How do you factorise quadratic equations with more than one coefficient?

2x^2 + 2xy +3xy +3y^2

2x ( x +y) +3y ( x+y)

( 2x +3y) ( x+y) ANSWER

First, you have misunderstood the word "coefficient" - this is the number in front of the x (or y) so in your example they are 2, 5 & 3. The x & y are the variables, and to deal with one of these the method is basically the same as something like 2x^2+ 5x + 3. The only possible factors of 2x^2 are 2x & x, so the brackets start off as (2x ....)(x....). Now we look at the other end - the only possible factors of 3y^2 are y & 3y, so we have to try them both ways round in the brackets to get a middle term of 5xy, just as if there were no ' Y ' involved, to get (2x + 3y)(x + y). Hope that helped.

I like to imagine that one of the variables in eliminated. Let's say that y is not there.

2x^2 + 5x + 3

= ( 2x + 3 ) ( x + 1 )

now, put the y back:

( 2x + 3y ) ( x + y )