Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 4 - Quadratic Functions and Equations - 4-8 Complex Numbers - Practice and Problem-Solving Exercises - Page 253: 34



Work Step by Step

The given expression, $ x^2+1 ,$ is equivalent to \begin{align*} & x^2-(-1) .\end{align*} Since $i^2=-1,$ the expression above is equivalent to \begin{align*} & x^2-i^2 .\end{align*} Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of $2$ squares, the expression above is equivalent to \begin{align*} & (x)^2-(i)^2 \\&= (x+i)(x-i) .\end{align*} Hence, the factored form of the given expression is $ (x+i)(x-i) .$ Checking: Multiplying the factors $ (x+i)$ and $(x-i) $ results to \begin{align*} & (x+i)(x-i) \\&= (x)^2-(i^2) &\left(\text{use }(a+b)(a-b)=a^2-b^2\right) \\&= x^2-i^2 \\&= x^2-(-1) &\left(\text{use }i^2=-1\right) \\&= x^2+1 \text{ (same as original expression)} .\end{align*}
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