MATH SOLVE

5 months ago

Q:
# PLEASE 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 thetrinomials.

Accepted Solution

A:

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)

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)