Factoring another EquationDate: 11/14/95 at 19:26:13 From: Anonymous Subject: Factoring 40y squared + yz - 6z squared Date: 11/14/95 at 21:47:12 From: Doctor Ethan Subject: Re: Factoring Good question. For y squared I am going to use y^2 So your question can be written 40y^2 - zy - 6z^2 When you do these problems, you are looking for two binomials in the form: (ay+bz)(cy+dz) = acy^2 + (ad+bc)zy + bdz^2 = 40y^2 - zy - 6z^2 This means that ac = 40 and bd = -6 Let us look at what a and c can be: a * c = 40 1 * 40 = 40 2 * 20 = 40 4 * 10 = 40 5 * 8 = 40 8 * 5 = 40 10 * 4 = 40 20 * 2 = 40 40 * 1 = 40 and b and d can be: b * d = -6 -1 * 6 = -6 1 * -6 = -6 -2 * 3 = -6 2 * -3 = -6 -3 * 2 = -6 3 * -2 = -6 -6 * 1 = -6 6 * -1 = -6 Whew, that is a lot of choices. But now we can narrow them down. We need choices so that ad + bc = -1, so now we just go through by trial and error. The idea is to pick values for a and c from our possible values list and then see if we can find a, b, and d that will work. In order to speed this up I am going to choose 5 for a and 8 for c. Then let us try values for b and d. Let b = -1 and d = 6 Then ad + bc = 30 - 8 = 22. Well, that isn't it. So try b = -2 and d = 3 - then ad + bc = 15 - 16 = -1 Hey, that worked. Yeah! So a = 5, b = -2, c = 8, d = 3 is what we want. Then (5y - 2z)(8y - 3z) = 40y^2 - zy -6z^2 Just like we wanted. Yeah! Write back if this doesn't make sense to you. -Doctor Ethan, The Geometry Forum |
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