Chapter SH, Problem 5.23EP

To determine

Find all the rationalzeros of the given function.

Answer to Problem 5.23EP

The rational zeros of the given function are $x=-\frac{1}{2},x=\frac{3}{4}\text{and}x=2.5$ .

Explanation of Solution

Given:

The given function is $f\left(x\right)=16x^3-44x^2+4x+15$ .

Calculation:

Since $f\left(x\right)$ has degree $3$ , the function has $3$ zeros.

Find the type of zeros.

Examine the number of sign changes for $f\left(x\right)$ and $f\left(-x\right)$ .

  $\begin{array}{l}f\left(x\right)=16x^3-44x^2+4x+15\\ \text{yes}\text{yes}\text{no}\end{array}$

Because there are two sign changes for the coefficients of $f\left(x\right)$ , the function has two or zero positive real zeros.

  $\begin{array}{l}f\left(-x\right)=16{\left(-x\right)}^3-44{\left(-x\right)}^2+4\left(-x\right)+15\\ =-16x^3-44x^2-4x+15\\ \text{no}\text{no}\text{yes}\end{array}$

Because there is one sign changes for the coefficients of $f\left(-x\right)$ , the function has one or zero negative real zeros.

The possible rational zeros are $\pm 1,\pm 3,\pm 5,\pm 15,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{15}{2},$

  $\pm \frac{1}{4},\pm \frac{3}{4},\pm \frac{5}{4},\pm \frac{15}{4},\pm \frac{1}{8},\pm \frac{3}{8},\pm \frac{5}{8},\pm \frac{15}{8},\pm \frac{1}{16},\pm \frac{3}{16},\pm \frac{5}{16},\pm \frac{15}{16}$ .

Draw a table and test some possible rational zeros.

$\frac{p}{q}$	$16$	$-44$	$4$	$15$
$-1$	$16$	$-60$	$64$	$-49$
$1$	$16$	$-28$	$-24$	$-9$
$2.5$	$16$	$-4$	$-6$	$0$

Because $f\left(2\right)=0$ , there is a zero at $x=2.5$ . Factor the depressed polynomial $16x^2-4x-6$ .

Use factorization method.

  $\begin{array}{l}16x^2-4x-6=0\\ \Rightarrow 8x^2-2x-3=0\\ \Rightarrow 8x^2-6x+4x-3=0\\ \Rightarrow 2x\left(4x-3\right)+\left(4x-3\right)=0\\ \Rightarrow \left(4x-3\right)\left(2x+1\right)=0\\ \Rightarrow 4x-3=0\text{and}2x+1=0\\ \Rightarrow 4x-3+3=0+3\text{and}2x+1-1=0-1\\ \Rightarrow 4x=3\text{and}2x=-1\\ \Rightarrow \frac{4}{4}x=\frac{3}{4}\text{and}\frac{2}{2}x=-\frac{1}{2}\\ \Rightarrow x=\frac{3}{4}\text{and}x=-\frac{1}{2}\end{array}$

Hence the rational zeros of the given function are $x=-\frac{1}{2},x=\frac{3}{4}\text{and}x=2.5$ .