# Zeros of Polynomials

*This post is part of a series.*

The **zeros** of a polynomial are the inputs that cause it to evaluate to zero.

For example, a zero of the polynomial $x^3-2x^2-5x+6$ is $x=1$ because $(1)^3-2(1)^2-5(1)+6=0$. Another zero is $x=-2$ because $(-2)^3-2(-2)^2-5(-2)=0$. Can you find the rest?

## Finding Zeros by Factoring

One trick for finding the zeros of polynomials is to write the polynomial in factored form.

Since we know that $x=1$ and $x=-2$ are zeros of the polynomial, we know the polynomial has to have factors $x-1$ and $x+2$. If we multiply these factors together, we get a polynomial whose highest-exponent term is $x^2$.

But our original polynomial has a highest-exponent term of $x^3$, so we need to multiply by one more factor. Consequently, the factored polynomial will take the form $(x-1)(x+2)(x-a)$ for some other zero $x=a$.

Let’s multiply out the factors and group like terms into the form of the original polynomial.

From here, we can proceed in any of several different ways to discover that $a=3$.

- • The $x^2$ coefficient of the right-hand side is $-(a-1)$, and the $x^2$ coefficient of the left-hand side is $-2$, so we need $-(a-1)=2$, which means $a=3$.
- • The $x$ coefficient of the right-hand side is $-(a+2)$, and the $x$ coefficient of the left-hand side is $-5$, so we need $-(a+2)=-5$, which means $a=3$.
- • The constant coefficient of the right-hand side is $2a$, and the constant coefficient of the left-hand side is $6$, so we need $2a=6$, which means $a=3$.

Indeed, checking our answer, we find that substituting $x=3$ makes the polynomial evaluate to $0$.

## Fundamental Theorem of Algebra

Through this example, we’ve learned an important thing about the zeros of polynomials: **the number of zeros of a polynomial is no more than its degree**.

Each zero comes from a factor, and the degree of a polynomial limits the amount of factors it has, which in turn limits the amount of zeros it has. A third-degree polynomial can’t have more than $3$ factors, so it has at most $3$ zeros. A tenth-degree polynomial can’t have more than $10$ factors, so it has at most $10$ zeros.

Some polynomials look like they have fewer zeros than their degree – for example, the polynomial $x^2+1$ doesn’t appear to have any zeros, because there is no solution to $x^2=-1$. But if we allow the use of the imaginary unit $i=\sqrt{-1}$, then it does have two zeros: $x=i$ and $x=-i$.

Likewise, the polynomial $x^2+2x+1$ factors to $(x+1)^2$ and thus appears to have only one zero, $x=-1$. But since this factor is squared, we can think of counting the $x=-1$ zero twice, i.e. it has a **multiplicity** of two.

This is the **fundamental theorem of algebra**: the number of zeros of a polynomial is equal to its degree, provided we allow the use of the imaginary unit and count zeros according to their multiplicity.

## Solving a Polynomial Equation

Finding zeros of polynomials is important because of its generality: every polynomial equation reduces to finding the zeros of some polynomial.

For example, consider the polynomial equation $x^3+5x^2=11x-x^3-4$, for which we can see that $x=1$ is a solution because $1+5=11-1-4$. Subtracting $11x-x^3-4$ from both sides, we reach $2x^3+5x^2-11x+4=0$. Now, the problem is to find the zeros of the polynomial $2x^3+5x^2-11x+4$. .

The polynomial has degree $3$, so we are looking for $3$ zeros, each of which corresponds to a factor of the polynomial. We know one of the zeros is $x=1$, which corresponds to a factor $x-1$, and we know the other two factors need to multiply to a quadratic $2x^2+bx+c$.

By multiplying out $(x-1)(2x^2+bx+c)$ and comparing coefficients to the original polynomial, we can solve for $b$ and $c$. Then, we can solve the quadratic to find the remaining zeros.

Equating $x^2$ coefficients, we see that $5=b-2$, so $b=7$. Finally, by equating the constants $4$ and $-c$, we see that $c=-4$. The polynomial can then be written as

Solving the quadratic $2x^2+7x-4=0$ leads us to the two remaining zeros: $x=-4$ and $x=\frac{1}{2}$.

We check to ensure that these zeros are indeed solutions of the original equation:

## Exercises

For each of the following polynomials, use the given zero(s) to find the remaining zero(s). (You can view the solution by clicking on the problem.)

## $\begin{align*} 1) \hspace{.5cm} &p(x)=2x^3-x^2-2x+1 \\ &\text{given zeros: } 1, -1 \end{align*}$

*Solution:*

$\dfrac{1}{2}$

## $\begin{align*} 2) \hspace{.5cm} &p(x)=4x^3-8x^2-59x+63 \\ &\text{given zero: } 1 \end{align*}$

*Solution:*$-\dfrac{7}{2}, \dfrac{9}{2}$

## $\begin{align*} 3) \hspace{.5cm} &p(x)=3x^4-22x^3+41x^2+2x-24 \\ &\text{given zeros: } 1, 3 \end{align*}$

*Solution:*$-\dfrac{2}{3}, 4$

## $\begin{align*} 4) \hspace{.5cm} &p(x)=49x^5-133x^4+15x^3+145x^2-64x-12 \\ &\text{given zeros: } 1, -1, 2 \end{align*}$

*Solution:*$-\dfrac{1}{7}, \dfrac{6}{7}$

For each of the following equations, use the given solution(s) to find the remaining solution(s). (You can view the solution by clicking on the problem.)

## $\begin{align*} 5) \hspace{.5cm} &x^3+17x=8x^2+10 \\ &\text{given zeros: } 1, 2 \end{align*}$

*Solution:*$5$

## $\begin{align*} 6) \hspace{.5cm} &x(x^2-11)=2(x^2-6) \\ &\text{given zeros: } 1 \end{align*}$

*Solution:*$-3,4$

## $\begin{align*} 7) \hspace{.5cm} &4x^3+26x+5x^4=63x^2-88 \\ &\text{given zeros: } -1, 2 \end{align*}$

*Solution:*$-4, \dfrac{11}{5}$

## $\begin{align*} 8) \hspace{.5cm} &x(x^4+144x-47)=2(3x^4+13x^3+105) \\ &\text{given zeros: } -1, 2, 3 \end{align*}$

*Solution:*$-5, 7$

*This post is part of a series.*