Question

How do you find the product of $ 4n\left( 2{{n}^{3}}{{p}^{2}}-3n{{p}^{2}}+5n \right)+4p\left( 6{{n}^{2}}p-2n{{p}^{2}}+3p \right)$ ?

Answer 1

Hint: This is a very easy question to solve. So, in order to do this question, we just need to follow the BODMAs rule. First, we will multiply the terms outside the brackets with the terms in the brackets using the distributive law. So, we will multiply 4n and 4p with the terms inside the brackets. Then, we can check if we can add any terms in the final equation, so that we can further simplify the answer.

Complete step-by-step answer:
Let us begin the solution. We will use BODMAS - bracket, of, division, multiplication, addition and subtraction. Suppose if we have similar terms inside the bracket, we should first add the terms in the bracket and then multiply with the outside terms. So, here, we have brackets, but nothing to simplify inside it.
$ \Rightarrow 4n\left( 2{{n}^{3}}{{p}^{2}}-3n{{p}^{2}}+5n \right)+4p\left( 6{{n}^{2}}p-2n{{p}^{2}}+3p \right)$
Here, we can see that there are no similar terms inside the brackets, so we can carry on with our multiplication. So, we multiply 4n and 4p with the terms inside the brackets using the distributive law.
$ \Rightarrow 8{{n}^{4}}{{p}^{2}}-12{{n}^{2}}{{p}^{2}}+20{{n}^{2}}+24{{n}^{2}}{{p}^{2}}-4n{{p}^{3}}+12{{p}^{2}}$
Then, we now have to see if we can add any terms in the final equation, so that we can further simplify the answer. From the above equation, we can see that we can add two terms. Therefore, after adding we get the final answer as
$ \Rightarrow 8{{n}^{4}}{{p}^{2}}+12{{n}^{2}}{{p}^{2}}+20{{n}^{2}}-4n{{p}^{3}}+12{{p}^{2}}$

Note: First of all to do this question, you need not know any formulas. You just need to know basic multiplication where you multiply the terms outside the brackets with the terms in the brackets. To do this question, you also can first take out n from the first bracket and then p from the other bracket but then you again have to multiply all the terms.