Chapter 3.6, Problem 48E

To determine

To find: the zeroes of the polynomial

Answer to Problem 48E

The zeroes of polynomial $P\left(x\right)=x^3+7x^2+18x+18$ is $1,-2-\sqrt{2}i,-2+\sqrt{2}i$

Explanation of Solution

Given information: The polynomial is $P\left(x\right)=x^3+7x^2+18x+18$

Calculation: From the rational zeroes theorem the possible rational zeroes of $P\left(x\right)$ are $\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 9.\pm 18$

By substitution of $x$ by $-3$ in the form of $P$

  $\begin{array}{l}P\left(-3\right)=-3^2+7{\left(-3\right)}^2+18\left(-3\right)+18\\ \Rightarrow P\left(-3\right)=-27+63-54+18\\ \Rightarrow P\left(-3\right)=0\left[\text{ the value -3 is a zero of P}\left(x\right)\right]\end{array}$

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

Divide $P\left(x\right)$ by $\left(x+3\right)$

  $\frac{x^3+7x^2+18x+18}{x+3}=x^2+4x+6$

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

  $P\left(x\right)=\left(x+3\right)\left(x^2+4x+6\right)$

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

  $\begin{array}{l}P\left(x\right)=0\\ \Rightarrow \left(x+3\right)\left(x^2+4x+6\right)=0\\ x+3=0\Rightarrow \boxed{x=-3}\\ x^2+4x+6=0\\ x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x_{1,2}=\frac{-4\pm \sqrt{{\left(4\right)}^2-4\left(1\right)\left(6\right)}}{2\left(1\right)}\\ x_{1,2}=\frac{-4\pm \sqrt{-8}}{2}\\ x_{1,2}=\frac{-4\pm 2\sqrt{2}i}{2}\\ \Rightarrow \boxed{x=-2-\sqrt{2}i}\&\boxed{x=-2+\sqrt{2}i}\end{array}$