Factorise 3x^2 - 14xy + 15y^2 + 13x - 23y + 4.

Step-by-step solution

1. 1

   Factor the degree-2 part first. Any factorisation into two linear factors must have leading parts multiplying to

   $3x^{2}-14xy+15y^{2}=(3x-5y)(x-3y)$

   (Check: $3x\cdot(-3y)+(-5y)\cdot x=-9xy-5xy=-14xy$. ✓)

2. 2

   Write the unknown constants. The factorisation must be

   $(3x-5y+p)(x-3y+q)$

   for constants $p,q$ with $pq=4$.

3. 3

   Match the coefficient of x. Expanding, the $x$ terms are $3qx+px$, so $3q+p=13$.

4. 4

   Match the coefficient of y. The $y$ terms are $-5yq-3py$, so $-5q-3p=-23$, i.e. $5q+3p=23$.

5. 5

   Solve the small system. From $p=13-3q$, substituting gives $5q+39-9q=23$, so $-4q=-16$ and $q=4$, $p=1$. Both fit $pq=4$. Hence

   $\boxed{(3x-5y+1)(x-3y+4)}$

6. 6

   Expand to confirm. $(3x-5y+1)(x-3y+4)=3x^2-9xy+12x-5xy+15y^2-20y+x-3y+4=3x^2-14xy+15y^2+13x-23y+4$. ✓