Zeros of Polynomial Functions

We have methods to find the zeros of linear and quadratic functions. We can take this a step further with methods that use polynomial division and allow us to find the zeros of any polynomial function regardless of the degree.

Remainder Theorem

The Remainder Theorem states that if a polynomial $f(x)$ is divided by $x - k$, then the remainder is the value $f(k)$.

$\underline{Example}$

Use the Remainder Theorem to evaluate $f(x) = 2x^5 - 3x^4 - 9x^3 + 8x^2 + 2$ at $x = -3$.

Divide $2x^5 - 3x^4 - 9x^3 + 8x^2 + 2$ by $x + 3$ because $k = -3$ using synthetic division...

$$\begin{array}{c|rrrrrr} -3 & 2 & -3 & -9 & 8 & 0 & 2 \\ \: & & -6 & 27 & -54 & 138 & -414 \\ \hline & 2 & -9 & 18 & -46 & 138 & -412 \\ \end{array}$$

The remainder is $-412$ and so $f(3) = -412$. We can verify this...

$$\begin{array}{l} f(-3) = 2(-3)^5 - 3(-3)^4 - 9(-3)^3 + 8(-3)^2 + 2 \\ \phantom{f(-3)} = 2(-243) - 3(81) - 9(-27) + 8(9) + 2 \\ \phantom{f(-3)} = -486 - 243 + 243 + 72 + 2 \\ \phantom{f(-3)} = -412 \end{array}$$

note

We can arrive at the remainder theorem from the division theorem. The division algorithm states that given a polynomial dividend $f(x)$ and a non-zero polynomial divisor $d(x)$, there exist unique polynomials $q(x)$ and $r(x)$ such that $f(x) = d(x)q(x) + r(x)$ and $r(x)$ is less than the degree of $d(x)$. If we divide by the linear factor $x - k$ then we get $f(x) = (x - k)q(x) + r$ where $r$ is a constant because $r(x)$ must be a degree less than $1$ as $x - k$ is linear. Finally, if we solve for $k$, we get $f(k) = (k - k)q(k) + r = 0q(k) + r = r$ which shows that $f(k)$ equals the remainder of $f(x)$ divided by $x - k$.

Factor Theorem

The Factor Theorem states that $k$ is a zero of $f(x)$ if and only if $(x - k)$ is a factor of $f(x)$. This means that if $f(x)$ divided by $(x - k)$ results in a remainder of $0$ then $k$ is a zero of $f(x)$.

$\underline{Example}$

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

We can verify that $(x - 2)$ is a factor of $f(x)$...

$$\begin{array}{c|rrrrrr} 2 & 1 & 4 & -4 & -16 \\ \: & & 2 & 12 & 16 \\ \hline & 1 & 6 & 8 & 0 \\ \end{array}$$

This means that $q(x) = x^2 + 6x + 8$ and $f(x) = (x^2 + 6x + 8)(x - 2)$. We can find the remaining linear factors by factoring $x^2 + 6x + 8$...

$$\begin{array}{l} q(x) = x^2 + 6x + 8 \\ \phantom{q(x)} = x^2 + 4x + 2x + 8 \\ \phantom{q(x)} = x(x + 4) + 2(x + 4) \\ \phantom{q(x)} = (x + 2)(x + 4) \end{array}$$

This means that $f(x) = (x + 2)(x + 4)(x - 2)$ which means the zeros of $f(x)$ are $-4, -2, 2$.

Rational Zero Theorem

The Rational Zero Theorem states that, if the polynomial $f(x) = a_nx^n + a_{n - 1}x^{n - 1} + ... + a_1x + a_0$ has integer coefficients and $a_n \neq 0$, then every rational zero of $f(x)$ has the form $\frac{p}{q}$ where $p$ is a factor of the constant term $a_0$ and $q$ is a factor of the leading coefficient $a_n$. This means we can determine all the possible rational zeros of $f(x)$ by finding all the possible values of $\frac{p}{q}$.

$\underline{Example}$

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

The factors of the leading coefficient are $\pm 1$ and $\pm 2$ and the only factor of the constant term is $\pm 1$.

This means $\frac{p}{q} = \pm \frac{1}{1}, \pm \frac{1}{2}$. The possible zeros are $-1$, $-\frac{1}{2}$, $\frac{1}{2}$ and $1$.

We can determine the actual zeros by substituting these possible zeros into the function.

$$\begin{array}{l} f(-1) = 2(-1)^3 + (-1)^2 - 4(-1) + 1 = 4 \\ f(-\frac{1}{2}) = 2(-\frac{1}{2})^3 + (-\frac{1}{2})^2 - 4(-\frac{1}{2}) + 1 = 3 \\ f(\frac{1}{2}) = 2(\frac{1}{2})^3 + (\frac{1}{2})^2 - 4(\frac{1}{2}) + 1 = -\frac{1}{2} \\ f(1) = 2(1)^3 + (1)^2 - 4(1) + 1 = 0 \end{array}$$

The only zero is $1$.

note

Both the Factor Theorem and the Rational Zero Theorem are theorems built on the concepts of the Remainder Theorem.

Zeros of Polynomial Functions

Given a polynomial function $f$, we can find the zeros using the following steps...

1. Use the Rational Zero Theorem to list all possible rational zeros of $f$.
2. Use synthetic division to evaluate a given possible zero by dividing the candidate into the polynomial. If the remainder is $0$, the candidate is a zero. If the remainder is not a $0$, discard the candidate.
3. Repeat step two using the quotient found with synthetic division. Repeat till the quotient is a quadratic.
4. Find the zeros of the quadratic function.

