Chapter 3.6, Problem 50E

To determine

To find: the zeroes of the polynomial

Answer to Problem 50E

The zeroes of polynomial $P\left(x\right)=x^3-x-6$ is $2,-1-\sqrt{2}i,-1+\sqrt{2}i$

Explanation of Solution

Given information: The polynomial is $P\left(x\right)=x^3-x-6$

Calculation: From the rational zeroes theorem the possible rational zeroes of $P\left(x\right)$ are $\pm 1,\pm 2,\pm 3,\pm 6$

By substitution of $x$ by $2$ in the form of $P$

  $\begin{array}{l}\left(2\right)=2^3-\left(2\right)-6\\ \Rightarrow P\left(2\right)=8-8\\ \Rightarrow P\left(2\right)=0\left[\text{ the value 2 is a zero of P}\left(x\right)\right]\end{array}$

From the factor theorem if $3$ is a zero of $P\left(x\right)$ , the binomial $\left(x-2\right)$ is a factor of $P\left(x\right)$

Divide $P\left(x\right)$ by $\left(x-2\right)$

  $\frac{x^3-x-6}{x-2}=x^2+2x+3$

The polynomial function $P\left(x\right)$ can be written as

  $P\left(x\right)=\left(x-2\right)\left(x^2+2x+3\right)$

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

  $\begin{array}{l}P\left(x\right)=0\\ \Rightarrow \left(x-2\right)\left(x^2+2x+3\right)=0\\ x-2=0\Rightarrow \boxed{x=2}\\ x^2+2x+3=0\\ x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x_{1,2}=\frac{-2\pm \sqrt{{\left(2\right)}^2-4\left(1\right)\left(3\right)}}{2\left(1\right)}\\ x_{1,2}=\frac{-2\pm \sqrt{-8}}{2}\\ x_{1,2}=\frac{-2\pm 2\sqrt{2}i}{2}\\ \Rightarrow \boxed{x=-1-\sqrt{2}i}\&\boxed{x=-1+\sqrt{2}i}\end{array}$