Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 6 - Radical Functions and Rational Exponents - 6-4 Rational Exponents - Practice and Problem-Solving Exercises - Page 388: 111

Answer

$(4a + 3b)(4a - 3b)$

Work Step by Step

We can solve this quadratic expression by factoring. We see that this expression can be factored using the formula to factor the difference of two squares. According to this formula: $x^2 - y^2 = (x + y)(x - y)$ In the polynomial $16a^2 - 9b^2$, $x$ is the $\sqrt {16a^2}$ or $4a$, and $b$ is $\sqrt {9b^2}$ or $3b$. We plug these values into the formula: $(4a + 3b)(4a - 3b)$
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