Chapter 2.6, Problem 20E

Solution

To find: all real zeros of the function.

The real zeros of the function $\text{-1,1}\text{.5,2}\text{.5}$ .

Given information: 

The function $f\left(x\right)=4x^3-12x^2-x+15$ .

  

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

  $f\left(x\right)=4x^3-12x^2-x+15$

The constant term is $15$ , and its factors are

  $\pm 1,\pm 3,\pm 5,\pm 15$

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

  $\pm 1,\pm 2,\pm 4$

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

factors of $4$ ):

  $\pm \frac{1}{1},\pm \frac{3}{1},\pm \frac{5}{1},\pm \frac{15}{1},\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{15}{2},\pm \frac{1}{4},\pm \frac{3}{4},\pm \frac{5}{4},\pm \frac{15}{4}$

Simplify all fractions:

  $\pm 1,\pm 3,\pm 5,\pm 15,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{15}{2},\pm \frac{1}{4},\pm \frac{3}{4},\pm \frac{5}{4},\pm \frac{15}{4}$

From the graph, we see that all zeros are between $-1$ and $3$ . So, all possible rational zeros are:

  $-1,-\frac{3}{4},-\frac{1}{2},-\frac{1}{4},\frac{1}{4},\frac{1}{2},\frac{3}{4},1,\frac{5}{4},\frac{3}{2},\frac{5}{2},3$

Also, from the graph, we see that $-1$ might be a good candidate to test first.

Use synthetic substitution:

  $\begin{array}{ccccc}-1& 4& -12& -1& 15\\ & & -4& 16& -15\\ & 4& -16& 15& 0\end{array}$

So, $f(-1)=0$ , so $-1$ is indeed a zero of this function. Also,

  $f\left(x\right)=\left(x+1\right)\left(4x^2-16x+15\right)$

(Read the coefficients of the quotient from the final row.)

To find other zeros, solve the equation

  $4x^2-16x+15=0$

The solution of the equation

  $a{x}^2+bx+c=0$ is

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here $a=4,b=-16$ and $c=15$ , so

  $\begin{array}{c}x=\frac{-(-16)\pm \sqrt{{(-16)}^2-4(4)(15)}}{2(4)}\\ =\frac{16\pm \sqrt{256-240}}{8}\\ =\frac{16\pm \sqrt{16}}{8}\\ =\frac{16\pm 4}{8}\\ =\frac{4\pm 1}{2}\end{array}$

So, other zeros are

  $\begin{array}{c}x=\frac{4+1}{2}\\ =2.5\end{array}$ and $\begin{array}{c}x=\frac{4-1}{2}\\ =1.5\end{array}$ .