Chapter 2.6, Problem 10Q

Solution

To find: all real zero of the function.

The only real zero of this function is $8$ .

Given information: 

The function $f\left(x\right)=x^3-4x^2-25x-56$ .

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

  $f\left(x\right)=x^3-4x^2-25x-56$

The constant term is $-56$ , and its factors are

  $\pm 1,\pm 2,\pm 4,\pm 7,\pm 8,\pm 14,\pm 28,\pm 56$

The leading coefficient is $1$ , and its factors are

  $\pm 1$

So, all possible rational zeros are:

  $\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{4}{1},\pm \frac{7}{1},\pm \frac{8}{1},\pm \frac{14}{1},\pm \frac{28}{1},\pm \frac{56}{1}$

Simplify fractions:

  $\pm 1,\pm 2,\pm 4,\pm 7,\pm 8,\pm 14,\pm 28,\pm 56$

Try it $1$ is a zero (use of synthetic substitution):

  $\begin{array}{l}1\overline{)\begin{array}{llll}1\hfill & -4\hfill & -25\hfill & -56\hfill \\ \hfill & 1\hfill & -3\hfill & -28\hfill \end{array}}\\ \begin{array}{lllll}\hfill & 1\hfill & -3\hfill & -28\hfill & -84\hfill \end{array}\end{array}$

So, $f\left(1\right)=-84$ , hence $1$ is not a zero of this function.

Try it $8$ is a zero:

  $\begin{array}{l}\overline{)\begin{array}{llll}1\hfill & -4\hfill & -25\hfill & -56\hfill \\ \hfill & 8\hfill & 32\hfill & 56\hfill \end{array}}\\ \begin{array}{lllll}\hfill & 1\hfill & 4\hfill & 7\hfill & 0\hfill \end{array}\end{array}$

So, $f\left(8\right)=0$ , hence $8$ is a zero! Furthermore,

  $f(x)=(x-8)\left(x^2+4x+7\right)$

(The coefficients of the last polynomial can be read from the last row of the table used for synthetic substitution.)

To find other zeros and solve the equation.

  $x^2+4x+7=0$

The solution of the equation

  $a{x}^2+bx+c=0$ is

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here $a=1,b=4$ and $c=7$ , so the solution is

  $\begin{array}{c}x=\frac{-4\pm \sqrt{{(4)}^2-4(1)(7)}}{2(1)}\\ =\frac{-4\pm \sqrt{16-28}}{2}\\ =\frac{-4\pm \sqrt{-12}}{2}\end{array}$

But these are imaginary numbers!

Thus, the only real zero of this function is $8$ .