Answer + brainliest (show work) find two different values that complete each expression so that the trinomial can be factored into the product of two binomials. factor the trinomials.

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)