Chapter 2.6, Problem 11Q

Solution

To find: all real zero of the function.

The only real zero of this function is $\text{1,-1,2,-6}$ .

Given information: 

The function $f\left(x\right)=x^4+4x^3-13x^2-4x+12$ .

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

  $f\left(x\right)=x^4+4x^3-13x^2-4x+12$

The constant term is $12$ , and its factors are

  $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$

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

  $\pm 1$

So, all possible rational zeros are:

  $\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{3}{1},\pm \frac{4}{1},\pm \frac{6}{1},\pm \frac{12}{1}$

Simplify fractions:

  $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$

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

  $\begin{array}{l}1\overline{)\begin{array}{lllll}1\hfill & 4\hfill & -13\hfill & -4\hfill & 12\hfill \\ \hfill & 1\hfill & 5\hfill & -8\hfill & -12\hfill \end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 1& \hfill 5& \hfill -8& \hfill -12& \hfill 0\end{array}\end{array}$

So, $f\left(1\right)=0$ , hence $1$ is not a zero of this function.

Try it $8$ is a zero:

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

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

  $f(x)=(x-1)\left(x^3+5x^2-8x-12\right)$

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

Define $g\left(x\right)=x^3+5x^2-8x-12$ .

Any zero of $g\left(x\right)$ is also a zero of $f\left(x\right)$ . So, possible zeros of $g\left(x\right)$ are

  $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$

Try it $-1$ is a zero of $g\left(x\right)$ :

  $\begin{array}{l}-1\overline{)\begin{array}{llll}1\hfill & 5\hfill & -8\hfill & -12\hfill \\ \hfill & -1\hfill & -4\hfill & 12\hfill \end{array}}\\ \begin{array}{lllll}\hfill & 1\hfill & 4\hfill & -12\hfill & 0\hfill \end{array}\end{array}$

So, $g(-1)=0$ , hence $-1$ is a zero of $g\left(x\right)$ ! Also, it is a zero of $f\left(x\right)$ .

Furthermore,

  $g(x)=(x+1)\left(x^2+4x-12\right),$

and

  $f(x)=(x-1)(x+1)\left(x^2+4x-12\right)$

To find other zeros and solve the equation

  $x^2+4x-12=0$

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 $c=-12$ , so the solution is

  $\begin{array}{c}x=\frac{-4\pm \sqrt{{(4)}^2-4(1)(-12)}}{2(1)}\\ =\frac{-4\pm \sqrt{16+48}}{2}\\ =\frac{-4\pm \sqrt{64}}{2}\\ =\frac{-4\pm 8}{2}\\ =-2\pm 4\end{array}$

So, the solutions of the equation $x^2+4x-12=0$ are real and they are:

  $-2+4=2\text{ and }-2-4=-6$

Thus, all real zeros of $f\left(x\right)$ are:

Thus, the only real zero of this function is $\text{1,-1,2,-6}$ .