Chapter 2.2, Problem 51E

To determine

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

Answer to Problem 51E

The polynomial function $h\left(t\right)=t^2-6t+9$ has a repeated zero $x=3$ of multiplicity 2. The result is verified with graph.

Explanation of Solution

Given:

The polynomial function is $h\left(t\right)=t^2-6t+9$ .

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 $h\left(t\right)=t^2-6t+9$ may be written as

  $h\left(t\right)=t^2-2\cdot 3\cdot t+{\left(3\right)}^2$

  $\Rightarrow h\left(t\right)={\left(t-3\right)}^2$$\because {\left(a-b\right)}^2=a^2-2\cdot a\cdot b+b^2$

Thus, $x=3$ is the zero of function $h\left(t\right)=t^2-6t+9$ of multiplicity 2.

Now, we verify the result with graph:

The graph of function $h\left(t\right)=t^2-6t+9$ is shown in Figure-1 here.

  

The graph of function $h\left(t\right)=t^2-6t+9$ touches x-axis at points $\left(3,0\right)$ where two t- intercepts of $h\left(t\right)$ are equal. Thus, we have $f\left(3\right)=0$ and $f\left(3\right)=0$ .

Hence, $x=3$ is the zeros of polynomial function is $h\left(t\right)=t^2-6t+9$ of multiplicity 2.

Conclusion:

The polynomial function $h\left(t\right)=t^2-6t+9$ has a repeated zero $x=3$ of multiplicity 2. The result is verified with graph.