Chapter 2.6, Problem 3GP

Solution

To find: all real zeros of the function.

The real zeros of the function $\text{1}$ .

Given information: 

The function $f\left(x\right)=x^3-4x^2-15x+18$ .

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

  $f\left(x\right)=x^3-4x^2-15x+18$

The constant term is $18$ , and its factors are

  $\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18$

The leading coefficient is $2$

Therefore, each and every rational zero is combinations of factors of $15$ and

factors of $1$ :

  $\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{3}{1},\pm \frac{6}{1},\pm \frac{9}{1},\pm \frac{18}{1}$

Simplify all fractions:

  $\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18$

Use synthetic division:

  $\begin{array}{ccccc}1& 1& -4& -15& 18\\ & & 1& -3& -18\\ & 1& -3& -18& 0\end{array}$

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

  $f\left(x\right)=\left(x-1\right)\left(x^2-3x+18\right)$

Read the coefficients of the quotient from the final row.

To find other zeros, solve the equation

  $x^2-3x+18=0$

The solution of the equation is,

  $a{x}^2+bx+c=0$ is

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here $a=1,b=-3$ and $c=18$ , so

  $\begin{array}{c}x=\frac{-(-3)\pm \sqrt{{(-3)}^2-4(1)(18)}}{2(1)}\\ =\frac{9\pm \sqrt{9-72}}{2}\end{array}$

Hence, the only real zero of the function is $1$ .