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: 35

Answer

$3(s+5i)(s-5i)$

Work Step by Step

The given expression, $ 3s^2+75 ,$ is equivalent to \begin{align*} & 3(s^2+25) \\&= 3[s^2-(-25)] \\&= 3[s^2-(-1\cdot25)] .\end{align*} Since $i^2=-1,$ the expression above is equivalent to \begin{align*} & 3[s^2-(i^2\cdot25)] \\&= 3[s^2-(25i^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*} & 3[(s)^2-(5i)^2] \\&= 3[(s+5i)(s-5i)] \\&= 3(s+5i)(s-5i) .\end{align*} Hence, the factored form of the given expression is $ 3(s+5i)(s-5i) .$ Checking: Multiplying the factors above results to \begin{align*} & 3[(s+5i)(s-5i)] \\&= 3[(s)^2-(5i)^2] &\left(\text{use }(a+b)(a-b)=a^2-b^2\right) \\&= 3(s^2-25i^2) \\&= 3(s^2)-3(25i^2) &\left(\text{use Distributive Property}\right) \\&= 3s^2-75i^2 \\&= 3s^2-75(-1) &\left(\text{use }i^2=-1\right) \\&= 3s^2+75 \text{ (same as original expression)} .\end{align*}
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