Chapter 2.2, Problem 44E

a.

To determine

To Find zeros of a Polynomial function algebraically.

Explanation of Solution

Given:

The equation is $y(t)=t^5-6t^3+9t$

Approach:

1. Solve the Polynomial algebraically.
2. Find the values of t.

Calculation:

Consider,

  $t^5-6t^3+9t=0$

Putting the value of $y(t)=0$

  $t^5-6t^3+9t=0$

  $\begin{array}{l}{t}^5-3t^3-3t^3+9t=0\\ (t^2-3)(t^3-3t)=0\\ (t-\sqrt{3})(t+\sqrt{3})t(t-\sqrt{3})(t+\sqrt{3})=0\\ t=-\sqrt{3},\sqrt{3},\text{}0,\sqrt{3},-\sqrt{3}\end{array}$

Therefore, the real zeros of a polynomial function are $t=-\sqrt{3},\sqrt{3},\text{}0,\sqrt{3},-\sqrt{3}$

b.

To determine

Use a graphing utility to graph the function.

Explanation of Solution

Given:

The equation are given as, $y(t)=t^5-6t^3+9t$

Approach:

1. Draw the given equations on the graph.

Graph:

The graph of the above equations are given below

  

Since the above graph intersect x-axis at many points.

Therefore, the polynomial have many zeros.

c.

To determine

Use the graph to compare them with those from part (a).

Explanation of Solution

Given:

The equation is $y(t)=t^5-6t^3+9t$

Approach:

1. Draw the given equations on the graph.
2. Compare them with the part -a.

Graph:

The graph of the above equations are given below

  

This shows that the zeros of the polynomial are $t=-\sqrt{3},\sqrt{3},\text{}0,$

While here, all the zeros were represented by the graph actually crossing through the x-axis.