Quadratic Factorization

You could try the quadratic formula. x = (-b (+/-) sqrt(b^2 - 4ac))/2a. If 4ac > b^2 then you have imaginary roots, if 4ac = b^2 you have a repeated real root and if 4ac < b^2 you have two unique real roots.

Sounds suspiciously like homework.....

Anyway, here's a hint. Rearrange both equations (ie. ax²+bx+c=0 and (x+m)(x+n)=0) until they are of similar form. With this you will be able to create 2 equations that equate some combination of m and n with some combination of a,b and c. Solve these equations simultaneously and you'll have your answer.

Well, it's probably homework.
Anyway.
[google] clickster with title of this thread entered into search field.(hint: 3rd result)
Also, you'd better find roots using general formula and plug 'em as m and n

Quote:

Original post by Anonymous Poster Quote: Original post by joanusdmentia Anyway, here's a hint. Rearrange both equations (ie. ax2+bx+c=0 and (x+m)(x+n)=0) until they are of similar form. With this you will be able to create 2 equations that equate some combination of m and n with some combination of a,b and c. Solve these equations simultaneously and you'll have your answer. Could you, please, show how you plan to solve it like that? If x^2 + px + q = (x-m)(x-n) you get: m + n = -p mn = q But solving this system is equivalent to solving a quadratic equation, so this does not help the original problem (i.e. it's the same). That is if I haven't missed something.

Yes, it is still solving a quadratic. The important difference is that you're no longer solving the quadratic for x, but for n and m.

Here's another way to do the same thing. For (x+m)(x+n)=0 to be true then one of (x+m) or (x+n) must also equal 0 (or both if m=n). This happens when m=-x and n=-x respectively (remember that x has 2 different roots, even if they are equal). Now all you have to do is solve for x.....

I think it depends upon what you are doing. The original question was about factorization of a polynomial. mn=q isn't real helpful when m and n can by any real number. The set of all such pairs is infinite. When the set of all such pairs is finite it is a differant story. An example is when m and n are limited to integers. There are only two such pairs then, i.e. (x,y) and (-x,-y). Further x and y have to be factors of q. It may seem contrived to limit it to integers, but you have no such reservation about limiting it to reals.

Homework.