Algebra 1

Published by Prentice Hall

Chapter 10 - Radical Expressions and Equations - 10-2 Simplifying Radicals - Practice and Problem-Solving Exercises - Page 610: 29

Answer

$16y^{3}$

Work Step by Step

We first combine this into one square root: $\sqrt {2y \times 128 y^{5}} =\sqrt {256 y^{6}}$ We now separate the number and the variable into two separate square roots: $\sqrt {256} \times \sqrt {y^{6}} =\sqrt {256} \times y^{3}$ We see if any of the factors of a radical are perfect squares (meaning that their square root will be an integer) to see if the radical is in its most simplified form. We see that 256 is a perfect square, so we know that we can simplify: $\sqrt {256} \times y^{3}= 16y^{3}$

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