Chapter 2.6, Problem 26E

Solution

To find: all real zeros of the function.

The real zeros of the function $x=1,3,-\frac{5}{2}$ .

Given information: 

The function $h\left(x\right)=2x^3-3x^2-14x+15$ .

Calculation:

Consider the function

  $h\left(x\right)=2x^3-3x^2-14x+15$

Rational zero theorem:

Factors of constant term $15$ :

  $p=1,3,5,15$

Leading coefficient factors $2$ :

  $q=1,2$

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

The function's zeros could be all or none of them.

  $\pm \frac{p}{q}=\pm \left(1,3,5,15,\frac{1}{2},\frac{3}{2},\frac{5}{2},\frac{15}{2}\right)$

Test the possible rational zero combinations.

  $/$ Polly Horner scheme $\left[x=1\right]\left\{2x^3-3x^2-14x+15\right\}$

  $x=1$ is zero, so $\left(x-1\right)$ is a factor and using the results of the division.

  $\begin{array}{c}h\left(x\right)=2x^3-3x^2-14x+15\\ =\left(x-1\right)\left(2x^2-x-15\right)\end{array}$

The quadratic part can be factored

  $\left(x-1\right)\left(2x^2-x-15\right)=\left(x-1\right)\left(x-3\right)\left(2x+5\right)$

So, the zeros are $x=1,3,-\frac{5}{2}$ .