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: 15

Answer

$p=\left\{ -4,0 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ 4p^2+16p=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} 4p(p+4)=0 .\end{array} Equating each factor to zero (Zero-Factor Property), then \begin{array}{l}\require{cancel} 4p=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} 4p=0 \\\\ p=\dfrac{0}{4} \\\\ p=0 \\\\\text{ OR }\\\\ p+4=0 \\\\ p=-4 .\end{array} Hence, the solutions are $ p=\left\{ -4,0 \right\} .$ $\bf{\text{Supplementary Solution/s:}}$ In the expression $ 4p^2+16p ,$ the $GCF$ of the constants of the terms $\{ 4, 16 \}$ is $ 4 .$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{ p^2,p \}$ is $ p .$ Hence, the entire expression has $GCF= 4p .$ Factoring the $GCF= 4p ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 4p \left( \dfrac{4p^2}{4p}+\dfrac{16p}{4p} \right) .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} 4p \left( p^{2-1}+4p^{1-1} \right) \\\\= 4p \left( p^{1}+4p^{0} \right) \\\\= 4p \left( p+4(1) \right) \\\\= 4p \left( p+4 \right) .\end{array}
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