Chapter 2.6, Problem 1GP

Solution

To find: the possible rational zeros of the function.

The possible rational zeros of the function $\left\{\pm 1,\pm 3,\pm 5,\pm 15\right\}$ .

Given information: 

The function $f\left(x\right)=x^3+9x^2+23x+15$ .

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

  $f\left(x\right)=x^3+9x^2+23x+15$

Factors of constant term $15$ :

  $p=1,3,5,15$

Leading coefficient factors $1$ :

  $q=1$

All possible combinations of $\pm \frac{p}{q}$ .

Some or none may be zeros of the function.

Simplify and eliminate duplicate entries.

  $\begin{array}{c}\pm \frac{p}{q}=\pm \left\{\frac{1}{1},\frac{3}{1},\frac{5}{1},\frac{15}{1}\right\}\\ =\left\{\pm 1,\pm 3,\pm 5,\pm 15\right\}\end{array}$