College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.5 - Page 74: 88

Answer

16$a^{2}$$x$ - 25$y$ - 25$x$ + 16$a^2$$y$ = 16$a^{2}$$x$ - 25$x$ + 16$a^{2}$$y$ - 25$y$ = $x$(16$a^{2}$ - 25) + $y$(16$a^{2}$ - 25) = ($x$ + $y$)(16$a^{2}$ - 25) = ($x$ + $y$)(4$a$ + 5)(4$a$ - 5)

Work Step by Step

To factor this polynomial, first you must change the order of the terms so that the first group, 16a^2x - 25x, shares the x variable, and the second group, 16a^2y - 25y, shares the y variable, as shown in step 2. Then, factor out the x from the first group and the y from the second in step 3. From there, you can combine the numbers into two pairs in step 4. Then factor the second pair by using the formula A^2 - B^2 = (A+B)(A-B) in step 5.
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