Coefficient

Steven G said:
Jeff,
The way I did it was basically counting the number of ways that you get an x^8 term (or any other term) by multiplying nine terms together (they happened to be the same nine factors but that doesn't change anything). In multiplication of polynomials (or even non-polynomials) you choose one term from each factor and multiply them together. This method does not require you to know the multinomial theorem. I got the answer without even thinking of the multinomial theorem.

There are, for example, 9!/(6!2!) ways to get 1*1*1*1*1*1x^2*(-x^3)(-x^3) (= 1x^8). Is this clear?
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Steven

Absolutely, and again more efficient than my way. The point I think is that there are multiple ways to think this problem through. But they all come down to recognizing that there are a limited number of ways to generate x^8 by multiplying powers of x^2 and x^3. But the key is to think.
 
JeffM said:
Steven

Absolutely, and again more efficient than my way. The point I think is that there are multiple ways to think this problem through. But they all come down to recognizing that there are a limited number of ways to generate x^8 by multiplying powers of x^2 and x^3. But the key is to think.
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But the key is to think should be my new signature.
 
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