Chapter 3.6, Problem 13E

a.

To determine

To find: the zeroes of polynomial $P\left(x\right)=x^3+8$

Answer to Problem 13E

The zeroes of polynomial $P\left(x\right)=x^3+8$ is $-2,1-\sqrt{3i},1+\sqrt{3i}$

Explanation of Solution

Given information: The polynomial is $P\left(x\right)=x^3+8$

Calculation: For the zeroes of $P\left(x\right)$ solve the equation $P\left(x\right)=0$ for $x$

  $\begin{array}{l}P\left(x\right)=0\\ \Rightarrow x^3+8=0\\ \Rightarrow x^3+2^3=0\\ \left(x+2\right)\left(x^2-2x+2^2\right)=0\\ \left(x+2\right)\left(x^2-2x+4\right)=0\\ x+2=0\\ x=-2\\ x^2-2x+4=0\\ x^2-2x=-4\\ x^2-2x+1=-4+1\\ {\left(x-1\right)}^2=-3\\ \sqrt{{\left(x-1\right)}^2}=\sqrt{-3}\\ \left|{\left(x-1\right)}^2\right|=i\sqrt{3}\\ x-1=\pm \sqrt{3i}\\ x=1\pm \sqrt{3i}\end{array}$

The zeroes of the polynomial function $P\left(x\right)$ are $-2,1-\sqrt{3i},1+\sqrt{3i}$

b.

To determine

To calculate: the factored form of polynomial $P\left(x\right)=x^3+8$

Answer to Problem 13E

The factored form of polynomial $P\left(x\right)=x^3+8$ is $P\left(x\right)=\left(x+2\right),\left(x-1+\sqrt{3i}\right),\left(x-1-\sqrt{3i}\right)$

Explanation of Solution

Given information: The polynomial is $P\left(x\right)=x^3+8$

Calculation: According to the factor theorem if $c$ is a zero of a polynomial function $P\left(x\right)$ the binomial $\left(x-c\right)$ is a factor of this polynomial

The zeroes of the polynomial is expressed as

  $P\left(x\right)=\left(x+2\right),\left(x-1+\sqrt{3i}\right),\left(x-1-\sqrt{3i}\right)$

Hence,

The factored form of polynomial $P\left(x\right)$ is

  $P\left(x\right)=\left(x+2\right)\left(x-1+\sqrt{3i}\right)\left(x-1-\sqrt{3i}\right)$