Chapter 3, Problem 67RE

To determine

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

Answer to Problem 67RE

the zeros of the polynomial is $\pm 1\pm \sqrt{3}i,\pm 2$ .

Explanation of Solution

Given information:

the polynomial is $P\left(x\right)=x^6-64$

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 $64$ and the leading coefficient is 1 so

  $\text{possible rational zero of }P=\frac{\text{factor of 64}}{\text{factor of 1}}$

The factors of 64are $\pm 1,\pm 2,\pm 4,\pm 8,\pm 16,\pm 32,\pm 64$ and the factors of 1 are $\pm 1$ .

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

  $\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{4}{1},\pm \frac{8}{1},\pm \frac{16}{1},\pm \frac{32}{1},\pm \frac{64}{1}$

the possible zeros of $P$ are $\pm 1,\pm 2,\pm 4,\pm 8,\pm 16,\pm 32,\pm 64$ .

Using synthetic division,

-2	1	0	0	0	0	0	-64
		-2	4	-8	16	-32	64
	1	-2	4	-8	16	-32	0

2	1	0	0	0	0	0	-64
		2	4	8	16	32	64
	1	2	4	8	16	32	0

Factors,

  $x^6-64=\left(x+2\right)\left(x-2\right)\left(x^4+4x^2+16\right)$

  $\begin{array}{l}{x}^2=\frac{-4\pm \sqrt{4^2-4·1·16}}{2·1}\\ x^2=\frac{-4\pm \sqrt{16-64}}{2}\text{ [simplify]}\\ x^2=-2\pm 2\sqrt{3}i\text{ [substitute }\sqrt{-1}=i]\\ x=\pm 1\pm \sqrt{3}i\end{array}$

The zeros of the polynomial is $\pm 1\pm \sqrt{3}i,\pm 2$ .