Chapter 2.2, Problem 52E

To determine

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

Answer to Problem 52E

The polynomial function $f\left(x\right)=x^2+10x+25$ has a repeated zero $x=-5$ of multiplicity 2. The result is verified with graph.

Explanation of Solution

Given:

The polynomial function is $f\left(x\right)=x^2+10x+25$ .

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 $f\left(x\right)=x^2+10x+25$ may be written as

  $f\left(x\right)=x^2+10x+25$

  $\Rightarrow f\left(x\right)={\left(x+5\right)}^2$$\because {\left(a+b\right)}^2=a^2+2\cdot a\cdot b+b^2$

Thus, $x=-5$ is the zero of function $f\left(x\right)=x^2+10x+25$ of multiplicity 2,

Now, we verify the result with graph:

The graph of function $f\left(x\right)=x^2+10x+25$ is shown in Figure-1 here.

  

The graph of function $f\left(x\right)=x^2+10x+25$ touches x-axis at points $\left(-5,0\right)$ where two x- intercepts of $f\left(x\right)$ are equal. Thus, we have $f\left(-5\right)=0$ and $f\left(-5\right)=0$ .

Hence, $x=-5$ is the zeros of polynomial function is $f\left(x\right)=x^2+10x+25$ of multiplicity 2.

Conclusion:

The polynomial function $f\left(x\right)=x^2+10x+25$ has a repeated zero $x=-5$ of multiplicity 2. The result is verified with graph.