# Zeros of Polynomials

The **zeros** of a polynomial are the inputs which 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?

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 $0$.

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.

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 $(x-1)(2x^2+7x-4)$. 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:

*Practice Problems*

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:*$-5, 7$

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