Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.5 - Solving Equations by the Zero-Factor Property - 5.5 Exercises - Page 355: 19


$p=\left\{ -4,4 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ 4p^2-16=0 ,$ in factored form. Then, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation. $\bf{\text{Solution Details:}}$ The factored form of the equation above is \begin{array}{l}\require{cancel} 4(p^2-16)=0 \\\\ 4(p+4)(p-4)=0 \\\\ \dfrac{4(p+4)(p-4)}{4}=\dfrac{0}{4} \\\\ (p+4)(p-4)=0 .\end{array} Equating each factor to zero (Zero-Factor Property), then \begin{array}{l}\require{cancel} p+4=0 \text{ OR } p-4=0 .\end{array} Using the properties of equality to solve each of the equation above results to \begin{array}{l}\require{cancel} p+4=0 \\\\ p=-4 \\\\\text{ OR }\\\\ p-4=0 \\\\ p=4 .\end{array} Hence, the solutions are $ p=\left\{ -4,4 \right\} .$ $\bf{\text{Supplementary Solution/s:}}$ Factoring the $GCF= 4 ,$ the expression, $ 4p^2-16 ,$ above is equivalent to \begin{array}{l}\require{cancel} 4 \left( \dfrac{4p^2}{4}-\dfrac{16}{4} \right) \\\\= 4 \left( p^2-4 \right) .\end{array} Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} 4(p+4)(p-4) .\end{array}
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