Chapter 2.2, Problem 58E

To determine

To find: The zeros and their multiplicities of given polynomial function, and verify the results with graphing utility.

Answer to Problem 58E

The zeros of polynomial function $f\left(x\right)=\frac{5}{3}{x}^2+\frac{8}{3}x-\frac{4}{3}$ are $x=-2$ and $x=0.4$ each of multiplicity 1. Theresults are verified with the graph.

Explanation of Solution

Given:

The polynomial function is $f\left(x\right)=\frac{5}{3}{x}^2+\frac{8}{3}x-\frac{4}{3}$ .

Concept used:

The zeros of a polynomial function $f\left(x\right)$ are the values of x for which $f\left(x\right)=0$ . If monomial $\left(x-a\right),\left(x-b\right),\left(x-c\right)\dots$ , are the factors of function then $x=a,x=b,\dots$ , are the zeros of $f\left(x\right)$ . For a polynomial function, a factor of ${\left(x-a\right)}^m$

  $m>0$ , yields a repeated zero

  $x=a$ , of multiplicity m.

Calculations:

The zeros of a polynomial function $f\left(x\right)=\frac{5}{3}{x}^2+\frac{8}{3}x-\frac{4}{3}$ may be written as $f\left(x\right)=\frac{1}{3}\left(5x^2+8x-4\right)$

  $\Rightarrow f\left(x\right)=\frac{1}{3}\left(5x^2+10x-2x-4\right)$

  $\Rightarrow f\left(x\right)=\frac{1}{3}\left(5\left(x+2\right)x-2\left(x+2\right)\right)$

  $\Rightarrow f\left(x\right)=\frac{1}{3}{\left(x+2\right)}^1{\left(5x-2\right)}^1$

The zeros of function are values of x for which $f\left(x\right)=0$

  $\Rightarrow \frac{1}{3}\left(x+2\right)\left(5x-2\right)=0$

  $\Rightarrow \left(x+2\right)\left(5x-2\right)=0$...............(1)

  $\left(x+2\right)=0\Rightarrow x=2,\left(5x-2\right)=0\Rightarrow x=\frac{2}{5}=0.4$

  $\Rightarrow x=-5.541,x=0.541$

Thus, the function $f\left(x\right)=\left(x+2\right)\left(5x-2\right)$

Thus, $x=-2$ and $x=0.4$ are zeros of function $f\left(x\right)=\frac{5}{3}{x}^2+\frac{8}{3}x-\frac{4}{3}$ each of multiplicity1.

Now, we verify the results with graph:

The graph of function $f\left(x\right)=\frac{5}{3}{x}^2+\frac{8}{3}x-\frac{4}{3}$ is shown in Figure-1 here.

  

The graph of function $f\left(x\right)=\frac{5}{3}{x}^2+\frac{8}{3}x-\frac{4}{3}$ intersects x-axis at points $\left(-2,0\right)$ and $\left(0.4,0\right)$ . Thus, we have $f\left(-2\right)=0$ , and $f\left(0.4\right)=0$ .

Hence, $x=-2$ and $x=0.4$ are zero of function $f\left(x\right)=\frac{5}{3}{x}^2+\frac{8}{3}x-\frac{4}{3}$ each of multiplicity 1.

Conclusion:

The zeros of polynomial function $f\left(x\right)=\frac{5}{3}{x}^2+\frac{8}{3}x-\frac{4}{3}$ are $x=-2$ and $x=0.4$ each of multiplicity 1. Theresultsare verified with the graph.