Factor completely using the difference of squares along with factoring by grouping.

7n-7p+n^2-p^2

Question

Factor completely using the difference of squares along with factoring by grouping.
   
7n-7p+n^2-p^2

anonymous · Answer

So far I factored the first two terms by using factoring by grouping and got 7(n-p). Then I factored the last two terms by using the difference of squares and got. (n-p)(n^2-np+p^2) So altogether it's 7(n-p)(n-p)(n^2-np+p^2) So then it ends up being 7(n-p)^2(n^2-np+p^2) right?

anonymous · Answer

You will have the form (n + x)^2 - (p + x)^2 for a particular value of "x".

anonymous · Answer

cause the only choices are
a.prime
b. 7(n-p)+n^2+p^2
c. 7(n-p)(n+p)
d. (n-p)(7+n+p)

anonymous · Answer

which would be the answer then?

anonymous · Answer

So it would be D

anonymous · Answer

how so?

anonymous · Answer

Because if you add and subtract pn you get -pn + pn +7n - 7p + n^2 - p^2 so you can work with those 6 terms.  That's 2 in addition to the original 4.