# 5.5 Zeros of polynomial functions  (Page 5/14)

 Page 5 / 14

Let $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ be a polynomial function with real coefficients, and suppose is a zero of $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$ Then, by the Factor Theorem, $\text{\hspace{0.17em}}x-\left(a+bi\right)\text{\hspace{0.17em}}$ is a factor of $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$ For $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ to have real coefficients, $\text{\hspace{0.17em}}x-\left(a-bi\right)\text{\hspace{0.17em}}$ must also be a factor of $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$ This is true because any factor other than $\text{\hspace{0.17em}}x-\left(a-bi\right),\text{\hspace{0.17em}}$ when multiplied by $\text{\hspace{0.17em}}x-\left(a+bi\right),\text{\hspace{0.17em}}$ will leave imaginary components in the product. Only multiplication with conjugate pairs will eliminate the imaginary parts and result in real coefficients. In other words, if a polynomial function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ with real coefficients has a complex zero $\text{\hspace{0.17em}}a+bi,\text{\hspace{0.17em}}$ then the complex conjugate $\text{\hspace{0.17em}}a-bi\text{\hspace{0.17em}}$ must also be a zero of $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$ This is called the Complex Conjugate Theorem .

## Complex conjugate theorem

According to the Linear Factorization Theorem     , a polynomial function will have the same number of factors as its degree, and each factor will be in the form $\text{\hspace{0.17em}}\left(x-c\right)$ , where $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ is a complex number.

If the polynomial function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ has real coefficients and a complex zero in the form $\text{\hspace{0.17em}}a+bi,\text{\hspace{0.17em}}$ then the complex conjugate of the zero, $\text{\hspace{0.17em}}a-bi,\text{\hspace{0.17em}}$ is also a zero.

Given the zeros of a polynomial function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ and a point ( c , f ( c )) on the graph of $\text{\hspace{0.17em}}f,\text{\hspace{0.17em}}$ use the Linear Factorization Theorem to find the polynomial function.

1. Use the zeros to construct the linear factors of the polynomial.
2. Multiply the linear factors to expand the polynomial.
3. Substitute $\text{\hspace{0.17em}}\left(c,f\left(c\right)\right)\text{\hspace{0.17em}}$ into the function to determine the leading coefficient.
4. Simplify.

## Using the linear factorization theorem to find a polynomial with given zeros

Find a fourth degree polynomial with real coefficients that has zeros of –3, 2, $\text{\hspace{0.17em}}i,\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}f\left(-2\right)=100.\text{\hspace{0.17em}}$

Because $\text{\hspace{0.17em}}x=i\text{\hspace{0.17em}}$ is a zero, by the Complex Conjugate Theorem $\text{\hspace{0.17em}}x=–i\text{\hspace{0.17em}}$ is also a zero. The polynomial must have factors of $\text{\hspace{0.17em}}\left(x+3\right),\text{\hspace{0.17em}}\left(x-2\right),\text{\hspace{0.17em}}\left(x-i\right),\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(x+i\right).\text{\hspace{0.17em}}$ 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.

$\begin{array}{ccc}\hfill f\left(x\right)& =& a\left(x+3\right)\left(x-2\right)\left(x-i\right)\left(x+i\right)\hfill \\ \hfill f\left(x\right)& =& a\left({x}^{2}+x-6\right)\left({x}^{2}+1\right)\hfill \\ \hfill f\left(x\right)& =& a\left({x}^{4}+{x}^{3}-5{x}^{2}+x-6\right)\hfill \end{array}$

We need to find a to ensure $\text{\hspace{0.17em}}f\left(–2\right)=100.\text{\hspace{0.17em}}$ Substitute and $\text{\hspace{0.17em}}f\left(2\right)=100\text{\hspace{0.17em}}$ into $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$

$\begin{array}{ccc}\hfill 100& =& a\left({\left(-2\right)}^{4}+{\left(-2\right)}^{3}-5{\left(-2\right)}^{2}+\left(-2\right)-6\right)\hfill \\ \hfill 100& =& a\left(-20\right)\hfill \\ \hfill -5& =& a\hfill \end{array}$

So the polynomial function is

$f\left(x\right)=-5\left({x}^{4}+{x}^{3}-5{x}^{2}+x-6\right)$

or

$f\left(x\right)=-5{x}^{4}-5{x}^{3}+25{x}^{2}-5x+30$

If $\text{\hspace{0.17em}}2+3i\text{\hspace{0.17em}}$ were given as a zero of a polynomial with real coefficients, would $\text{\hspace{0.17em}}2-3i\text{\hspace{0.17em}}$ 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.

Find a third degree polynomial with real coefficients that has zeros of 5 and $\text{\hspace{0.17em}}-2i\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}f\left(1\right)=10.\text{\hspace{0.17em}}$

$f\left(x\right)=-\frac{1}{2}{x}^{3}+\frac{5}{2}{x}^{2}-2x+10$

## Using descartes’ rule of signs

There is a straightforward way to determine the possible numbers of positive and negative real zeros for any polynomial function. If the polynomial is written in descending order, Descartes’ Rule of Signs tells us of a relationship between the number of sign changes in $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ and the number of positive real zeros. For example, the polynomial function below has one sign change. This tells us that the function must have 1 positive real zero.

root under 3-root under 2 by 5 y square
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