Chapter 2.6, Problem 37E

Solution

To describe: correct the error.

Possible zeros: $\pm 1,\pm 5,\pm \frac{1}{2},\pm \frac{5}{2},\pm \frac{1}{3},\pm \frac{5}{3},\pm \frac{1}{6},\pm \frac{5}{6}$

Error description: numerator and denominator candidates were reversed.

Given information: 

The function $f\left(x\right)=6x^3-3x^2+12x+5$ .

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

  $f\left(x\right)=6x^3-3x^2+12x+5$

Here, the constant term is $5$ , and the leading coefficient is $6$ .

Factors of $6$ :

  $\pm 1,\pm 2,\pm 3,\pm 6$

Factors of $5$ :

  $\pm 1,\pm 5$

So, possible rational zeros are (write all combinations of factors of $6$ and factors of $5$ ):

  $\pm \frac{1}{1},\pm \frac{5}{1},\pm \frac{1}{2},\pm \frac{5}{2},\pm \frac{1}{3},\pm \frac{5}{3},\pm \frac{1}{6},\pm \frac{5}{6}$

Simplify fractions:

  $\pm 1,\pm 5,\pm \frac{1}{2},\pm \frac{5}{2},\pm \frac{1}{3},\pm \frac{5}{3},\pm \frac{1}{6},\pm \frac{5}{6}$

In this exercise, $p$ and $q$ (from step $1$ ) were reversed.