Chapter 3, Problem 66RE

To determine

To find:the rational and irrational zeros of the polynomial.

Answer to Problem 66RE

the zeros of the polynomial is $-3,3,3i$ (multiplicities 2).

Explanation of Solution

Given information:

the polynomial is $P\left(x\right)=x^4-81$

concept used:

rational zero theorem,

  $\text{possible rational zero of }P=\frac{\text{factor of constant term}}{\text{factor of leading coefficient}}$

Quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

The constant term is $81$ and the leading coefficient is 1 so

  $\text{possible rational zero of }P=\frac{\text{factor of 81}}{\text{factor of 1}}$

The factors of 81are $\pm 1,\pm 3,\pm 9,\pm 27,\pm 81$ and the factors of 1 are $\pm 1$ .

Thus, the possible rational zeros of $P$ are,

  $\pm \frac{1}{1},\pm \frac{3}{1},\pm \frac{9}{1},\pm \frac{27}{1},\pm \frac{81}{1}$

the possible zeros of $P$ are $\pm 1,\pm 3,\pm 9,\pm 27,\pm 81$ .

Using synthetic division,

-3	1	0	0	0	-81
		-3	9	-27	81
	1	-3	9	-27	0

3	1	0	0	0	-81
		3	9	27	81
	1	3	9	27	0

Factors,

  $x^4-81=\left(x+3\right)\left(x-3\right)\left(x^2+9\right)$

  $\begin{array}{l}x=\frac{-0\pm \sqrt{0^2-4·1·9}}{2·1}\\ x=\frac{\sqrt{-36}}{2}\text{ [simplify]}\\ x=3i\text{ [substitute }\sqrt{-1}=i]\end{array}$

The zeros of the polynomial is $-3,3,3i$ .