Chapter 2.6, Problem 17E

Solution

To find: all real zeros of the function.

The real zeros of the function $x=-1,-4,1,2$ .

Given information: 

The function $f\left(x\right)=x^4+2x^3-9x^2-2x+8$ .

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

  $f\left(x\right)=x^4+2x^3-9x^2-2x+8$

Factors of constant term $8$ :

  $p=1,2,4,8$

Factors of leading coefficient $1$ :

  $q=1$

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

Some or none may be zeros of the function.

Simplify and eliminate duplicate entries.

  $\pm \frac{p}{q}=\pm \left(1,2,4,8\right)$

Test the possible rational zeros.

  

  $1$ and $2$ are zeros, so factors are $\left(x-1\right)\left(x-2\right)=x^2-3x+2$ .

Divide $f\left(x\right)$ by this

  

Complete the factoring

  $\begin{array}{c}f\left(x\right)=x^4+2x^3-9x^2-2x+8\\ =\left(x-1\right)\left(x-2\right)\left(x^2+5x+4\right)\\ =\left(x-1\right)\left(x-2\right)\left(x+1\right)\left(x+4\right)\\ x=-1,-4,1,2\end{array}$