Chapter SH, Problem 5.24EP

To determine

Find all the rationalzeros of the given function.

Answer to Problem 5.24EP

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

Explanation of Solution

Given:

The given function is $f\left(x\right)=4x^4-81x^2-110x+75$ .

Calculation:

Since $f\left(x\right)$ has degree $4$ , the function has $4$ zeros.

Find the type of zeros.

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

  $\begin{array}{l}f\left(x\right)=4x^4-81x^2-110x+75\\ \text{yes}\text{no}\text{yes}\end{array}$

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

  $\begin{array}{l}f\left(-x\right)=4{\left(-x\right)}^4-81{\left(-x\right)}^2-110\left(-x\right)+75\\ =4x^4-81x^2+110x+75\\ \text{yes}\text{yes}\text{no}\end{array}$

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

The possible rational zeros are $\pm 1,\pm 3,\pm 5,\pm 15,\pm 25,\pm 75,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{15}{2},$

  $\pm \frac{25}{2},\pm \frac{75}{2},\pm \frac{1}{4},\pm \frac{3}{4},\pm \frac{5}{4},\pm \frac{15}{4},\pm \frac{25}{4},\pm \frac{75}{4}$ .

Draw a table and test some possible rational zeros.

$\frac{p}{q}$	$4$	$0$	$-81$	$-110$	$75$
$-3$	$4$	$-12$	$-45$	$25$	$0$
$-2$	$4$	$-8$	$-65$	$20$	$35$
$-1$	$4$	$-4$	$-77$	$-33$	$108$

Because $f\left(-3\right)=0$ , there is a zero at $x=-3$ . Factor the depressed polynomial $4x^3-12x^2-45x+25$ .

Use synthetic substitution.

$\frac{p}{q}$	$4$	$-12$	$-45$	$25$
$3$	$4$	$0$	$-45$	$-110$
$4$	$4$	$4$	$-29$	$-91$
$5$	$4$	$8$	$-5$	$0$

Because $f\left(5\right)=0$ , there is a zero at $x=5$ . Factor the depressed polynomial $4x^2+8x-5$ .

Use factorization method.

  $\begin{array}{l}4x^2+8x-5=0\\ \Rightarrow 4x^2+10x-2x-5=0\\ \Rightarrow 4x\left(2x+5\right)-\left(2x+5\right)=0\\ \Rightarrow \left(4x-1\right)\left(2x+5\right)=0\\ \Rightarrow \left(4x-1\right)\left(2x+5\right)=0\\ \Rightarrow 4x-1=0\text{and}2x+5=0\\ \Rightarrow 4x-1+1=0+1\text{and}2x+5-5=0-5\\ \Rightarrow 4x=1\text{and}2x=-5\\ \Rightarrow \frac{4}{4}x=\frac{1}{4}\text{and}\frac{2}{2}x=-\frac{5}{2}\\ \Rightarrow x=\frac{1}{4}\text{and}x=-\frac{5}{2}\end{array}$

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