Intermediate Algebra (12th Edition)

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

Chapter 4 - Section 4.4 - Multiplying Polynomials - 4.4 Exercises: 101


$A=\dfrac{9x^2-4y^2}{2}\text{ square units}$

Work Step by Step

Using $A=\dfrac{1}{2}bh$, the area of the triangle with height equal to $ 3x-2y $ and base equal to $ 3x+2y ,$ is \begin{array}{l}\require{cancel} A=\dfrac{1}{2}(3x-2y)(3x+2y) .\end{array} Using $(a+b)(a-b)=a^2-b^2$ or the product of the sum and difference of like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} A=\dfrac{1}{2}[(3x)^2-(2y)^2] \\\\ A=\dfrac{1}{2}(9x^2-4y^2) \\\\ A=\dfrac{9x^2-4y^2}{2}\text{ square units} .\end{array}
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