Chapter 2.2, Problem 57E

To determine

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

Answer to Problem 57E

The zeros of polynomial function $f\left(x\right)=\frac{1}{2}{x}^2+\frac{5}{2}x-\frac{3}{2}$ are $x=-5.541$ and $x=0.541$ each of multiplicity 1. Theresults are verified with the graph.

Explanation of Solution

Given:

The polynomial function is $f\left(x\right)=\frac{1}{2}{x}^2+\frac{5}{2}x-\frac{3}{2}$ .

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 $f\left(x\right)=\frac{1}{2}{x}^2+\frac{5}{2}x-\frac{3}{2}$ may be written as $f\left(x\right)=\frac{1}{2}\left(x^2+5x-3\right)$ .

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

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

  $\Rightarrow x^2+5x-3=0$...............(1)

The quadratic equation (1) can be solved formula for equation $a{x}^2+bx+c=0$ that is

  $x=\frac{-5\pm \sqrt{5^2-4\times 1\times \left(-3\right)}}{2\times 1}=\frac{-5\pm \sqrt{37}}{2}$

  $\Rightarrow x=-5.541,x=0.541$

Thus, the function $f\left(x\right)={\left(x+5.551\right)}^1{\left(x-0.541\right)}^1$

Thus, $x=-5.541$ and $x=0.541$ are zeros of function $f\left(x\right)=\frac{1}{2}{x}^2+\frac{5}{2}x-\frac{3}{2}$ each of multiplicity1.

Now, we verify the results with graph:

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

  

The graph of function $f\left(x\right)=\frac{1}{2}{x}^2+\frac{5}{2}x-\frac{3}{2}$ intersects x-axis at points $\left(-5.541,0\right)$ and $\left(0.541,0\right)$ . Thus, we have $f\left(-5.541\right)=0$

  $f\left(0.541\right)=0$ .

Hence, $x=-5.541$ and $x=0.541$ are zero of function $f\left(x\right)=\frac{1}{2}{x}^2+\frac{5}{2}x-\frac{3}{2}$ each of multiplicity 1.

Conclusion:

The zeros of polynomial function $f\left(x\right)=\frac{1}{2}{x}^2+\frac{5}{2}x-\frac{3}{2}$ are $x=-5.541$ and $x=0.541$ each of multiplicity 1. Theresultsare verified with the graph.