Chapter 2.2, Problem 49E

To determine

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

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Answer to Problem 49E

The zeros of the polynomial function $f\left(x\right)=x^2-25$ are $x=5$ of multiplicity 1 and $x=-5$ also of multiplicity 1. The results are verified with graph.

Explanation of Solution

Given:

The polynomial function is $f\left(x\right)=x^2-25$ .

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 $f\left(x\right)=x^2-25$ may be written as

  $f\left(x\right)=x^2-25$

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

Now, ${\left(x-5\right)}^1=0\Rightarrow x=5$ , i.e., $x=5$ is the zero of function $f\left(x\right)$ of multiplicity 1.

  ${\left(x+5\right)}^1=0\Rightarrow x=-5$ , i.e., $x=5$ is the zero of function $f\left(x\right)$ of multiplicity 1.

The graph of function $f\left(x\right)=x^2-25$ is shown in Figure-1 here.

  

The graph of function $f\left(x\right)=x^2-25$intersectsx-axis at points $\text{A}\left(-5,0\right)$ and $\text{B}\left(\text{5,0}\right)$ . Thus, we have $f\left(-5\right)=0$ and $f\left(5\right)=0$ .

Hence, $x=-5$ and $x=5$ are the zeros of polynomial function is $f\left(x\right)=x^2-25$ .

Conclusion:

The zeros of the polynomial function $f\left(x\right)=x^2-25$ are $x=5$ of multiplicity 1 and $x=-5$ also of multiplicity 1. The results are verified with graph.