Factor 2x2 - x - 6xy + 3y completely.

Step-by-step solution

1. 1

   Split the four terms into two useful pairs. Grouping works only if each pair has a common factor and the leftover brackets match. Pairing the first two and the last two:

   $\left(2x^2-x\right)+\left(-6xy+3y\right)$

2. 2

   Factor each pair. From the first pair take out $x$; from the second take out $-3y$ (the minus is deliberate, so the bracket comes out the same way round):

   $2x^2-x=x(2x-1),\qquad -6xy+3y=-3y(2x-1)$

   Had we taken out $+3y$ we would have got $3y(-2x+1)$ — correct, but it hides the match.

3. 3

   Confirm the brackets agree. Both pairs produced $(2x-1)$:

   $x(2x-1)-3y(2x-1)$

   This agreement is the signal that the grouping was chosen correctly.

4. 4

   Pull the common binomial out.

   $x(2x-1)-3y(2x-1)=(2x-1)(x-3y)$

5. 5

   Expand back and test numerically. $(2x-1)(x-3y)=2x^2-6xy-x+3y$, which is the original after reordering ✓. At $x=2$, $y=1$: the original is $8-2-12+3=-3$, and $(4-1)(2-3)=3(-1)=-3$ ✓. Neither factor breaks down further, so the factoring is complete.