\end{array}\nonumber\]So the polynomial function is\[f(x)=-5\left(x^4+x^3-5 x^2+x-6\right)\nonumber\]or\[f(x)=-5 x^4-5 x^3+25 x^2-5 x+30.\nonumber\]

In Example \(\PageIndex{8}\), we found that both \(i\) and \(-i\) were zeros, but only one of these zeros needed to be given. If \(i\) is a zero of a polynomial with real coefficients, then \(-i\) must also be a zero of the polynomial because \(-i\) is the complex conjugate of \(i\).

Q & A

If \(2+3i\) were given as a zero of a polynomial with real coefficients, would \(2-3i\) also need to be a zero?

Yes. When any complex number with an imaginary component is given as a zero of a polynomial with real coefficients, the conjugate must also be a zero of the polynomial.

Checkpoint \(\PageIndex{8}\)

Find a third degree polynomial with real coefficients that has zeros of \(5\) and \(-2i\) such that \(f(1)=10\).

It's probably good to summarize the theorems and corollaries we have encountered so far in this section, along with their requirements.

Table \(\PageIndex{1}\)

Theorem or Corollary	Requirement	Result
Rational Zeros Theorem	Coefficients must be integers	Rational zeros of the polynomial are always ratios of factors of the constant term and the leading coefficient.
Fundamental Theorem of Algebra	Coefficients can be any number	The polynomial has at least one complex zero.
Complete Factorization Theorem	Coefficients can be any number	The polynomial can be factored into \(n\) linear factors over the complex number system.
Conjugate Zeros Theorem	Coefficients must be real numbers	If \(a + bi\) is a zero, then \(a - bi\) is also a zero.

Solving Real-World Applications

We have now introduced a variety of tools for solving polynomial equations. Let's use these tools to solve the bakery problem from the beginning of the section.

Example \(\PageIndex{9}\)

A new bakery offers decorated, multi-tiered cakes for display and cutting at Quinceañera and wedding celebrations, as well as sheet cakes to serve most of the guests. The bakery wants the volume of a small sheet cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be?

Solution

Begin by writing an equation for the volume of the cake. The volume of a rectangular solid is given by \(V=lwh\). We were given that the length must be four inches longer than the width, so we can express the length of the cake as \(l=w+4\). We were given that the height of the cake is one-third of the width, so we can express the height of the cake as \(h=\frac{1}{3} w\). Let's write the volume of the cake in terms of width of the cake.\[\begin{array}{rcl}
V & = & (w+4)(w)\left(\dfrac{1}{3} w\right) \\[6pt]
& = & \dfrac{1}{3} w^3+\dfrac{4}{3} w^2 \\[6pt]
\end{array}\nonumber\]Substitute the given volume into this equation.\[\begin{array}{rrclcl}
& 351 & = & \dfrac{1}{3} w^3+\dfrac{4}{3} w^2 & \quad & \left(\text{substituting} \right) \\[6pt]
\implies & 1053 & = & w^3+4 w^2 & \quad & \left( \text{multiplying both sides by } 3 \right) \\[6pt]
\implies & 0 & = & w^3+4 w^2-1053 & \quad & \left(\text{subtracting } 1053 \text{ from both sides} \right) \\[6pt]
\end{array}\nonumber\]The Rational Zeros Theorem tells us that the possible rational zeros are \(\pm 1\), \(\pm 3\), \(\pm 9\), \(\pm 13\), \(\pm 27\), \(\pm 39\), \(\pm 81\), \(\pm 117\), \(\pm 351\), and \(\pm 1053\). We can use synthetic division to test these possible zeros. Only positive numbers make sense as dimensions for a cake, so we need not test any negative values. Let's begin by testing values that make the most sense as dimensions for a small sheet cake. Use synthetic division to check \(x=1\).\[\begin{array}{r}
\underline{1} \mid & 1 & 4 & 0 & -1053 \\
& & 1 & 5 & 5 \\
\hline & 1 & 5 & 5 & -1048 \\
\end{array}\nonumber\]Since 1 is not a solution, we will check \(x=3\).\[\begin{array}{r}
\underline{3} \mid & 1 & 4 & 0 & -1053 \\
& & 3 & 21 & 63 \\
\hline & 1 & 7 & 21 & -990 \\
\end{array}\nonumber\]Since 3 is not a solution either, we will test \(x=9\).\[\begin{array}{r}
\underline{9} \mid & 1 & 4 & 0 & -1053 \\
& & 9 & 117 & 1053 \\
\hline & 1 & 13 & 117 & 0 \\
\end{array}\nonumber\]Synthetic division gives a remainder of \(0\), so \(9\) is a solution to the equation. We can use the relationships between the width and the other dimensions to determine the length and height of the sheet cake pan.\[l=w+4=9+4=13 \text{ and } h=\dfrac{1}{3} w = \dfrac{1}{3}(9)=3.\nonumber\]The sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches.

Checkpoint \(\PageIndex{9}\)

A shipping container in the shape of a rectangular solid must have a volume of 84 cubic meters. The client tells the manufacturer that, because of the contents, the length of the container must be one meter longer than the width, and the height must be one meter greater than twice the width. What should the dimensions of the container be?

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Success in College: Relying On Others and Fooling Yourself

It is a common mistake that students make:

• You do homework with your friend, and she constantly is the one with the correct answers. Instead of talking about the mistakes, you copy the answer and move on.
• Your classmates have formed a Discord, and you copy the solutions from one of them.
• You only have a couple hours before the homework is due and you use your preferred A.I. model to do most of the work, copying frantically to submit the work before it's too late.

Is this helping you learn? Do you truly think it is helping your grade?

If you answered yes, you're in the majority of students who have been fooled by this method of "get the right answer by any means and move on."

When I was a student at UC Davis, I concentrated on understanding every concept that came my way. I studied with other students, but more often than not, they relied on my answers and explanations to succeed in the course. The problem is that they were just "parroting" my methods and not understanding them. In the end, most of them did poorly due to their lack of understanding of the material.

- Professor Simpson, Cosumnes River College

This sort of "copying" also leads to another damaging process - a lack of self-confidence. Because they never solve anything by themselves, the students who copy other students' work end up freaking out more before exams and, as a result of their heightened state of panic and their fundamental lack of understanding of the material, they fail the exams.

Think about what kind of student you are, and, without fear of judgement, consider the following questions:

1. A student was having trouble with his homework. For the next few weeks, a friend helped him do his homework. The friend answered his questions and told him what to do when he was stuck on a problem. Why do you think this student may have trouble doing well on a test?
2. A student can do about 75% of the problems on most of her homework assignments without help. She thinks that this is good enough. She thinks she should be able to score about 75% on a test and pass the test. Do you think she is right?

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Success in Mathematics: Why Study Mathematics? (Part 2)

Tracy Moody graduated in 2006 with a bachelor's degree in mathematics. She then went to graduate school, studying to become a physical therapist. She writes,

"Many people cannot seem to fathom how math and physical therapy overlap. Well, one of the obvious things is that physical therapists use statistics to see if new treatments work. However, the main thing I bring as a mathematician to my graduate studies is my ability to problem solve. A degree in mathematics taught me how to look at a problem from several angles. Furthermore, one of the main classes that carried over into my career choice is, believe it or not, real analysis. I can remember working with my fellow classmates on problems for hours. However, through this process, I learned the importance of perseverance, and I learned how to communicate the problem and solutions in several ways."

Source: www.maa.org, Career Profiles.

1. Do you think studying mathematics has helped you become a better problem solver? Explain.
2. Do you think studying mathematics has helped you learn to "persevere," to keep working on a difficult problem even though you are struggling? Explain.