How do you factor 81 n2 - 100?

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2006-10-10 15:14:19 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

You have to recognize these terms as perfect squares.

81n^2 - 100 = (9n + 10)(9n - 10)

2006-10-10 15:36:26 · answer #1 · answered by Anonymous · 0⤊ 0⤋

for any a^2-b^2 factors are (a+b) & (a-b) so
81*n^2-100=(9n+10)(9n-10)

2006-10-10 22:16:14 · answer #2 · answered by yupchagee 7 · 0⤊ 0⤋

81n^2 - 100 = (9n - 10)(9n + 10)

2006-10-10 22:34:01 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋

81n2-100
=(9n)^2-(10)^2
=(9n-10)(9n+10)(formula is a^2-b^2=(a+b)(a-b)

2006-10-11 08:54:22 · answer #4 · answered by srirad 2 · 0⤊ 0⤋

(9n-10)(9n+10)

2006-10-10 22:16:49 · answer #5 · answered by First L 1 · 0⤊ 0⤋