√20

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4 5

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2 2

The bolded numbers are the leaves that you want to pick. There has to be two leaves that are the same before you can pick them. Since there are two 2’s that’s one numbers we can pick. However there is only one 5 leaf so that one has to stay on the tree. So the final answer is 2√5. This is just a warm up to the lesson ahead. The first example on the noted was this ^4√48p^11. This may look difficult but you need to just use the jailbreak method on √48.

√48

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6 8

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3 2 2 4

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2 2

The little 4 above the square root means that we need to find 4 pairs of a single number of the square root of 48. There are exactly 2 pairs of 2 so that and a 3 left over that we cannot pick from the tree so the answer will look like this so far 2√3. This is still not done because we still need to get rid of the p^11. Now write out 11 p’s.

Ppppppppppp

Now we need to find all the 2 pairs of p. So we will circle(bold) 4 p’s.

**Pppppppp**ppp

There are two 2 pairs of 4 p’s and 3 p’s left over. Then we will write the final answer out like this. 2p^2 ^4√3p^3. This is read as 2p squared times 4 square roots of 3p cubed.

Another sample problem of using products with radicals is ^5√32y^9. First we need to use the factor tree method on √32.

√32

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8 4

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4 2 2 2

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2 2

There has to be 5 of a certain number and luckily there is five 2's. That means so far we have only one 2 in the answer. Next we find out how many y's are in the problem. There are 9 so there is only one group of five y's. There is 4 y's that have to stay so the final answer will be 2y^5√y^4. This is read two y to the fifth times the fifth sqaure root of y to the fourth.

Hopes this helps.

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