Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 5 - Section 5.5 - Factoring Special Forms - Exercise Set - Page 372: 96

Answer

$(0.3x-0.2)^2$

Work Step by Step

The formula for factoring the difference of two cubes is: $A^3-B^3=(A-B)(A^2+AB+B^2)$. The square of a binomial sum is: $(A+B)^2=A^2+2AB+B^2$. The square of a binomial difference is: $(A-B)^2=A^2-2AB+B^2$. Hence here: $0.09x^2-0.12x+0.04=(0.3x)^2-2\cdot0.3x\cdot0.2+(0.2)^2=(0.3x-0.2)^2$
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