Chapter 2.5, Problem 32E

To determine

Find the zeros of the function $f(x)=3x^3-2x^2+75x-50$ and write the polynomial as a product of linear factors and verify zeros using graphical method.

Answer to Problem 32E

  $\frac{2}{3},-5i,5i$

Explanation of Solution

Given:

Function: $f(x)=3x^3-2x^2+75x-50$

Calculation:

The leading coefficient of given polynomial is 3 and the constant term is -50.

So, possible rational zeros are:

  $\frac{\text{Factors of}-50}{\text{Factors of 3}}=\frac{\pm 1,\pm 2,\pm 5,\pm 10,\pm 25,\pm 50}{\pm 1,\pm 3}=\pm 1,\pm 2,\pm 5,\pm 10,\pm 25,\pm 50$

  $\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{5}{3},\pm \frac{10}{3},\pm \frac{25}{3},\pm \frac{50}{3}$

Using synthetic division to check whether $x=\frac{2}{3}$ , is a zero or not.

  $\begin{array}{ccccc}\frac{2}{3}& 3& -2& 75& -50\\ & & 2& 0& 50\\ & 3& 0& 75& \boxed{0}\end{array}\begin{array}{c}\\ \\ \hspace{1em}\leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, $x=\frac{2}{3}$ is a zero of the given polynomial.

So, $f(x)$ factors as

  $f(x)=\left(x-\frac{2}{3}\right)(3x^2+75)$

Now, consider $3x^2+75.$

The above polynomial is a second degree polynomial.

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

  $\begin{array}{l}\Rightarrow x=\frac{-(0)\pm \sqrt{{(0)}^2-4(3)(75)}}{2(3)}\\ \Rightarrow x=\frac{\pm \sqrt{0-900}}{6}\\ \Rightarrow x=\frac{\pm \sqrt{-900}}{6}\\ \Rightarrow x=\frac{\pm 30\sqrt{-1}}{6}\\ \Rightarrow x=\pm 5i\\ \Rightarrow x=5i,-5i\end{array}$

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

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

Calculation for graph:

Consider $f(x)=3x^3-2x^2+75x-50$

Values of x	Values of f (x)
0	-50
1	26
-1	-130
2	116
-2	-232

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

  $x=\frac{2}{3}.$

Hence, the real zeros of the function are $x=\frac{2}{3}.$

Conclusion:

Therefore, the zeros of given function are $x=\frac{2}{3},-5i\text{and}5i.$