Question 181557
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Actually, you shouldn't expect your method to work because:


*[tex \LARGE \text{          }\math sqrt{162} = sqrt{(6)(27)} = sqrt{6}sqrt{27} \neq 6 sqrt{27}]


What you need to do first is find the prime factorization of the radicand.  162 is small enough that it won't be too tedious to do this with the brute force method.


*[tex \LARGE \frac {162}{2} = 81].  Integer result, so one factor of 2.


81 is odd, so there are no more factors of 2, but 8 + 1 = 9, divisible by 3, so 81 is divisible by 3.


*[tex \LARGE \frac {81}{3} = 27].  Integer result, so one factor of 3.


2 + 7 = 9, divisible by 3, so 27 is divisible by 3.


*[tex \LARGE \frac {27}{3} = 9].  Integer result, so two factors of 3.


9 is divisible by 3.


*[tex \LARGE \frac {9}{3} = 3].  Integer result, so three factors of 3, and the result is 3 so four factors of three.


Therefore, the prime factorization of *[tex \LARGE 162 = 2 \times 3 \times 3 \times 3 \times 3].


Now let's go back and look at your problem:


*[tex \LARGE \text{          }\math sqrt{162} = sqrt{2 * 3 * 3 * 3 * 3}]


For every pair of prime factors, you can take one of the factors out of the radicand.  That's because for a pair of prime factors, one of the factors is the square root of the product of the pair.


You have two pairs of 3s, so eliminate all of the 3s from under the radicand and take two 3s outside of the radical sign.


*[tex \LARGE \text{          }\math sqrt{162} = sqrt{2 * 3 * 3 * 3 * 3} = 3*3*sqrt{2} = 9 sqrt{2}]

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*[tex \LARGE 4 sqrt{48} + 5 sqrt{108}]


Same process as above.  Find the prime factorization of 48.  Then find the prime factorization of 108.  You should end up with both terms having a radicand of 3.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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