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: 38

Answer

5(4g+3)(4g-3)

Work Step by Step

$80g^{2}$ - 45 We see that both the terms have a 5 common thus we factor a 5 out. 5($16g^{2}$ - 9) We use the formula for the difference of squares to apply to this question. The difference of squares formula is: $(a-b) (a+b) = a^{2} - b^{2}$ = 5($16g^{2}$ - 9) *** Take the square root of $16g^{2}$ which is 4g. Becuase 4g × 4g= $16g^{2}$ *** Take the square root of 9 which is 3. Becuase 3 × 3= 9 = 5($(4g)^{2}−3^{2}$) In the given formula let 4g represents a and 3 represents b. 5($(4g)^{2}−3^{2}$) = 5(4g+3)(4g-3)
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