Chapter 2, Problem 32RE

a

To determine

To calculate the zeros of the given polynomial using algebraic method

Answer to Problem 32RE

  $x=0,0,-2,2$

Explanation of Solution

Given Information:

  $f(t)=t^4-4t^2$

Formula:

The zeros of the quadratic equation of the form $a{x}^2+bx+c=0$ can be found by using the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

Consider the expression $t^4-4t^2$ and factor the common term.

  $t^4-4t^2=t^2(t^2-4)$

Let $f(t)=0$

  $\begin{array}{c}{t}^4-4t^2=0\\ t^2(t^2-4)=0\end{array}$

Now either $t^2=0$ or $(t^2-4)=0$

If $t^2=0$ then, $t=0$

If $(t^2-4)=0$ then solve the quadratic equation.

  $t^2-4=(t-2)(t+2)\text{ [}\because a^2-b^2=(a+b)(a-b)]$

Hence, $f(t)=t^4-4t^2=t^2(t-2)(t+2)$ .

Therefore, the zeros of the polynomial are $x=0,0,-2,2$ .

b

To determine

To sketch the graph of the given function

Explanation of Solution

Given information:

  $f(t)=t^4-4t^2$

Graph:

  

Interpretation:

Using a graphing utility, the graph of the function occurs as above.

c

To determine

To approximate the zeros of the given function using graph and comparing with the zeros in subpart (a)

Explanation of Solution

Given information:

  $f(t)=t^4-4t^2$

Concept used:

The zeros of the function are the points in which the curve of the function passes through the $x$ axis.

Graph:

  

Interpretation:

From the above figure,

The curve of the given function passes through three points $x=0,0,-2,2$ in the $x$ axis.

Conclusion:

Therefore, the zeros found algebraically in subpart (a) is exactly the same when found graphically.