How to factor 6w^2 - 5wt - 4t^2

[zeion]

Homework Statement

Factor 6w^2 - 5wt - 4t^2

Homework Equations

The Attempt at a Solution

= 6w^2 + 3wt - 8wt - 4t^2
= 3w(2w + t) - 4t(2w + t)
= (3w - 4t)(2w + t)

I was able to figure out what to use for splitting up the -5wt after a lot of trial and errors, but I was just wondering if there was a general rule for finding out what to use to factor things in this form?

I noticed that I needed them to be opposite signs and is half of the coefficient of the first term and twice of the last.

 

[Bohrok]

For quadratics where the leading coefficient isn't 1, you'll often use the ac method where you multiply a and c, then find the factors of the result that add up to b. Notice that you split up the middle coefficient -5 into 3 and -8, which are two factors of 24, or the product of 6 and -4.

If that didn't make sense, just take a look on Google or read up on it http://people.richland.edu/james/misc/acmeth.html" .

 

[zeion]

Is there a proof for this method?

 

[Mentallic]

If you're given some quadratic ax²+bx+c it can be proven that it can't always be factorized into the form a(x-\alpha)(x-\beta) where \alpha and \beta are rational numbers, which is why factorizing can only be done through trial and error, because if they are either irrational or complex then the usual factorizing process won't work and you need to use the quadratic formula.