Algebra 1 Factoring completely problem

Factor Completely: 12x^3-32v^2+6v-16
The answer is 2(2v^2+1)(3v-8)
I got to 4v^2(3v-8) 2(3v-8).
Can't figure out how to get the answer from that. Can someone tell me what I'm missing here please? Thanks.

distributive property:
(a + b)(x + y) = a(x + y) + b(x + y)
distributive property backwards:
a(x + y) + b(x + y) = (a + b)(x + y)

12v^3 - 32v^2 + 6v - 16
factor by grouping
= (12v^3 - 32v^2) + (6v - 16)
= 4v^2(3v - 8) + 2(3v - 8)
^^ you got up to here
use the distributive property backwards
= (4v^2 + 2)(3v - 8)
factor out 2 from 4v^2 + 2
= 2(2v^2 + 2)(3v - 8)

12v^3-32v^2+6v-16

2(6v^3 - 16v^2 +3v - 8)

group (6v^3 +3v) - 16v^2 -8
you can see that 3v comes out of the first group and -8 comes out of the secodn

that leaves 2v^2+1

2(3v-8)(2v^2+1)

12v^3 - 32v^2 + 6v -16

4v^2(3v-8) + 2(3v-8)

SO, (4v^2+2)(3v-8)