- #26

Dick

Science Advisor

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p^6 contains a term x1^6. None of your other forms do, and neither does the original polynomial. Hence it's coefficient must be 0. Same reasoning with p^4q. It contains an x1^5*x2 term.I don't think I understand. On what basis do you exclude the possibility of a p^{6}term? You say that when expanding the expression we get a_{ijk}Ʃx_{1}^{i}x_{2}^{j}x_{3}^{k}where a_{ijk}is a coefficient (integer) and i+j+k = 6. But that doesn't exclude the possibility of the case where i=6, j=0, k=0?