Factor E = 4x^2 y^3 + 4x^3 y^2 + a^2 x + a^2 y + 12x + 12y.

Step-by-step solution

1. 1

   Look for a repeated pattern rather than a single common factor. No factor divides all six terms, so grouping is the tool. The last four terms are already suggestive: $a^2x+a^2y$ and $12x+12y$ both contain $x+y$.

2. 2

   Factor the easy pairs first.

   $a^2x+a^2y=a^2(x+y),\qquad 12x+12y=12(x+y)$

   Two of the three pairs now share the binomial $x+y$, so the remaining pair must be made to match.

3. 3

   Force the first pair into the same shape. Both leading terms contain $4x^2y^2$:

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

   This is the step that decides the problem — noticing that $y^3$ and $x^3$ leave behind exactly $y$ and $x$ after pulling out $4x^2y^2$.

4. 4

   Collect the common binomial. All three groups now carry $x+y$:

   $E=(x+y)\left(4x^2y^2+a^2+12\right)$

5. 5

   Check numerically and confirm this is complete. At $x=1$, $y=2$, $a=3$: the original is $4(1)(8)+4(1)(4)+9+18+12+24=32+16+63=111$, and the factored form gives $3\left(16+9+12\right)=3\cdot 37=111\;\checkmark$. The second factor is a sum of non-negative terms plus $12$, so it never vanishes and contributes no further real factors.