Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 5 Polynomials and Polynomial Functions - 5.6 Find Rational Zeroes - 5.6 Exercises - Skill Practice - Page 374: 19


$1, \frac{-1-\sqrt{17}}{2},\frac{-1+\sqrt{17}}{2}$

Work Step by Step

We are given the polynomial function: $$f(x)=4x^3-20x+16.$$ $\bf{Step\text{ }1}$ First we will list the possible rational zeros. The leading coefficient is $4$ and the constant term is $16$. So the possible rational zeros are: $$\pm 1,\pm2,\pm 4,\pm 8,\pm 16,\pm \dfrac{1}{2},\pm \dfrac{1}{4}.$$ $\bf{Step\text{ }2}$ Choose reasonable values from the list using the graph of the function. We find the only reasonable value is: $$x=1.$$ $\bf{Step\text{ }3}$ Check this value using synthetic division: Test $x=1$. Because we got that the remainder is $0$, $1$ is a zero of $f$. $\bf{Step\text{ }4}$ Factor out the binomial using the result of the synthetic division: $$\begin{align*} f(x)&=(x-1)(4x^2+4x-16)\\ &=4(x-1)(x^2+x-4).\\ \end{align*}$$ $\bf{Step\text{ }5}$ Find the remaining zeros of $f$ by solving $x^2+x-4=0$ using the quadratic formula: $$\begin{align*} x&=\dfrac{-1\pm\sqrt{1^2-4(1)(-4)}}{2(1)}\\ &=\dfrac{-1\pm\sqrt{17}}{2}. \end{align*}$$ The real zeros of $f$ are $1, \frac{-1-\sqrt{17}}{2},\frac{-1+\sqrt{17}}{2}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.