albertlee Posted September 7, 2006 Share Posted September 7, 2006 say, on a graphic of cubic function, there is a root at 2, a y point on 8 when x is 1, and another on 8 also when x is -1, and when x is 0, y = 8. 1) if I solve this by: (x-2)(ax^2+bx+c).... I get an answer because I can solve a, b, c by 3 different simul equations. but 2) if I solve this by: (x-2)(ax+b)(x+c), I get more answers because when I try to solve c, I get a quadratic equation for it, which means c can have 2 values...... but this cannot be possible, since c can only be one value as a root. plz help Link to comment Share on other sites More sharing options...

Dave Posted September 7, 2006 Share Posted September 7, 2006 Well the reason why we try to find a factor of the form ax^2+bx+c is because we don't necessarily know that this quadratic factor is even reducible*. That is, when we solve for a, b and c the answer is not guaranteed to then factor into two linear factors. For example, say you were to obtain the quadratic term x^2+x+1. It is impossible to reduce this into two linear factors*. So by assuming the form of the non-linear term is (ax+b)(x+c) you may run into problems. * I'm assuming we're talking about the reals, any real polynomial is reducible over C. Link to comment Share on other sites More sharing options...

albertlee Posted September 7, 2006 Author Share Posted September 7, 2006 alright...thanks dave Link to comment Share on other sites More sharing options...

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