Chapter 2.6, Problem 4GP

Solution

To find: all real zeros of the function.

The real zeros of the function are $-1,7,2$ .

Given information: 

The function $f\left(x\right)=x^3-8x^2+5x+14$ .

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

  $f\left(x\right)=x^3-8x^2+5x+14$

The constant term is $14$ , and its factors are

  $\pm 1,\pm 2,\pm 7,\pm 14$

The leading coefficient is $1$

Therefore, each and every rational zero is combinations of factors of $14$ and

factors of $1$ :

  $\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{7}{1},\pm \frac{14}{1}$

Simplify all fractions:

  $\pm 1,\pm 2,\pm 7,\pm 14$

Use synthetic division:

  $\begin{array}{ccccc}1& 1& -8& 5& 14\\ & & 1& -3& -18\\ & 1& -3& -18& 0\end{array}$

So, $f\left(-1\right)=0$ , so $-1$ is indeed a zero of this function.

  $f\left(x\right)=\left(x+1\right)\left(x^2-9x+14\right)$

Read the coefficients of the quotient from the final row.

To find other zeros, solve the equation

  $x^2-9x+14=0$

The solution of the equation is,

  $\begin{array}{c}{x}^2-7x-2x+14=0\\ x\left(x-7\right)-2\left(x-7\right)=0\\ \left(x-7\right)\left(x-2\right)=0\\ x=7,2\end{array}$

Hence, the real zero of the function are $-1,7,2$ .