Chapter 2.2, Problem 39E

a.

To determine

To find: The zeros of polynomial function algebraically.

Answer to Problem 39E

The real zeros of the polynomialfunction $g\left(t\right)=\frac{1}{2}{t}^4-\frac{1}{2}$ are $t=-1$ and $t=1$ .

Explanation of Solution

Given:

The polynomial function is $g\left(t\right)=\frac{1}{2}{t}^4-\frac{1}{2}$ .

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 $\left(x-a\right)$ , $\left(x-b\right)$ , … are the factors of function then $x=a$ , $x=b$ …., are the zeros of $f\left(x\right)$ .

Calculations:

The zeros of a polynomial function $g\left(t\right)$ may be written as

  $g\left(t\right)=\frac{1}{2}{t}^4-\frac{1}{2}=\frac{1}{2}\left(t^4-1\right)$

The zeros of $g\left(t\right)$ are the values of t for which $g\left(t\right)=0$

  $\Rightarrow \frac{1}{2}\left(t^4-1\right)=0$

  $\Rightarrow \frac{1}{2}\left(t^2-1\right)\left(t^2+1\right)=0$

  $\Rightarrow \frac{1}{2}\left(t-1\right)\left(t+1\right)\left(t^2+1\right)=0$

Now, $\left(t-1\right)=0\Rightarrow t=1$ , $\left(t+1\right)=0\Rightarrow t=-1$ , and $\left(t^2+1\right)=0\Rightarrow t=\sqrt{-1}$ an imaginary number.

Thus, algebraically calculated zeros of the function $g\left(t\right)=\frac{1}{2}{t}^4-\frac{1}{2}$ are $t=-1$ and $t=1$ other two zeros are imaginary.

Conclusion:

The real zeros of the polynomialfunction $g\left(t\right)=\frac{1}{2}{t}^4-\frac{1}{2}$ are $t=-1$ and $t=1$ .