Factorise x^2 + 5xy + 6y^2 - 2x - 7y - 3.

Step-by-step solution

1. 1

   Split off the degree-2 part. If the whole expression factors into two linear factors, those factors' leading parts must multiply to give $x^2+5xy+6y^2$. Factoring that piece alone fixes the shape of the answer.

2. 2

   Factor the quadratic part.

   $x^{2}+5xy+6y^{2}=(x+2y)(x+3y)$

   so the factorisation must look like $(x+2y+p)(x+3y+q)$ for constants $p$ and $q$.

3. 3

   Match the constant term. Multiplying the two constants gives $pq=-3$, so $\{p,q\}$ is $\{1,-3\}$ or $\{3,-1\}$.

4. 4

   Match the x and y coefficients to choose between them. Expanding $(x+2y+p)(x+3y+q)$ gives an $x$-coefficient of $p+q$ and a $y$-coefficient of $3p+2q$. We need $p+q=-2$ and $3p+2q=-7$. The pair $p=-3$, $q=1$ satisfies both: $-3+1=-2$ and $-9+2=-7$.

5. 5

   Write the answer.

   $\boxed{(x+2y-3)(x+3y+1)}$

6. 6

   Expand as a check.

   $(x+2y-3)(x+3y+1)=x^2+3xy+x+2xy+6y^2+2y-3x-9y-3=x^2+5xy+6y^2-2x-7y-3$