Chapter 2.6, Problem 12Q

Solution

To find: all real zero of the function.

The only real zero of this function is $\frac{1}{2},-\frac{5}{2}$ .

Given information: 

The function $f\left(x\right)=4x^4-5x^2+42x-20$ .

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

  $f\left(x\right)=4x^4-5x^2+42x-20$

The constant term is $-20$ , and its factors are

  $\pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20$

The leading coefficient is $4$ , and its factors are

  $\pm 1,\pm 2,\pm 4$

So, all possible rational zeros are:

  $\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{4}{1},\pm \frac{5}{1},\pm \frac{10}{1},\pm \frac{20}{1},\pm \frac{1}{2},\pm \frac{2}{2},\pm \frac{4}{2},\pm \frac{5}{2},\pm \frac{10}{2},\pm \frac{20}{2},\pm \frac{1}{4},\pm \frac{2}{4},\pm \frac{4}{4},\pm \frac{5}{4},\pm \frac{10}{4},\pm \frac{20}{4}$

Simplify fractions:

  $\pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20,\pm \frac{1}{2},\pm 1,\pm 2,\pm \frac{5}{2},\pm 5,\pm 10,\pm \frac{1}{4},\pm \frac{1}{2},\pm 1,\pm \frac{5}{4},\pm \frac{5}{2},\pm 5$

Remove duplicates:

  $\pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20,\pm \frac{1}{2},\pm 2,\pm \frac{5}{2},\pm \frac{1}{4},\pm \frac{5}{4}$

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

  $\begin{array}{l}1\overline{)\begin{array}{lllll}4\hfill & 0\hfill & -5\hfill & 42\hfill & -20\hfill \\ \hfill & 4\hfill & 4\hfill & -1\hfill & 41\hfill \end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 4& \hfill 4& \hfill -1& \hfill 41& \hfill 21\end{array}\end{array}$

So, $f\left(1\right)=21$ , hence $1$ is not a zero of this function.

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

  $\begin{array}{l}\frac{1}{2}\overline{)\begin{array}{lllll}4\hfill & 0\hfill & -5\hfill & 42\hfill & -20\hfill \\ \hfill & 2\hfill & 2\hfill & -2\hfill & 20\hfill \end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 4& \hfill 2& \hfill -4& \hfill 40& \hfill 0\end{array}\end{array}$

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

  $f(x)=\left(x-\frac{1}{2}\right)\left(4x^3+2x^2-4x+40\right)$

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

Define $g(x)=4x^3+2x^2-4x+40$ .

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

  $\pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20,\pm \frac{1}{2},\pm 2,\pm \frac{5}{2},\pm \frac{1}{4},\pm \frac{5}{4}$

Try if $-\frac{5}{2}$ is a zero of $g(x)$ :

  $-\frac{5}{2}\begin{array}{rrrr}\hfill 4& \hfill 2& \hfill -4& \hfill 40\\ \hfill & \hfill -10& \hfill 20& \hfill -40\\ \hfill 4& \hfill -8& \hfill 16& \hfill 0\end{array}$

So, $g\left(-\frac{5}{2}\right)=0$ , hence $-\frac{5}{2}$ is a zero of $g(x)$ ! Also, it is a zero of $f(x)$ .

Furthermore,

  $g(x)=\left(x+\frac{5}{2}\right)\left(4x^2-8x+16\right),$

and

  $f(x)=\left(x-\frac{1}{2}\right)\left(x+\frac{5}{2}\right)\left(4x^2-8x+16\right)$

To find other zeros and solve the equation

  $4x^2-8x+16=0$

The solution of the equation

  $a{x}^2+bx+c=0$ is

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here $a=4,b=-8$ and $c=16$ , so the solution is

  $\begin{array}{c}x=\frac{-(-8)\pm \sqrt{{(-8)}^2-4(4)(16)}}{2(4)}\\ =\frac{8\pm \sqrt{84-256}}{8}\\ =\frac{8\pm \sqrt{-172}}{8}\end{array}$

However, these are imaginary numbers!

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

Thus, the only real zero of this function is $\frac{1}{2},-\frac{5}{2}$ .