some problems solving this equation....

[albertlee]

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

[Dave]

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.

[albertlee]

alright...thanks dave