Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 0 - Section 0.3 - Multiplying and Factoring Algebraic Expressions - Exercises: 20

Answer

$x^{2}+y^{2}+2xy-2x^{2}y-2xy^{2}+x^{2}y^{2}$

Work Step by Step

We can rewrite this as a special formula (square of a difference) $[(x+y)-xy]^{2}=...$ $(I-II)^{2}= I^{2}-2\cdot I\cdot II+II^{2}$, where $I=(x+y),\quad II=xy$ $I^{2}=(x+y)^{2}$=special f.$=x^{2}+2xy+y^{2}$ $2\cdot I\cdot II=2(x+y)xy=2xy(x+y)=2x^{2}y+2xy^{2}$ $II^{2}=(xy)^{2}=x^{2}y^{2}$ $...=x^{2}+2xy+y^{2}-(2x^{2}y+2xy^{2})+x^{2}y^{2}$ $=x^{2}+2xy+y^{2}-2x^{2}y-2xy^{2}+x^{2}y^{2}$ ... look for like terms to add ... none
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