The missing value could be 16 or 19; if it is 16, the binomials would be
4(x+3)(x+1). If it is 19, the binomials would be (x+4)(4x+3).
To factor these, we want factors of a*c that sum to b; in this trinomial, a*c is 4(12)=48. If we use 12 and 4 to make 48, this would give a b value of 16:
4x²+16x+12
These two factors will be how we "split up" bx:
4x²+12x+4x+12
Now we group together the first two and last two:
(4x²+12x)+(4x+12)
Factor out the GCF of each group:
4x(x+3)+4(x+3)
Factor out the GCF of this new expression:
(x+3)(4x+4)
In the last binomial, they are both divisible by 4, so factor this out:
4(x+1)(x+3)
If we use 16*3 to make 48, this would give us a b value of 19; we would "split up" bx as follows:
4x²+16x+3x+12
Group:
(4x²+16x)+(3x+12)
Factor out the GCF:
4x(x+4)+3(x+4)
Factor out the GCF again:
(x+4)(4x+3)
