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: 13

Answer

$(3x+1)(5x-4)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ 15x^2-7x=4 ,$ 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} 15x^2-7x-4=0 \\\\ (3x+1)(5x-4)=0 .\end{array} Equating each factor to zero (Zero-Factor Property), then \begin{array}{l}\require{cancel} 3x+1=0 \text{ OR } 5x-4=0 .\end{array} Using the properties of equality to solve each of the equation above results to \begin{array}{l}\require{cancel} 3x+1=0 \\\\ 3x=-1 \\\\ x=-\dfrac{1}{3} \\\\\text{ OR }\\\\ 5x-4=0 \\\\ 5x=4 \\\\ x=\dfrac{4}{5} .\end{array} Hence, the solutions are $ x=\left\{ -\dfrac{1}{3},\dfrac{4}{5} \right\} .$ $\bf{\text{Supplementary Solution:}}$ To factor the expression, $ 15x^2-7x-4 ,$ find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping. In the expression above the value of $ac$ is $ 15(-4)=-60 $ and the value of $b$ is $ -7 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-60\}, \{2,-30\}, \{3,-30\}, \{4,-15\}, \{5,-12\}, \{6,-10\}, \{-1,60\}, \{-2,30\}, \{-3,30\}, \{-4,15\}, \{-5,12\}, \{-6,10\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 5,-12 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 15x^2+5x-12x-4 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (15x^2+5x)-(12x+4) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 5x(3x+1)-4(3x+1) .\end{array} Factoring the $GCF= (3x+1) $ of the entire expression above results to \begin{array}{l}\require{cancel} (3x+1)(5x-4) .\end{array}
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