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Exercises Polynomials and factoring

Example 4 from Algebra and Trigonometry, 5.5 Zeros of Polynomial Functions

Using the Rational Zero Theorem to Find Rational Zeros

Use the Rational Zero Theorem to find the rational zeros of $f (x) = 2 x^{3} + x^{2} - 4 x + 1.$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
The Rational Zero Theorem tells us that if $\frac{p}{q}$ is a zero of $f (x),$ then $p$ is a factor of 1 and $q$ is a factor of 2.

$\begin{matrix} \frac{p}{q} & = & \frac{factor of constant term}{factor of leading coefficient} \\ = & \frac{factor of 1}{factor of 2} \end{matrix}$

The factors of 1 are $±1$ and the factors of 2 are $±1$ and $±2.$ The possible values for $\frac{p}{q}$ are $±1$ and $\pm \frac{1}{2} .$ 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 $x$ in $f (x) .$

$\begin{matrix} f (−1) & = & 2 {(−1)}^{3} + {(−1)}^{2} - 4 (−1) + 1 = 4 \\ f (1) & = & 2 {(1)}^{3} + {(1)}^{2} - 4 (1) + 1 = 0 \\ f (- \frac{1}{2}) & = & 2 {(- \frac{1}{2})}^{3} + {(- \frac{1}{2})}^{2} - 4 (- \frac{1}{2}) + 1 = 3 \\ f (\frac{1}{2}) & = & 2 {(\frac{1}{2})}^{3} + {(\frac{1}{2})}^{2} - 4 (\frac{1}{2}) + 1 = - \frac{1}{2} \end{matrix}$
Of those, $−1, - \frac{1}{2}, and \frac{1}{2}$ are not zeros of $f (x) .$ 1 is the only rational zero of $f (x) .$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Use the Rational Zero Theorem to find the rational zeros of $f (x) = x^{3} - 5 x^{2} + 2 x + 1.$

How did it go?

PreviousListing All Possible Rational Zeros Next Finding the Zeros of a Polynomial Function with Repeated Real Zeros

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 5.5, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.