Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 4 - Polynomials - 4.6 Special Products - 4.6 Exercise Set: 115

Answer

$x=-7$

Work Step by Step

RECALL: $(a+b)(c+d)=ac+ad+bc+bd$ (more popularly known as the FOIL method) Use the FOIL method to obtain: $(x+2)(x-5)=(x+1)(x-3) \\x(x) + x(-5)+2(x) + 2(-5)=x(x)+x(-3)+1(x)+(1)(-3) \\x^2-5x+2x-10=x^2-3x+x-3 \\x^2-3x-10=x^2-2x-3$ Put all the terms with a variable on the left side and the constants on the right side. Note that when you move a term from one side of the equation to the other, it changes to its opposite. $x^2-3x-x^2+2x=-3+10 \\(x^2-x^2)+(-3x+2x)=7 \\0+(-x)=7 \\-x=7 \\-1(-x)=-1(7) \\x=-7$
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