find the product of(2m+n-p)(2m+n+p)?

(2m + n - p)(2m + n + p)
Let (2m + n) = a
(2m + n - p)(2m + n + p)
= (a – p)(a + p)
= a² – p²
= (2m + n)² – p²
= 4m² + 4mn + n² – p²
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this can be solve as a polynomial function in which one term in the bracket multiply all of the term in the other bracket separately and then sum all of the outcomes with respects to like terms . it would be also be nice to allowed your answer in the order of ascending or descending power of n(n=degree)
solution.
(2m+n-p)(2m+n+p) let each term in the other bracket multiple each in (2m+n+p)
then =2m(2m+n+p)+n(2m+n+p)-p(2m+n+p)
=4(m)^2+2mn+2mp+2mn+(n)^2+np-2mp-np-(p)^2
=4(m)^2+4mn+(n)^2-(p)^2
=(n)^2+4mn+4(m)^2-(p)^2 and this is the final stage .you should not that the following assumptions where made
^=let to the power, and m and p has been consider as constants ie why we respect but to the power of n for the second degree .

(2m + n - p)(2m + n + p)
= 2m*2m + n*2m - p*2m + 2m*n + n*n - p*n + 2m*p + n*p - p*p
= 4m^2 + 2mn - 2mp + 2mn + n^2 - np + 2mp + np - p^2
= 4m^2 + 2mn + 2mn + n^2 - 2mp + 2mp - np + np - p^2
= 4m^2 + 4mn + n^2 - p^2

Step by step:

(2m+n-p)(2m+n+p) =
= 2m(2m+n+p) + n(2m+n+p) - p(2m+n+p) =
= 4m^2 + 2mn + 2mp + 2mn + n^2 + np - 2mp - np - p^2 =
= 4m^2 + n^2 - p^2 + 4mn

That would be it. x^2 means square x

If I remember this correctly:
Just distribute each part to each other. multiply 2m to 2m, then to n, then to p. Then n to 2m, n, p, respectively. Do the same with p. It would look like this:

4m^2 + 2mn + 2mp + 2mn + n^2 + np - 2mp - np - p^2

Add those that are similar:
2mn + 2mn
2mp - 2mp
np - np

So the product will now be:
4m^2 + 4mn + n^2 - p^2

Well, I'm sure you know that (a + b)(a - b) = a^2 - b^2. Your question is exactly that same pattern with the added trickiness that you have to substitute [2m+n] for a.

(2m+n)^2-p^2

or

4m^2+4mn+n^2-p^2