Factorise 6x^2 - 7ax + 2a^2 - 6x + 5a - 12.

Step-by-step solution

1. 1

   Treat a like a second variable. The expression is quadratic in $x$ and quadratic in $a$, and there is no term of degree higher than $2$ in the pair, so it can plausibly factor into two linear factors in $x$ and $a$.

2. 2

   Factor the degree-2 part.

   $6x^{2}-7ax+2a^{2}=(3x-2a)(2x-a)$

   (Check the cross terms: $3x\cdot(-a)+(-2a)\cdot 2x=-3ax-4ax=-7ax$. ✓)

3. 3

   Introduce the unknown constants. The full factorisation must be

   $(3x-2a+p)(2x-a+q),\qquad pq=-12$

4. 4

   Match the x and a coefficients. Expanding gives an $x$-coefficient of $3q+2p$ and an $a$-coefficient of $-2q-p$. The targets are $-6$ and $+5$:

   $3q+2p=-6,\qquad -2q-p=5$

5. 5

   Solve for p and q. From the second equation $p=-5-2q$; substituting into the first: $3q-10-4q=-6$, so $-q=4$ and $q=-4$, $p=3$. Both satisfy $pq=3(-4)=-12$. ✓

   $\boxed{(3x-2a+3)(2x-a-4)}$

6. 6

   Expand as a final check. $(3x-2a+3)(2x-a-4)=6x^2-3ax-12x-4ax+2a^2+8a+6x-3a-12=6x^2-7ax+2a^2-6x+5a-12$. ✓