A lot of students look at this problem and they don’t see a difference of perfect squares because of that x to the 6th term. That’s kind of scary. But keep in mind it’s a difference of perfect squares if we have even exponents.
Let me remind you of something, x to the 6th is the same this as x to the 3rd times x to the 3rd or x to the 3rd squared. So what I’m really working with is x to the 3rd squared take away 4y². That might help you recognize this as a difference of perfect squares. So when I write it in the factor form, first I’m going to have x to the 3rd as my first term in both the binomials, I need a plus sign and a minus sign, doesn’t matter which goes where and then I need the square root of 4y².
Well the square root of 4 is 2 and the square root of y² is y. So this is my factored form of that original difference of perfect squares. It looks kind of funky when you see problems like this, remember you’re looking for square numbers that show up as the coefficients or constants and then you’re looking for even exponents. 6 is an even exponent and that’s why I was able to write it as x to the 3rd power squared.
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M.A. in Secondary Mathematics, Stanford University B.S., Stanford University
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