5.5 Zeros of polynomial functions  (Page 3/14)

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Given a polynomial function $\text{\hspace{0.17em}}f\left(x\right),$ use the Rational Zero Theorem to find rational zeros.

1. Determine all factors of the constant term and all factors of the leading coefficient.
2. Determine all possible values of $\text{\hspace{0.17em}}\frac{p}{q},\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ is a factor of the constant term and $\text{\hspace{0.17em}}q\text{\hspace{0.17em}}$ is a factor of the leading coefficient. Be sure to include both positive and negative candidates.
3. Determine which possible zeros are actual zeros by evaluating each case of $\text{\hspace{0.17em}}f\left(\frac{p}{q}\right).\text{\hspace{0.17em}}$

Listing all possible rational zeros

List all possible rational zeros of $\text{\hspace{0.17em}}f\left(x\right)=2{x}^{4}-5{x}^{3}+{x}^{2}-4.$

The only possible rational zeros of $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ are the quotients of the factors of the last term, –4, and the factors of the leading coefficient, 2.

The constant term is –4; the factors of –4 are $\text{\hspace{0.17em}}p=±1,±2,±4.$

The leading coefficient is 2; the factors of 2 are $\text{\hspace{0.17em}}q=±1,±2.$

If any of the four real zeros are rational zeros, then they will be of one of the following factors of –4 divided by one of the factors of 2.

Note that $\text{\hspace{0.17em}}\frac{2}{2}=1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\frac{4}{2}=2,\text{\hspace{0.17em}}$ which have already been listed. So we can shorten our list.

Using the rational zero theorem to find rational zeros

Use the Rational Zero Theorem to find the rational zeros of $\text{\hspace{0.17em}}f\left(x\right)=2{x}^{3}+{x}^{2}-4x+1.\text{\hspace{0.17em}}$

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 1 and $\text{\hspace{0.17em}}q\text{\hspace{0.17em}}$ is a factor of 2.

The factors of 1 are $±1\text{\hspace{0.17em}}$ and the factors of 2 are $±1\text{\hspace{0.17em}}$ and $±2.\text{\hspace{0.17em}}$ The possible values for $\text{\hspace{0.17em}}\frac{p}{q}\text{\hspace{0.17em}}$ are $±1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}±\frac{1}{2}.\text{\hspace{0.17em}}$ These are the possible rational zeros for the function. We can determine which of the possible zeros are actual zeros by substituting these values for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ in $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$

$\begin{array}{ccc}\hfill f\left(-1\right)& =& 2{\left(-1\right)}^{3}+{\left(-1\right)}^{2}-4\left(-1\right)+1=4\hfill \\ \hfill f\left(1\right)& =& 1{\left(1\right)}^{3}+{\left(1\right)}^{2}-4\left(1\right)+1=0\hfill \\ f\left(-\frac{1}{2}\right)\hfill & =& 2{\left(-\frac{1}{2}\right)}^{3}+{\left(-\frac{1}{2}\right)}^{2}-4\left(-\frac{1}{2}\right)+1=3\hfill \\ \hfill f\left(\frac{1}{2}\right)& =& 2{\left(\frac{1}{2}\right)}^{3}+{\left(\frac{1}{2}\right)}^{2}-4\left(\frac{1}{2}\right)+1=-\frac{1}{2}\hfill \end{array}$

Of those, are not zeros of $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$ 1 is the only rational zero of $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$

Use the Rational Zero Theorem to find the rational zeros of $\text{\hspace{0.17em}}f\left(x\right)={x}^{3}-5{x}^{2}+2x+1.\text{\hspace{0.17em}}$

There are no rational zeros.

Finding the zeros of polynomial functions

The Rational Zero Theorem helps us to narrow down the list of possible rational zeros for a polynomial function. Once we have done this, we can use synthetic division    repeatedly to determine all of the zeros    of a polynomial function.

Given a polynomial function $\text{\hspace{0.17em}}f,$ use synthetic division to find its zeros.

1. Use the Rational Zero Theorem to list all possible rational zeros of the function.
2. Use synthetic division to evaluate a given possible zero by synthetically dividing the candidate into the polynomial. If the remainder is 0, the candidate is a zero. If the remainder is not zero, discard the candidate.
3. Repeat step two using the quotient found with synthetic division. If possible, continue until the quotient is a quadratic.
4. Find the zeros of the quadratic function. Two possible methods for solving quadratics are factoring and using the quadratic formula.

Finding the zeros of a polynomial function with repeated real zeros

Find the zeros of $\text{\hspace{0.17em}}f\left(x\right)=4{x}^{3}-3x-1.\text{\hspace{0.17em}}$

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 –1 and $\text{\hspace{0.17em}}q\text{\hspace{0.17em}}$ is a factor of 4.

The factors of $\text{\hspace{0.17em}}–1\text{\hspace{0.17em}}$ are $±1\text{\hspace{0.17em}}$ and the factors of $\text{\hspace{0.17em}}4\text{\hspace{0.17em}}$ are $±1,±2,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}±4.\text{\hspace{0.17em}}$ The possible values for $\text{\hspace{0.17em}}\frac{p}{q}\text{\hspace{0.17em}}$ are $±1,\text{\hspace{0.17em}}±\frac{1}{2},\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}±\frac{1}{4}.\text{\hspace{0.17em}}$ These are the possible rational zeros for the function. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with 1. Dividing by $\text{\hspace{0.17em}}\left(x-1\right)\text{\hspace{0.17em}}$ gives a remainder of 0, so 1 is a zero of the function. The polynomial can be written as

$\left(x-1\right)\left(4{x}^{2}+4x+1\right)$

The quadratic is a perfect square. $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ can be written as

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

We already know that 1 is a zero. The other zero will have a multiplicity of 2 because the factor is squared. To find the other zero, we can set the factor equal to 0.

$\begin{array}{ccc}\hfill 2x+1& =& 0\hfill \\ \hfill x& =& -\frac{1}{2}\hfill \end{array}$

The zeros of the function are 1 and $\text{\hspace{0.17em}}-\frac{1}{2}\text{\hspace{0.17em}}$ with multiplicity 2.

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