Factor the polynomial 100x^2^p - 121y^2^p, I think it is ( 10x^p + 11y^p)( 10x^p - 11y^p) but I am not sure can you help tell me if I am right? Thanks a lot
2007-05-25 03:48:19 · 4 answers · asked by ambi565 2 in Science & Mathematics ➔ Mathematics
RIGHT!
now you're getting difference of squares
2007-05-25 03:51:51 · answer #1 · answered by mark r 4 · 1⤊ 0⤋
yep yer right - you know that a^2 - b^2 = (a+b)(a-b)
take a = (10x)^p and b = (11y)^p
this is assuming yer original polynomial was (100x^2)^p - (121y^2)^p, but if its 100x^2^p - 121y^2^p then the factorization is different - you have to use bionomial theorem to get the answer - her a = 100x^2 and b = 121y^2
then it becomes a^p - b^p = (a+b)[a^(p-1) -ab^(p-2)+ ------ +b^(p-1)]
please do check the bionomial theorem - its clearer there
2007-05-25 10:57:30 · answer #2 · answered by beedotkiran 1 · 0⤊ 0⤋
You are right but your notation is a little off: it should be 100x^(2p)-121y^(2p).
2007-05-25 10:53:23 · answer #3 · answered by bruinfan 7 · 0⤊ 0⤋
You are completely right.
2007-05-25 10:52:48 · answer #4 · answered by gartfield72 2 · 0⤊ 0⤋