Question

How do you find the product of $ (2h + 3)(2{h^2} + 3h + 4) $ ?

Answer 1

Hint: When we multiply a binomial and a trinomial, we get another kind of polynomial multiplication problem. Even though the FOIL approach cannot be used since a trinomial has more than two terms, the Distributive Property is used to arrange the individual items. Each term in the binomial must be multiplied by each term in the trinomial using the distributive property.

Complete step-by-step answer:
We have a binomial and a trinomial expression.
 $ (2h + 3) $ is the binomial expression since it has $ 2 $ terms in the equation. $ (2{h^2} + 3h + 4) $ is the trinomial expression since it has $ 3 $ terms in the equation.
We begin solving it, by multiplying each term of the binomial with all the terms of the trinomial.
 $ (2h)(2{h^2} + 3h + 4) + (3)(2{h^2} + 3h + 4) $
On multiplying $ 2h $ with $ 2{h^2} $ , we will get $ 2h \times 2{h^2} = 4{h^3} $
On multiplying $ 2h $ with $ 3h $ , we will get $ 2h \times 3h = 6{h^2} $
On multiplying $ 2h $ with $ 4 $ , we will get $ 2h \times 4 = 8h $
On multiplying $ 3 $ with $ 2{h^2} $ , we will get $ 3 \times 2{h^2} = 6{h^2} $
On multiplying $ 3 $ with $ 3h $ , we will get $ 3 \times 3h = 9h $
On multiplying $ 3 $ with $ 4 $ , we will get $ 3 \times 4 = 12 $
So finally we get,
 $ (4{h^3} + 6{h^2} + 8h) + (6{h^2} + 9h + 12) $
Now we group the like terms,
 $ 4{h^3} + 6{h^2} + 6{h^2} + 8h + 9h + 12 $
On solving the above equation, we get,
 $ 4{h^3} + 12{h^2} + 17h + 12 $
Therefore, the product of $ (2h + 3 $ and $ (2{h^2} + 3h + 4) $ is $ 4{h^3} + 12{h^2} + 17h + 12 $
So, the correct answer is “ $ 4{h^3} + 12{h^2} + 17h + 12 $ ”.

Note: We simply multiply each term of the first polynomial by each term of the second polynomial to find the product of two polynomials, then we simplify it. When we need to find the result of two binomials, we can use the FOIL form. First, Outside, Within, and Last FOIL stands for First, Outside, Inside, and Last.