Chapter 2, Problem 92CR

To determine

Find the zeros of the given function.

Explanation of Solution

Given:

The given function is,

  $f\left(x\right)=2x^3-5x^2-9x+40$

Method:

By using factorization method and quadratic formula.

Calculation:

Consider,

  $f\left(x\right)=2x^3-5x^2-9x+40$

To find zeros of given function, put $f\left(x\right)=0$

  $\begin{array}{l}2x^3-5x^2-9x+40=0\\ \Rightarrow \left(x+\frac{5}{2}\right)\left(2x^2-10x+16\right)=0\\ \Rightarrow \left(x+\frac{5}{2}\right)=0\text{and}\left(2x^2-10x+16\right)=0\end{array}$

Consider,

  $\begin{array}{l}\left(x+\frac{5}{2}\right)=0\\ \Rightarrow x=-\frac{5}{2}\end{array}$

Now Consider,

  $\begin{array}{l}\left(2x^2-10x+16\right)=0\\ \end{array}$

By using quadratic formula,

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \Rightarrow x=\frac{-\left(-10\right)\pm \sqrt{{\left(-10\right)}^2-4\times 2\times 16}}{2\times 2}\\ \Rightarrow x=2.5\pm 1.3228i\end{array}$

Hence, the roots of given function are $x=-\frac{5}{2}$ and $x=2.5\pm 1.3228i$ .

Now the polynomial function can be written as a product of linear factors as follows:

  $x^3-7x^2+18x-24$ = $\left(x+\frac{5}{2}\right)\left[\left(x-\left(2.5-1.3228i\right)\right)\left(x+\left(2.5-1.3228i\right)\right)\right]$

The Graph of the given function is given below:

  

From the graph we can clearly see that $x=-\frac{5}{2}$ is a real root of the given function.