Place under one square root and simplify: align* 1535 &= 1535 &=35^27 &=5^237 &= 521 align*

jmc

prealgebra intermediate

Problem

Simplify

15 \cdot 35

Solution

Place under one square root and simplify:

15 \cdot 35 = 15 \cdot 35 = 3 \cdot 5^{2} \cdot 7 = 5^{2} \cdot 3 \cdot 7 = 521

Final answer

5\sqrt{21}