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.4: 89

Answer

$4x^2-28x+49$

Work Step by Step

$$\frac{(2x-7)^5}{(2x-7)^3}$$ When dividing two exponential expressions with the same non-zero base, subtract the exponent of the denominator from the exponent of the numerator. Note: $2x-7\ne0$. $$=(2x-7)^{(5-3)}$$ $$=(2x-7)^2$$ To square a binomial, use the formula: $(a-b)^2=a^2-2ab+b^2$. $$=(2x)^2-2(2x)(7)+7^2$$ $$=4x^2-28x+49$$
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