Chapter 2.5, Problem 40E

a.

To determine

To find: The possible rational zeroes of the function $f\left(x\right)=4x^3+7x^2-11x-18$

Answer to Problem 40E

Possible rational zeroes of the function are $\pm 1,\pm \frac{1}{2},\pm \frac{1}{4},\pm 2,\pm 3,\pm \frac{3}{2},\pm \frac{3}{4},\pm 6,\pm 9,\pm \frac{9}{2},\pm \frac{9}{4},\pm 18$

Explanation of Solution

Given information:

Given function is $f\left(x\right)=4x^3+7x^2-11x-18$

Formula used:

Rational root test: According to this test, the list of all possible rational zeroes of a polynomial is,

Possible zeroes $=\frac{\text{Factor of constant term}}{\text{Factor of leading coefficient}}$

Consider the function as, $f\left(x\right)=4x^3+7x^2-11x-18$

Here, leading coefficient is $4$ and constant term is $-18$

Factors of constant term 4 are $\pm 1,\pm 2,\pm 4$

Factors of leading coefficient $-18$ are $\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18$

Then possible rational zeroes are,

  $\begin{array}{c}\text{Possible zeroes}=\frac{\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18}{\pm 1,\pm 2,\pm 4}\\ =\pm 1,\pm \frac{1}{2},\pm \frac{1}{4},\pm 2,\pm 3,\pm \frac{3}{2},\pm \frac{3}{4},\pm 6,\pm 9,\pm \frac{9}{2},\pm \frac{9}{4},\pm 18\end{array}$

Therefore, possible rational zeroes of the function are

  $\pm 1,\pm \frac{1}{2},\pm \frac{1}{4},\pm 2,\pm 3,\pm \frac{3}{2},\pm \frac{3}{4},\pm 6,\pm 9,\pm \frac{9}{2},\pm \frac{9}{4},\pm 18$

b.

To determine

To draw: The graph of the function $f\left(x\right)=4x^3+7x^2-11x-18$ so that some of the possible zeroes in part (a) can be disregarded.

Explanation of Solution

Given information:

Given function is $f\left(x\right)=4x^3+7x^2-11x-18$

Possible rational zeroes of the function are $\pm 1,\pm \frac{1}{2},\pm \frac{1}{4},\pm 2,\pm 3,\pm \frac{3}{2},\pm \frac{3}{4},\pm 6,\pm 9,\pm \frac{9}{2},\pm \frac{9}{4},\pm 18$

Consider the function as $f\left(x\right)=4x^3+7x^2-11x-18$

Graph of the function can be drawn as,

  

From the graph of the function, it is clear that zeroes of the function are $-2,-1.38,1.63$ .

c.

To determine

All the real zeroes of the function $f\left(x\right)=4x^3+7x^2-11x-18$ with the help of part (a) and part (b).

Answer to Problem 40E

Real zeroes of the function are $x=-2,-1.38,1.63$

Explanation of Solution

Given information:

Given function $f\left(x\right)=4x^3+7x^2-11x-18$

Possible rational zeroes of the function are $\pm 1,\pm \frac{1}{2},\pm \frac{1}{4},\pm 2,\pm 3,\pm \frac{3}{2},\pm \frac{3}{4},\pm 6,\pm 9,\pm \frac{9}{2},\pm \frac{9}{4},\pm 18$

Consider the given function as function $f\left(x\right)=4x^3+7x^2-11x-18$

From part (a), possible rational zeroes of the function are $\pm 1,\pm \frac{1}{2},\pm \frac{1}{4},\pm 2,\pm 3,\pm \frac{3}{2},\pm \frac{3}{4},\pm 6,\pm 9,\pm \frac{9}{2},\pm \frac{9}{4},\pm 18$

From part (b), it is clear is clear that the curve do not touch x − axis at $x=\pm 1,\pm \frac{1}{2},\pm \frac{1}{4},2,\pm 3,\pm \frac{3}{2},\pm \frac{3}{4},\pm 6,\pm 9,\pm \frac{9}{2},\pm \frac{9}{4},\pm 18$

Hence, real zeroes of the function are $x=-2,-1.38,1.63$