Question

How do I find the product of $\left( {2h + 3} \right)\left( {2{h^2} + 3h + 4} \right)$?

Answer 1

Hint: In this problem we have given the product of two brackets and each bracket contains addition of some terms and moreover the second bracket seems like a quadratic equation. And here we are asked to find the product's result. We can find the result of the given product by multiplying, expanding, grouping and using distributive property.

Complete step-by-step solution:
Given term is $\left( {2h + 3} \right)\left( {2{h^2} + 3h + 4} \right)$.
Each term in the second bracket must be multiplied by each term in the first bracket or using distributive property, we get, $\left( {2h + 3} \right)\left( {2{h^2} + 3h + 4} \right) = 2h\left( {2{h^2} + 3h + 4} \right) + 3\left( {2{h^2} + 3h + 4} \right)$
So by using distributive property we got this term.
Now multiply $2h$ with each terms of $\left( {2{h^2} + 3h + 4} \right)$ and multiply $3$ with each terms of $\left( {2{h^2} + 3h + 4} \right)$, Now, we get $\left( {2h + 3} \right)\left( {2{h^2} + 3h + 4} \right) = 4{h^3} + 6{h^2} + 8h + 6{h^2} + 9h + 12$,
So here we distributed each pair of brackets.
Next we need to group the equal powers of the variables and we get,
 $ \Rightarrow \left( {2h + 3} \right)\left( {2{h^2} + 3h + 4} \right) = 4{h^3} + (6{h^2} + 6{h^2}) + (8h + 9h) + 12$,
Collecting like terms gives this step.
Adding together equal powers of the variable, we get
$ \Rightarrow \left( {2h + 3} \right)\left( {2{h^2} + 3h + 4} \right) = 4{h^3} + 12{h^2} + 17h + 12$, this is the result of the product.

Hence, $4{h^3} + 12{h^2} + 17h + 12$ is the required solution.

Note: Mathematical brackets are symbols, such as parentheses, that are most often used to create groups or clarify the orders that operations are to be done in an algebraic. Writing two brackets next to each other means the brackets need to be multiplied together. The problem which is given for us is an example. When expanding double brackets, every term in the first bracket has to be multiplied by every term in the second bracket.