Chapter 2.6, Problem 13E

Solution

To find: all real zeros of the function.

The three zeroes of $g\left(x\right)=x^3-31x-30$ are $-5,-1$ and $6$ .

Given information: 

The function $g\left(x\right)=x^3-31x-30$ .

Formula used:

The Rational Zero Theorem:

If $f\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ consists of integer coefficients, then the shape of each rational zero of $f$ is as follows:

  $\frac{p}{q}=\frac{\text{ factor of constant term }{a}_0}{\text{ factor of leading coefficient }{a}_n}$

Calculation:

Consider the function

  $g\left(x\right)=x^3-31x-30$ .

Then

  $\begin{array}{c}g\left(-1\right)={\left(-1\right)}^3-31\left(-1\right)-30\\ =-1+31-30\\ =0\end{array}$

By factor theorem $x+1$ is a factor of $g\left(x\right)$ .

Divide $g\left(x\right)$ by $x+1$ as shown below:

Therefore, $x^3-31x-30=\left(x+1\right)\left(x^2-x-30\right)$

The zeroes of $g\left(x\right)$ are the roots of $g\left(x\right)=0$ .

  $\begin{array}{c}g\left(x\right)=0\\ x^3-31x-30=0\\ \left(x+1\right)\left(x^2-x-30\right)=0\end{array}$

By zero product property we have $x+1=0$ or $x^2-x-30=0$ .

  $x=-1$ is a zero and the zeroes of $x^2-x-30$ are the also the zeroes of $g\left(x\right)$ .

Finding roots for $x^2-x-30=0$ .

In order to factorize $x^2-x-30$ , we rewrite $-1$ as a sum of two numbers who multiply to give $-30$ .

Therefore, $x^2-x-30=0$ as $x^2+\left(5-6\right)x-30=0$

  $\begin{array}{c}{x}^2+5x-6x-30=0\\ \left(x^2+5x\right)+\left(-6x-30\right)=0\\ x\left(x+5\right)-6\left(x+5\right)=0\\ \left(x-6\right)\left(x+5\right)=0\end{array}$

By zero product property we have $x-5=0$ and $x+6=0$ .

  $x=6$ and $x=-5$ are two zeroes.

Hence, the three zeroes of $g\left(x\right)=x^3-31x-30$ are $-5,-1$ and $6$ .