# 5.5 Zeros of polynomial functions  (Page 4/14)

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## 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.

Suppose $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ is a polynomial function of degree four, and $\text{\hspace{0.17em}}f\left(x\right)=0.\text{\hspace{0.17em}}$ The Fundamental Theorem of Algebra states that there is at least one complex solution, call it $\text{\hspace{0.17em}}{c}_{1}.\text{\hspace{0.17em}}$ By the Factor Theorem, we can write $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ as a product of $\text{\hspace{0.17em}}x-{c}_{\text{1}}\text{\hspace{0.17em}}$ and a polynomial quotient. Since $\text{\hspace{0.17em}}x-{c}_{\text{1}}\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}{c}_{\text{2}}.\text{\hspace{0.17em}}$ So we can write the polynomial quotient as a product of $\text{\hspace{0.17em}}x-{c}_{\text{2}}\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$

## The fundamental theorem of algebra

The Fundamental Theorem of Algebra    states that, if $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ is a polynomial of degree n>0 , then $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ has at least one complex zero.

We can use this theorem to argue that, if $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ is a polynomial of degree $\text{\hspace{0.17em}}n>0,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is a non-zero real number, then $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ has exactly $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ linear factors

$f\left(x\right)=a\left(x-{c}_{1}\right)\left(x-{c}_{2}\right)...\left(x-{c}_{n}\right)$

where $\text{\hspace{0.17em}}{c}_{1},{c}_{2},...,{c}_{n}\text{\hspace{0.17em}}$ are complex numbers. Therefore, $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ has $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ roots if we allow for multiplicities.

Does every polynomial have at least one imaginary zero?

No. Real numbers are a subset of complex numbers, but not the other way around. A complex number is not necessarily imaginary. Real numbers are also complex numbers.

## Finding the zeros of a polynomial function with complex zeros

Find the zeros of $\text{\hspace{0.17em}}f\left(x\right)=3{x}^{3}+9{x}^{2}+x+3.\text{\hspace{0.17em}}$

The Rational Zero Theorem tells us that if $\text{\hspace{0.17em}}\frac{p}{q}\text{\hspace{0.17em}}$ is a zero of $\text{\hspace{0.17em}}f\left(x\right),\text{\hspace{0.17em}}$ then $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ is a factor of 3 and $\text{\hspace{0.17em}}q\text{\hspace{0.17em}}$ is a factor of 3.

The factors of 3 are $±1\text{\hspace{0.17em}}$ and $±3.\text{\hspace{0.17em}}$ The possible values for $\text{\hspace{0.17em}}\frac{p}{q},\text{\hspace{0.17em}}$ and therefore the possible rational zeros for the function, are 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.

Dividing by $\text{\hspace{0.17em}}\left(x+3\right)\text{\hspace{0.17em}}$ gives a remainder of 0, so –3 is a zero of the function. The polynomial can be written as

$\left(x+3\right)\left(3{x}^{2}+1\right)$

We can then set the quadratic equal to 0 and solve to find the other zeros of the function.

$\begin{array}{ccc}\hfill 3{x}^{2}+1& =& 0\hfill \\ \hfill {x}^{2}& =& -\frac{1}{3}\hfill \\ \hfill x& =& ±\sqrt{-\frac{1}{3}}=±\frac{i\sqrt{3}}{3}\hfill \end{array}$

The zeros of $f\left(x\right)$ are –3 and $\text{\hspace{0.17em}}±\frac{i\sqrt{3}}{3}.$

Find the zeros of $\text{\hspace{0.17em}}f\left(x\right)=2{x}^{3}+5{x}^{2}-11x+4.$

The zeros are

## Using the linear 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 $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ will have $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ zeros in the set of complex numbers, if we allow for multiplicities. This means that we can factor the polynomial function into $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ factors. The Linear 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 $\text{\hspace{0.17em}}\left(x-c\right),\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ is a complex number.

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