Chapter 2.3, Problem 92E

To determine

To find the real zeroes of the polynomial function $f(x)=4x^3+7x^2-11x-18$ .

Answer to Problem 92E

  $-2,\frac{1+\sqrt{145}}{8}\text{,}\frac{1-\sqrt{145}}{8}$

Explanation of Solution

Given:

Function: $f(x)=4x^3+7x^2-11x-18$

Calculation:

The leading coefficient of given polynomial is 4 and the constant term is -18.

So, possible zeroes are:

  $\frac{\text{Factors of}-18}{\text{Factors of 4}}=\frac{\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18}{\pm 1,\pm 2,\pm 4}=\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{9}{2},\pm \frac{1}{4},\pm \frac{3}{4},\pm \frac{9}{4}$

Using synthetic division to check whether x = -1, is a zero or not.

  $\begin{array}{ccccc}-1& 4& 7& -11& -18\\ & & -4& -3& 14\\ & 4& 3& -14& \boxed{-4}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

Here, remainder is not 0. So, x = -1 is not a zero of the given polynomial.

Using synthetic c division to check whether x = -2, is a zero or not.

  $\begin{array}{ccccc}-2& 4& 7& -11& -18\\ & & -8& 2& 18\\ & 4& -1& -9& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, x = -2 is a zero of the given polynomial.

So, $f(x)$ factors as

  $f(x)=(x+2)(4x^2-x-9)$

Now, consider $4x^2-x-9.$

The above polynomial is a second degree polynomial.

So, zeroes of a second degree polynomial can be found using the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$

  $\begin{array}{l}\Rightarrow x=\frac{-(-1)\pm \sqrt{{(-1)}^2-4(4)(-9)}}{2(4)}\\ \Rightarrow x=\frac{1\pm \sqrt{1+144}}{8}\\ \Rightarrow x=\frac{1\pm \sqrt{145}}{8}\end{array}$

Conclusion:

Therefore, the zeroes of given polynomial function are $x=-2,\frac{1+\sqrt{145}}{8}\text{and}\frac{1-\sqrt{145}}{8}.$