2y^2(y^2)

that's all i got down so far

2y^2(y^2)

that's all i got down so far

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Jaz S. | Math Made ManageableMath Made Manageable

Hi Maralyn,

Students often find this kind of problem hard because they have learned the FOIL (First, Outer, Inner, Last) method for simplifying problems like (x + 2)(x - 5) and don't know how to extend the FOIL method to problems like this one that have more than two terms in one or both of the expressions inside the parentheses. But let's think for a moment about what the FOIL method is doing: it is multiplying each term from the first group by each term of the second group. Well, that's exactly what you want to do for this problem: multiply each term of the first group by each term of the second group. Of course, you also need to know how to multiply things like 2y^{2} by y^{2} which requires you to know that y^{m}*y^{n} = y^{m+n}. So, 2y^{2} * y^{2} = 2y^{4}.

Hope this helps,

Roger

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