Question

factor by grouping.
$4y^3 + 3y^2 + 20y + 15$

Explanation:

Step1: Group the terms

Group the first two terms and the last two terms: $(4y^3 + 3y^2) + (20y + 15)$

Step2: Factor out GCF from each group

Factor out $y^2$ from the first group and $5$ from the second group: $y^2(4y + 3) + 5(4y + 3)$

Step3: Factor out the common binomial factor

Factor out $(4y + 3)$: $(4y + 3)(y^2 + 5)$

Answer:

$(4y + 3)(y^2 + 5)$