Chapter 2.6, Problem 12E

Solution

To find: all real zeros of the function.

The real zeros are $7,-4$ and $2$ .

Given information: 

The function $f\left(x\right)=x^3-5x^2-22x+56$ .

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

  $f\left(x\right)=x^3-5x^2-22x+56$

Since the leading coefficient is 1 and the constant term is $56$ , the possible rational zeros are:

  $\begin{array}{l}\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{4}{1},\pm \frac{7}{1},\pm \frac{8}{1},\pm \frac{14}{1},\pm \frac{28}{1},\pm \frac{56}{1},\text{ i}\text{.e}\text{.,}\\ \pm 1,\pm 2,\pm 4,\pm 7,\pm 8,\pm 14,\pm 28,\pm 56\end{array}$

Test these potential zeros by the synthetic division until a zero. If a test number is a zero, the last number of the last row of the synthetic division will be $0$ . If a

the test number is not a zero, the last number of the last row of the synthetic division won't be 0.

Let's test $1$ first:

lpolyhornerscheme $\left[x=1\right]\left\{x^3-5x^2-22x+56\right\}$

And $1$ is not a zero.

Let's test $2$

L polyhorner scheme $\left[x=2\right]\left\{x^3-5x^2-22x+56\right\}$

And $2$ is indeed a zero.

The degree of the polynomial is $3$ and using synthetic division, discovered one zero.

By using the last row of the synthetic division, $x^2-3x-28=0$ .

Solve this by factoring:

  $\begin{array}{c}{x}^2-3x-28=0\\ (x-7)(x+4)=0\\ x=7\text{ or }x=-4\end{array}$

Therefore, the real zeros are $7,-4$ and $2$ (from the synthetic division).