Factoring an EquationDate: 11/13/95 at 14:43:6 From: Anonymous Subject: factor polynomials The problem is to factor 7y^3 + 14y^2 - 7y. I factor out the 7y: 7y(y^2 + 7y - 1) I get stuck there. Can I factor any further? Date: 11/13/95 at 15:9:41 From: Doctors Ethan and Ian Subject: Re: factor polynomials Hi, You were right to factor out the 7y, but you may have rushed through it a little too quickly: 7y^3 + 14y^2 - 7y ----------------- 7y 7y^3 14y^2 7y = ---- + ----- - -- 7y 7y 7y = y^2 + 2y - 1 However, this isn't any easier than what you ended up with. You need a pair of roots whose sum is -2 and whose product is -1, and there aren't any integers that fit the bill. So there are two possibilities: Either the roots are real, or they're complex. To find out which is the case, we can look at the discriminant, b^2 - 4ac where ax^2 + bx + c = 0 If this is negative, the roots are complex. Otherwise, they're real. In this case, a = 1, b = 2, and c = -1, so b^2 - 4ac = 2^2 - 4(1)(-1) = 4 + 4 = 8 That's positive, so the roots are real. But what are they? We can use the quadratic formula to find out: -b +/- sqrt(b^2 - 4ac) x = ---------------------- whenever ax^2 + bx + c = 0 2a -2 +/- sqrt(8) = -------------- 2 -2 +/- 2sqrt(2) = --------------- 2 = -1 + sqrt(2) OR -1 - sqrt(2) Now, if these are the roots of the quadratic expression, then 0 = (y - (-1 + sqrt(2))(y - (-1 - sqrt(2)) Can that be right? Let's multiply it out and see what we get. (I'm going to use 'sq2' instead of 'sqrt(2)' to save space and typing.) 0 = y^2 - [(-1 + sq2) + (-1 - sq2)]y + [(-1 + sq2) * (-1 - sq2)] = y^2 - [-1 + -1 + sq2 - sq2]y + [(-1)^2 - (sq2)^2] = y^2 - [-2]y + [-1] = y^2 + 2y - 1 So it ain't pretty! But it works. -Doctors Ethan and Ian, The Geometry Forum |
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