Chapter 2.6, Problem 16E

Solution

To find: all real zeros of the function.

The only real zeros are $-3$ and $4$ .

Given information: 

The function $f\left(x\right)=x^4-2x^3-9x^2+10x-24$ .

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

  $f\left(x\right)=x^4-2x^3-9x^2+10x-24$ .

Then

Since the leading coefficient is $1$ and the constant term is $-24$ , the possible rational zeros are:

  $\begin{array}{r}\hfill \pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{3}{1},\pm \frac{4}{1},\pm \frac{6}{1},\pm \frac{8}{1},\pm \frac{12}{1},\pm \frac{24}{1},\text{ i}\text{.e}\text{., }\\ \hfill \pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 8,\pm 12,\pm 24\end{array}$

Test these potential zeros by the synthetic division until we have found 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 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:

  $\backslash$ polyhornerscheme $[x=1]\left\{x^4-2x^3-9x^2+10x-24\right\}$

And $1$ is not a zero.

Let's test 2:

  $\backslash$ polyhornerscheme $[x=2]\left\{x^4-2x^3-9x^2+10x-24\right\}$

And $2$ is not a zero.

Let's test $3$ :

  $\backslash$ polyhornerscheme $[x=3]\left\{x^4-2x^3-9x^2+10x-24\right\}$

And $3$ is not a zero.

Let's test $4$ :

\polyhornerscheme $[x=4]\left\{x^4-2x^3-9x^2+10x-24\right\}$

And $4$ is indeed a zero.

Now, use the last row (without the last number $0$ ) of the synthetic division (for $4$ ) to test other numbers.

Let's test $-1$ :

  $\backslash$ poly horner scheme $[x=-1]\left\{1x^3+2x^2-x+6\right\}$

And $-1$ is not a zero.

Let's test $-2$ :

  $\backslash$ poly horner scheme $[x=-2]\left\{1x^3+2x^2-x+6\right\}$

And $-2$ is not a zero.

Let's test $-3$ :

  $\backslash$ poly horner scheme $[x=-3]\left\{1x^3+2x^2-x+6\right\}$

And $-3$ is indeed a zero.

Using synthetic division, the polynomial has a degree of $4$ and two zeros.

By using the last row of the synthetic division, with us $x^2-x+2=0$ .

However, this is not factorable, so we must use the quadratic formula:

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{1\pm \sqrt{1-8}}{2}\\ =\frac{1\pm i\sqrt{7}}{2}\end{array}$

However, these two zeros are imaginary.

Therefore, the only real zeros are $-3$ and $4$ .