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How many different solutions does (x+1)(x-1)(x+2)(x+1) have?

User Elliot Yap
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Answer:

Perhaps this will help. We have 11 identical candies, that we wish to distribute between 3 children C1. C2, and C3. (Possibly one or more of the children will get no candies.) For every solution of x1+x2+x3=11, we have a way of distributing the candies, x1 to C1, x2 to C2, and X3 to C3. conversely, every distribution of candies gives us a solution to x1+x2+x3=11.

Explanation:

Can u please answer my recent question!!!!

User Mustafa DOGRU
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