Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 8 - Polynomials and Factoring - 8-7 Factoring Special Cases - Practice and Problem-Solving Exercises: 50

Answer

(13)(7)

Work Step by Step

Given the number 91 we find two numbers that are perfect squares that add to give 91. We get +100 and -9 100 - 9 *** Take the square root of 100 which is 10. Becuase 10 × 10 = 100 *** Take the square root of 9 which is 3. Becuase 3 × 3 = 9 We rewrite it as the squares. $10^{2}$ - $3^{2}$ We use the formula for the difference of squares to apply to this question. The difference of squares formula is: $(a-b) (a+b) = a^{2} - b^{2}$ In the given formula let 10 represents a and 3 represents b. $10^{2}$ - $3^{2}$ = (10+3)(10-3) = (10+3)(10-3) = (13)(7)
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