Factor x2 + 2x - 8y - 3xy - 4y2 completely.

Step-by-step solution

1. 1

   Sort the terms by total degree. Three terms are degree two in $x$ and $y$ together, and two are degree one:

   $\underbrace{x^2-3xy-4y^2}_{\text{degree }2}+\underbrace{2x-8y}_{\text{degree }1}$

   This grouping is the key move: if the whole expression factors into two brackets, the degree-two part must be the product of their leading parts.

2. 2

   Factor the homogeneous quadratic part. Treat it as a quadratic in $x$ with $y$ as a constant: we need two terms multiplying to $-4y^2$ and adding to $-3y$, namely $-4y$ and $+y$:

   $x^2-3xy-4y^2=(x-4y)(x+y)$

3. 3

   Factor the linear part and look for the shared bracket.

   $2x-8y=2(x-4y)$

   The factor $(x-4y)$ appears in both pieces — that is the signal that the whole expression factors, and it tells us which of the two brackets from step 2 is the common one.

4. 4

   Pull out $(x-4y)$.

   $(x-4y)(x+y)+2(x-4y)=(x-4y)\bigl[(x+y)+2\bigr]=(x-4y)(x+y+2)$

5. 5

   Expand back and test numerically. Expanding: $(x-4y)(x+y+2)=x^2+xy+2x-4xy-4y^2-8y=x^2-3xy+2x-4y^2-8y$, which is the original after reordering ✓. At $x=3$, $y=1$: the original is $9+6-8-9-4=-6$, and $(3-4)(3+1+2)=(-1)(6)=-6$ ✓.