Chapter 2.6, Problem 9Q

Solution

To find: all real zero of the function.

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

Given information: 

The function $f\left(x\right)=2x^3-19x^2+50x+30$ .

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

  $f\left(x\right)=2x^3-19x^2+50x+30$

The constant term is $30$ , and its factors are

  $\pm 1,\pm 2,\pm 3,\pm 5,\pm 10,\pm 15,\pm 30$

The leading coefficient is $2$ , and its factors are

  $\pm 1,\pm 2$

So, all possible rational zeros are:

  $\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{3}{1},\pm \frac{5}{1},\pm \frac{10}{1},\pm \frac{15}{1},\pm \frac{30}{1},\pm \frac{1}{2},\pm \frac{2}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{10}{2},\pm \frac{15}{2},\pm \frac{30}{2}$

Simplify fractions:

  $\pm 1,\pm 2,\pm 3,\pm 5,\pm 10,\pm 15,\pm 30,\pm \frac{1}{2},\pm 1,\pm \frac{3}{2},\pm \frac{5}{2},\pm 5,\pm \frac{15}{2},\pm 15$

Remove duplicates:

  $\pm 1,\pm 2,\pm 3,\pm 5,\pm 10,\pm 15,\pm 30,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{15}{2}$

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

  $\begin{array}{l}1\overline{)\begin{array}{llll}12\hfill & -19\hfill & 50\hfill & 30\hfill \\ \hfill & 2\hfill & -17\hfill & 33\hfill \end{array}}\\ \begin{array}{lllll}\hfill & 2\hfill & -17\hfill & 33\hfill & 63\hfill \end{array}\end{array}$

So, $f\left(1\right)=30$ , hence $1$ is not a zero of this function.

Try it $-\frac{1}{2}=-0.5$ is a zero:

  $\begin{array}{l}-\frac{1}{2}\overline{)\begin{array}{llll}12\hfill & -19\hfill & 50\hfill & 30\hfill \\ \hfill & -1\hfill & 10\hfill & -30\hfill \end{array}}\\ \begin{array}{lllll}\hfill & 2\hfill & -20\hfill & 60\hfill & 0\hfill \end{array}\end{array}$

So, $f\left(-\frac{1}{2}\right)=0$ , hence $-\frac{1}{2}$ is a zero! Furthermore,

  $f(x)=\left(x+\frac{1}{2}\right)\left(2x^2-20x+60\right)$

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

To find other zerossolve the equation

  $2x^2-20x+60=0$

The solution of the equation

  $a{x}^2+bx+c=0$ is

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here $a=2,b=-20$ and $c=60$ , so the solution is

  $\begin{array}{c}x=\frac{-(-20)\pm \sqrt{{(-20)}^2-4(2)(60)}}{2(2)}\\ =\frac{20\pm \sqrt{400-480}}{4}\\ =\frac{20\pm \sqrt{-80}}{4}\end{array}$

But these are imaginary numbers!

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