Question

Find the product of the following pairs of monomials: $4{{p}^{3}},3p$.
A. 12p
B. $12{{p}^{4}}$
C. $12p+4$
D. ${{p}^{4}}$

Answer 1

Hint: The monomials are combinations of variables and constants. We multiply them separately to find the solution. For the variable part we apply the theorem of $\left( {{x}^{a}}\times {{x}^{b}} \right)={{x}^{a+b}}$ to find the multiplied value.

Complete step by step answer:
The given monomials are $4{{p}^{3}},3p$. The multiplication happens separately for constants and variables.
We first form the multiplication as $\left( 4{{p}^{3}} \right)\times \left( 3p \right)=\left( 4\times 3 \right)\times \left( {{p}^{3}}\times p \right)$.
We know the multiplication of variables with indices work for same base with the formula of $\left( {{x}^{a}}\times {{x}^{b}} \right)={{x}^{a+b}}$.
We take the constants’ multiplication first and get $\left( 4\times 3 \right)=12$ and the multiplication of the variables gives us $\left( {{p}^{3}}\times p \right)=\left( {{p}^{3}}\times {{p}^{1}} \right)={{p}^{4}}$.
So, the multiplication value is $\left( 4{{p}^{3}} \right)\times \left( 3p \right)=12{{p}^{4}}$.

So, the correct answer is “Option B”.

Note: The base value of the variables is similar here but if in some cases they are not the same then we can just keep the variables separated from but multiplied. For example: if we multiply $4{{p}^{3}},3q$, we get $\left( 4{{p}^{3}} \right)\times \left( 3q \right)=12{{p}^{3}}q$.