Factorising quadratic expressions in two variables

An expression such as \(x^2-2xy-15y^2\) is quadratic in \(x\), with \(y\) treated like a number. It factorises in the same way as \(x^2-2x-15\), with a \(y\) attached to each number.

Why: \(x^2-2xy-15y^2=x^2+(-2y)x+(-15y^2)\). The two "numbers" must multiply to \(-15y^2\) and add to \(-2y\); they are \(-5y\) and \(3y\).

Method

1. Arrange the terms with the \(x^2\)-term first, then the \(xy\)-term, then the \(y^2\)-term.
2. Factorise the matching one-variable expression (replace \(y\) by \(1\)).
3. Attach \(y\) to the number in each bracket.
4. Expand to check the \(xy\)-term.

Example

Factorise \(2p^2+5pq-12q^2\).

Matching one-variable expression: \(2p^2+5p-12=(2p-3)(p+4)\).

\(\therefore 2p^2+5pq-12q^2=(2p-3q)(p+4q)\)

Check: \((2p-3q)(p+4q)=2p^2+8pq-3pq-12q^2=2p^2+5pq-12q^2\). A common error is to put \(y\) in only one bracket, such as \((x-5y)(x+3)\), which creates a term \(3x\) that is not in the question.