Factoring into the Product of Two Trinomials
Date: 02/22/2003 at 00:27:36 From: Mark Subject: Factoring into the product of two trinomials Write as a product of two trinomials with integer coefficients: 6x^2-xy+23x-2y^2-6y+20 I tried many formulas to help me decode this, but none of them seemed to apply to this one. Please show this to me step by step, because I'm not sure if I'm starting correctly. Thanks for your help.
Date: 02/22/2003 at 14:05:01 From: Doctor Greenie Subject: Re: Factoring into the product of two trinomials Hi, Mark - Here is how I do it.... Let's start by supposing we have found the factorization we want: 6x^2-xy+23x-2y^2-6y+20 = (Ax+By+C)(Dx+Ey+F) And let's perform the multiplication on the right: (Ax+By+C)(Dx+Ey+F) = (AD)x^2+(AE+BD)xy+(BE)y^2+(AF+CD)x+(BF+CE)y+(CF) Now we equate the coefficients of like terms between this product and the original expression: AD = 6 AE+BD = -1 BE = -2 AF+CD = 23 BF+CE = -6 CF = 20 Since AD=6 is positive, A and D have the same sign; we can assume they are both positive. So we have the following possibilities for the coefficients A and D, since the coefficients are integers: A D -------- 1 6 2 3 3 2 6 1 And with BE = -2 and again the coefficients being integers, we have the following possibilities for B and E: B E -------- -2 1 -1 2 1 -2 2 -1 Combining each of the possibilities for A and D with each of the possibilities for B and E, and looking for a combination for which AE+BD = -1 we find the only combination that works is (A,D) = (2,3) and (B,E) = (1,-2). Using these values of A, B, D, and E in the two equations AF+CD = 23 BF+CE = -6 and solving the resulting pair of linear equations simultaneously, we find C = 5 and F = 4; and these values satisfy the final equation CF = 20 Checking our result, we have (Ax+By+C)(Dx+Ey+F) = (2x+y+5)(3x-2y+4) = 6x^2-4xy+8x+3xy-2y^2+4y+15x-10y+20 = 6x^2-xy-2y^2+23x-6y+20 So our calculations were correct. I hope this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/
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