I need to know how to factor these in A-C method form

Question

2a^2+1+3a

9w-w^3

Answer 1

I've never encountered this AC method before, but I've tried to google it. It's easy though it's crazy. XD
Anyway,
#1.
I'll change the variable into x so it won't be confusing:
2x^2 + 3x + 1
First step is to multiply a (the numerical coeff of x^2) by c (the constant). Then change a into 1. Here, a is equal to 2 and the c is 1:
x^2 + 3x + 1(2)
x^2 + 3x + 2
Second step is to factor this new expression:
(x+1)(x+2)
Third step is to change the numerical coeff of x by the original a (which is 2):
(2x+1)(2x+2)
Fourth step & final step is to further factor the expression above, and divide by the original a:
(2x+1)(2x+2) / 2
2(2x+1)(x+1) / 2
(2x+1)(x+1)
And it's actually correct (when you factor it normally).

Now try solving #2. We can rewrite the equation as
-w(w^2 - 9)
You'll use the AC method only in the w^2 - 9. Here, a = 1 and c = -9.

Hope this helps :3

Answer 2

2a^2+1+3a

A = 2
B = 3
C = 1

A* c
2* 1 = 2

1* 2
-1 -2

Sum
3
-3
B = 3
1 * 2
(a + 1) (2a+ 1)