Question

How do you factor a quadratic expression?

Answer 1

See explanation...

Explanation:

I will assume that you mean a quadratic expression in one variable. If it has more than one variable there are numerous cases to consider.

Given:

`ax^2+bx+c`

with `a, b, c` rational numbers.

Consider the discriminant `Delta` given by the formula:

`Delta = b^2-4ac`

Then we have several cases:

• `Delta > 0` with `Delta` a perfect square. Then the quadratic is factorable with rational coefficients.

• `Delta > 0` with `Delta` not a perfect square. Then the quadratic is factorable with irrational coefficients involving `sqrt(Delta)` or a rational multiple thereof.

• `Delta = 0`. The quadratic is factorable in the form `d(ex+f)^2`, where `d, e, f` are rational. If you allow irrational coefficients, then it is expressible in the form `(gx+h)^2` where `g, h` may be irrational.

• `Delta < 0`. The quadratic is not factorable "over the reals". That is, it has no factorisation using real coefficients. It is factorable "over the complex numbers".

We can express a general factorisation of our quadratic using the quadratic formula...

`ax^2+bx+c = a(x-(-b+sqrt(b^2-4ac))/(2a))(x-(-b-sqrt(b^2-4ac))/(2a))`

`color(white)(ax^2+bx+c) = a(x-(-b+sqrt(Delta))/(2a))(x-(-b-sqrt(Delta))/(2a))`

Such a factorisation will always "work", but it may involve the square root of a negative number - i.e. a non-real complex number.