Question

factor\
$4ab + 2a + 6b + 3$\
$\bigcirc$ $(4a+3)(b+1)$\
$\bigcirc$ $(2a+1)(2b+3)$\
$\bigcirc$ $(2a+3)(2b+1)$\
$\bigcirc$ $(a+3)(4b+1)$

Explanation:

Step1: Group the terms

Group the first two terms and the last two terms: $(4ab + 2a) + (6b + 3)$

Step2: Factor out the GCF from each group

From the first group, factor out $2a$: $2a(2b + 1)$
From the second group, factor out $3$: $3(2b + 1)$

Step3: Factor out the common binomial factor

Now we have $2a(2b + 1) + 3(2b + 1)$, factor out $(2b + 1)$: $(2a + 3)(2b + 1)$

We can also check by expanding each option:

• For option A: $(4a + 3)(b + 1) = 4ab + 4a + 3b + 3$, not equal to the original expression.
• For option B: $(2a + 1)(2b + 3) = 4ab + 6a + 2b + 3$, not equal to the original expression.
• For option C: $(2a + 3)(2b + 1) = 4ab + 2a + 6b + 3$, which matches the original expression.
• For option D: $(a + 3)(4b + 1) = 4ab + a + 12b + 3$, not equal to the original expression.

Answer:

C. (2a+3)(2b+1)