Can all quadratic equations be factorised?

[Kyoma]

1. The problem statement, all variables and given/known data

Is it true that all quadratic equations (ax2+bx+c) can be factorised into this form: (ax+b)(cx+d)?

I think so, but I'm not certain.

[gb7nash]

1. The problem statement, all variables and given/known data

Is it true that all quadratic equations (ax2+bx+c) can be factorised into this form: (ax+b)(cx+d)?

I think so, but I'm not certain.

If d is just some arbitrary number and you didn't accidentally duplicate your letters, then no.

Example:

y = x2-2x+1

So a = 1, b = -2, c = 1

So y = (x-1)(x-1)

a, c and d are fine, but you have a -1 instead of b.

[SammyS]

1. The problem statement, all variables and given/known data

Is it true that all quadratic equations (ax2+bx+c) can be factorised into this form: (ax+b)(cx+d)?

I think so, but I'm not certain.
As gb7nash said the coefficients a, b, and c, in (ax2+bx+c) are not, in general, the same as those in (ax+b)(cx+d).

So, can (ax2+bx+c) always be factored into the form, (px+q)(rx+s)?The answer is ... maybe.

If you allow q and s to be complex numbers, then the answer is yes.

If you restrict q and s to be real numbers, then you must have \sqrt{b^2-4ac}\ge0

If q and s must be rational numbers, then you must have \sqrt{b^2-4ac} be a perfect square.

This all assumes that a, b, and c are rational numbers.