Factoring with 3 Variables
Date: 4/8/96 at 11:49:19 From: Anonymous Subject: Factoring when either square is a square of a polynomial This one has been frustrating. Factor: 9a^2+4bc-4c^2-b^2 I re-arranged: 9a^2 -4c^2 +4bc-b^2 Then: (3a)^2+(-4c+b)(c-b), and that does not work, but it is the closest to being correct for the middle term. I have tried several possible combinations for the last factors. Surely you have seen this before and can help me over this hump. Thank you. LF
Date: 4/19/96 at 17:10:29 From: Doctor Cameron Subject: Re: Factoring when either square is a square of a polynomial Thanks for writing in. 9a^2 + 4bc - 4c^2 - b^2 I would rewrite it as 9a^2 - b^2 + 4bc -4c^2 then isolate a polynomial. 9a^2 - (b^2 - 4bc + 4c^2) The signs change when you subtract in the parentheses. Then factor the polynomial 9a^2 - (b - 2c)^2. Note that this polynomial is the difference of two squares, so it is in the form y^2 - z^2 where y = 3a and z = b - 2c. We know that we can factor the difference of two squares as follows: y^2 - z^2 = (y-z)(y+z) Therefore we get (3a - (b - 2c))(3a + (b - 2c)) as our finished answer. -Doctor Cameron, The Math Forum
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