## College Algebra (10th Edition)

Published by Pearson

# Chapter 5 - Section 5.6 - Complex Numbers; Quadratic Equations in the Complex Number System - 5.6 Assess Your Understanding: 31

#### Answer

The zeros of the given function are: $\color{blue}{\left\{1, \dfrac{-1}{2} - \dfrac{\sqrt3i}{2}, \dfrac{-1}{2} + \dfrac{\sqrt3i}{2}\right\}}$ The completely factored form of the given function is: $\\\color{blue}{P(x) = (x-1)\left(x+\frac{1}{2} + \frac{\sqrt3i}{2}\right)\left(x+\frac{1}{2} - \frac{\sqrt3i}{2}\right)}$

#### Work Step by Step

RECALL: (1) $a^3-b^3=(a-b)(a^2+ab+b^2)$ (2) The quadratic formula: $x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$ The given polynomial function can be written as $P(x) = x^3-1^3$. Factor the difference of two cubes using the formula in (1) above with $a=x$ and $b=1$ to obtain: $P(x) = (x-1)(x^2+x(1) + 1^2) \\P(x) = (x-1)(x^2+x+1)$ Equate each factor to zero then solve each equation to obtain: \begin{array}{ccc} &x-1=0 &\text{or} &x^2+x+1=0 \\&x=1 &\text{or} &x^2+x+1=0$\end{array} Solve the second equation using the formula in (2) above with$a=1, b=1$, and$c=1$to obtain:$x=\dfrac{-1 \pm \sqrt{1^2-4(1)(1)}}{2(1)} \\x=\dfrac{-1\pm\sqrt{1-4}}{2} \\x=\dfrac{-1\pm\sqrt{-3}}{2} \\x=\dfrac{-1\pm \sqrt{3(-1)}}{2} \\x=\dfrac{-1\pm \sqrt3 i}{2} \\x_1=\dfrac{-1}{2} - \dfrac{\sqrt3i}{2} \\x_2=\dfrac{-1}{2} + \dfrac{\sqrt3i}{2}$Thus, the zeros of the given function are:$\color{blue}{\left\{1, \dfrac{-1}{2} - \dfrac{\sqrt3i}{2}, \dfrac{-1}{2} + \dfrac{\sqrt3i}{2}\right\}}$The completely factored form of the given function is:$P(x) = \left(x-1\right)\left[x-\left(\frac{-1}{2} - \frac{\sqrt3i}{2}\right)\right]\left[x-\left(\frac{-1}{2} + \frac{\sqrt3i}{2}\right)\right] \\\color{blue}{P(x) = (x-1)\left(x+\frac{1}{2} + \frac{\sqrt3i}{2}\right)\left(x+\frac{1}{2} - \frac{\sqrt3i}{2}\right)}\$

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