How do I know when im looking at special case binomialproblem? Yes I understand

Question

How do I know when im looking at special case binomialproblem? Yes I understand that (a+b)(a-b) = a2 -b2 but for the problem (8p - 2)(6p + 2) it equals 48p2 + 4p - 4 Not 64p2 - 4(hence a2 -b2) which was what I though the answer was becausethe problem looked like it was in the form of (a+b)(a-b)! What is it that I need to look for so that I know that imsolving a special case problem and not just mutiplyingpolynomials? How do I know when im looking at special case binomialproblem? Yes I understand that (a+b)(a-b) = a2 -b2 but for the problem (8p - 2)(6p + 2) it equals 48p2 + 4p - 4 Not 64p2 - 4(hence a2 -b2) which was what I though the answer was becausethe problem looked like it was in the form of (a+b)(a-b)! What is it that I need to look for so that I know that imsolving a special case problem and not just mutiplyingpolynomials?

Explanation / Answer

(8p-2)(6p-2) is not that special case. It'd be better if Igive an example. x2-1 = (x - 1)(x+1). Do you see on the right handside that both a and b are perfect squares. Also on the left handside when you factor both first terms are x. x2-64 = (x-8)(x+8). Notice again that both a and bare perfect squares so this will only work for perfectsquares.

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