Chapter SH, Problem 5.22EP

To determine

Find all the rationalzeros of the given function.

Answer to Problem 5.22EP

The rational zeros of the given function are $x=-\frac{3}{4},x=\frac{1}{5}\text{and}x=2$ .

Explanation of Solution

Given:

The given function is $f\left(x\right)=20x^3-29x^2-25x+6$ .

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 $f\left(-x\right)$ .

  $\begin{array}{l}f\left(x\right)=20x^3-29x^2-25x+6\\ \text{yes}\text{no}\text{yes}\end{array}$

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

  $\begin{array}{l}f\left(-x\right)=20{\left(-x\right)}^3-29{\left(-x\right)}^2-25\left(-x\right)+6\\ =-20x^3-29x^2+25x+6\\ \text{no}\text{yes}\text{no}\end{array}$

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 $\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{1}{4},\pm \frac{3}{4},\pm \frac{1}{5},\pm \frac{2}{5},\pm \frac{3}{5},\pm \frac{6}{5},\pm \frac{1}{10},\pm \frac{3}{10},\pm \frac{1}{20},\pm \frac{3}{20}$ .

Draw a table and test some possible rational zeros.

$\frac{p}{q}$	$20$	$-29$	$-25$	$6$
$-1$	$20$	$-49$	$24$	$-18$
$1$	$20$	$-9$	$-34$	$-28$
$2$	$20$	$11$	$-3$	$0$

Because $f\left(2\right)=0$ , there is a zero at $x=2$ . Factor the depressed polynomial $20x^2+11x-3$ .

Use factorization method.

  $\begin{array}{l}20x^2+11x-3=0\\ \Rightarrow 20x^2+15x-4x-3=0\\ \Rightarrow 5x\left(4x+3\right)-\left(4x+3\right)=0\\ \Rightarrow \left(4x+3\right)\left(5x-1\right)=0\\ \Rightarrow 4x+3=0\text{and}5x-1=0\\ \Rightarrow 4x+3-3=0-3\text{and}5x-1+1=0+1\\ \Rightarrow 4x=-3\text{and}5x=1\\ \Rightarrow \frac{4}{4}x=-\frac{3}{4}\text{and}\frac{5}{5}x=\frac{1}{5}\\ \Rightarrow x=-\frac{3}{4}\text{and}x=\frac{1}{5}\end{array}$

Hence the rational zeros of the given function are $x=-\frac{3}{4},x=\frac{1}{5}\text{and}x=2$ .