$\underline{Example}$

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

The factors of the leading coefficient are $\pm 1$, $\pm 2$, and $\pm 4$. The only factor of the constant is $\pm 1$. This means $\frac{p}{q} = \pm\frac{1}{1}$, $\pm\frac{1}{2}$ and $\pm\frac{1}{4}$.

Lets test $x = -1$ by dividing $f(x)$ by $(x + 1)$...

$$\begin{array}{c|rrrrrr} -1 & 4 & 0 & -3 & -1 \\ \: & & -4 & 4 & -1 \\ \hline & 4 & -4 & 1 & -2 \\ \end{array}$$

We figured out that $x = 1$ is not a zero. So, we need to try another possible zero. Lets test $x = 1$ by dividing $f(x)$ by $(x - 1)$...

$$\begin{array}{c|rrrrrr} 1 & 4 & 0 & -3 & -1 \\ \: & & 4 & 4 & 1 \\ \hline & 4 & 4 & 1 & 0 \\ \end{array}$$

Due to the fact we got a remainder of $0$, $x = 1$ is a zero of $f(x)$. We are now left with $4x^2 + 4x + 1$ which is a quadratic function. If it was not a quadratic function, we would continue using synthetic division to find more zeros using this quotient.

We can find the remaining zeros using the quadratic function...

$$\begin{array}{l} f(x) = 4x^2 + 4x + 1 \\ \phantom{f(x)} = 4x^2 + 2x + 2x + 1 \\ \phantom{f(x)} = 2x(2x + 1) + 1(2x + 1) \\ \phantom{f(x)} = (2x + 1)(2x + 1) \\ \phantom{f(x)} = (2x + 1)^2 \end{array}$$

The zeros of $4x^3 - 4x - 1$ are $x = -\frac{1}{2}$ and $x = 1$.

Fundemental Theorem of Algebra

The Fundemental Theorem of Algebra states that, if $f(x)$ is a polynomial of degree $n > 0$, then $f(x)$ has at least one complex zero. This theorem can be used to also argue that, if $f(x)$ is a polynomial of degree $n > 0$, and $a$ is a non-zero real number, then $f(x)$ has exactly $n$ linear factors and $f(x) = a(x - c_1)(x - c_2)...(x - c_n)$ where $c_1$, $c_2$, ..., $c_n$ are complex numbers. Therefore, $f(x)$ has $n$ roots if we allow for multiplicities.

note

All real numbers are a subset of complex numbers and that means a real number can be the complex zero for a given function. However, remember that not all complex numbers are real numbers.

Complex Conjugate Theorem

The Linear Factorization Theorem states that a polynomial function will have the same number of factors as its degree, and each factor will be in the form $(x - c)$, where $c$ is a complex number. Building from this, the Complex Conjugate Theorem states that if the polynomial function $f$ has real coefficients and a complex zero in the form $a + bi$, then the complex conjugate of the zero, $a - bi$, is also a zero.

$\underline{Example}$

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

Based on the complex conjugate theorem, we know that the complex conjugate of $-2i$ must also be a zero. The complex conjugate of $-2i$ is $2i$ and so the zeros are $5$, $-2i$, and $2i$.

A polynomial of the third degree has at most $3$ zeros and we have found them all so we can use them to create the linear factors. We get the function...

$$\begin{array}{l} f(x) = a(x - 5)(x - 2i)(x + 2i) \\ \phantom{f(x)} = a(x - 5)(x^2 + 4) \\ \phantom{f(x)} = a(x^3 - 5x^2 + 4x - 20) \\ \end{array}$$

We can use the fact that $f(1) = 10$ to find $a$...

$$\begin{array}{l} f(x) = a(x^3 - 5x^2 + 4x - 20) \\ \phantom{()}10 = a((1)^3 - 5(1)^2 + 4(1) - 20) \\ \phantom{()}10 = a(1 - 5(1) + 4(1) - 20) \\ \phantom{()}10 = a(1 - 5 + 4 - 20) \\ \phantom{()}10 = -20a \\ \phantom{()0}a = -\frac{1}{2} \end{array}$$

We can simplify our polynomial with the value of $a$.

$$\begin{array}{l} f(x) = a(x^3 - 5x^2 + 4x - 20) \\ \phantom{f(x)} = -\frac{1}{2}(x^3 - 5x^2 + 4x - 20) \\ \phantom{f(x)} = -\frac{1}{2}x^3 + \frac{5}{2}x^2 - 2x + 10 \end{array}$$

Descartes' Rule of Signs

The Descartes' Rule of Signs is a method used to determine the possible numbers of positive and negative real zeros for any polynomial function. According to this rule, if we let $f(x) = a_nx^n + a_{n - 1}x^{n - 1} + ... + a_1x + a_0$ be a polynomial function with real coefficients:

1. The number of positive real zeros is either equal to the number of sign changes of $f(x)$ or is less than the number of sign changes by an even integer.
2. The number of negative real zeros is either equal to the number of sign changes of $f(-x)$ or is less than the number of sign changes by an even integer.

$\underline{Example}$

Use Descartes' Rule of Signs to determine the maximum possible numbers of positive and negative real zeros for $f(x) = 2x^4 - 10x^3 + 11x^2 - 15x + 12$.

There are $4$ sign changes for $f(x)$ which means there can only be $4$, $2$, or $0$ positive real zeros.

To find the negative real zeros, we need to solve for $f(-x)$ first...

$$\begin{array}{l} f(-x) = 2(-x)^4 - 10(-x)^3 + 11(-x)^2 - 15(-x) + 12 \\ \phantom{f(-x)} = 2x^4 + 10x^3 + 11x^2 + 15x + 12 \end{array}$$

There are no sign changes in $f(-x)$ which means there can only be $0$ negative real zeros.