Chapter 2.5, Problem 31E

To determine

Find the zeros of the function $f(x)=3x^3-5x^2+48x-80$ and write the polynomial as a product of linear factors and verify zeros using graphical method.

Answer to Problem 31E

  $\frac{5}{3},-4i,4i$

Explanation of Solution

Given:

Function: $f(x)=3x^3-5x^2+48x-80$

Calculation:

The leading coefficient of given polynomial is 3 and the constant term is -80.

So, possible rational zeros are:

  $\frac{\text{Factors of}-80}{\text{Factors of 3}}=\frac{\pm 1,\pm 2,\pm 4,\pm 5,\pm 8,\pm 10,\pm 16,\pm 20,\pm 40,\pm 80}{\pm 1,\pm 3}=\pm 1,\pm 2,\pm 4$

  $\pm 5,\pm 8,\pm 10,\pm 16,\pm 20,\pm 40,\pm 80,\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{5}{3},\pm \frac{8}{3},\pm \frac{10}{3},\pm \frac{16}{3},\pm \frac{20}{3},\pm \frac{40}{3},\pm \frac{80}{3}$

Using synthetic division to check whether $x=\frac{5}{3}$ , is a zero or not.

  $\begin{array}{ccccc}\frac{5}{3}& 3& -5& 48& -80\\ & & 5& 0& 80\\ & 3& 0& 48& \boxed{0}\end{array}\begin{array}{c}\\ \\ \hspace{1em}\leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, $x=\frac{5}{3}$ is a zero of the given polynomial.

So, $f(x)$ factors as

  $f(x)=\left(x-\frac{5}{3}\right)(3x^2+48)$

Now, consider $3x^2+48.$

The above polynomial is a second degree polynomial.

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

  $\begin{array}{l}\Rightarrow x=\frac{-(0)\pm \sqrt{{(0)}^2-4(3)(48)}}{2(3)}\\ \Rightarrow x=\frac{\pm \sqrt{0-576}}{6}\\ \Rightarrow x=\frac{\pm \sqrt{-576}}{6}\\ \Rightarrow x=\frac{\pm 24\sqrt{-1}}{6}\\ \Rightarrow x=\pm 4i\\ \Rightarrow x=4i,-4i\end{array}$

Now, writing the given function as product of linear factors,

  $\Rightarrow f(x)=\left(x-\frac{5}{3}\right)\left(x-4i\right)\left(x+4i\right)$

Calculation for graph:

Consider $f(x)=3x^3-5x^2+48x-80$

Values of x	Values of f (x)
0	-80
1	-34
-1	-136
2	20
-2	-220

By taking different values of x, the graph can be plotted.

Graph:

  

Interpretation:

Only real zeros of the function can be verified through graph.

By observing graph, it is clear that the curve of the function meets x-axis at

  $x=\frac{5}{3}.$

Hence, the real zeros of the function are $x=\frac{5}{3}.$

Conclusion:

Therefore, the zeros of given function are $x=\frac{5}{3},-4i\text{and}4i.$