How do you simplify $\sqrt{32}$ $sqrt32$ ?

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How do you simplify $\sqrt{32}$ $sqrt32$ ?

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Answer 1 · 2015-05-17T08:58:47

The answer is $4 \sqrt{2}$ $4sqrt2$ .

Let's decompose $32$ $32$ in its prime factors :

$32 \div 2 = 16$ $32 -: 2 = 16$ ,
$16 \div 2 = 8$ $16 -: 2 = 8$ ,
$8 \div 2 = 4$ $8-:2 = 4$ ,
$4 \div 2 = 2$ $4-:2 = 2$ ,
$2 \div 2 = 1$ $2-:2=1$ .

That gives us $32 = 2^{5}$ $32 = 2^5$ .

Therefore, $\sqrt{32} = \sqrt{2^{5}} = \sqrt{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2} = \sqrt{4} \cdot \sqrt{4} \cdot \sqrt{2} = 2 \cdot 2 \sqrt{2} = 4 \sqrt{2}$ $sqrt32 = sqrt(2^5) = sqrt(2*2*2*2*2) = sqrt4 * sqrt4 * sqrt2 = 2*2"sqrt2 = 4sqrt2$ .

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