Question

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

Answer 1

Hint: We first explain the meaning of the process ‘FOIL’. We multiply the terms according to their positions. There are four multiplications to be done. We complete all four multiplications according to the previously mentioned process.

Complete step-by-step answer:
We have been given multiplication of two linear equations. We have to do the breakings of the polynomials in order of FOIL. The word FOIL stands for First-Outside-Inside-Last. It is a technique to distribute the multiplication of polynomials.
There are two terms in the polynomial. We break it as $ {{\left( n+3 \right)}^{2}}=\left( n+3 \right)\left( n+3 \right) $ .
We start by multiplying the first terms of $ \left( n+3 \right) $ and $ \left( n+3 \right) $ . The terms are $ n $ and $ n $ .
The multiplication gives the result of $ n\times n={{n}^{2}} $ .
We now multiply the outside terms of $ \left( n+3 \right) $ and $ \left( n+3 \right) $ . The terms are $ n $ and 3.
The multiplication gives a result of $ n\times 3=3n $ .
Then we multiply the inside terms of $ \left( n+3 \right) $ and $ \left( n+3 \right) $ . The terms are 3 and $ n $ .
The multiplication gives a result of $ 3\times n=3n $ .
We end by multiplying the last terms of $ \left( n+3 \right) $ and $ \left( n+3 \right) $ . The terms are 3 and 3.
The multiplication gives a result of $ 3\times 3=9 $ .
Now we add all the terms and get the final solution as
 $ \left( n+3 \right)\left( n+3 \right)={{n}^{2}}+3n+3n+9={{n}^{2}}+6n+9 $ .
Therefore, the multiplied value of $ \left( n+3 \right)\left( n+3 \right) $ is $ {{n}^{2}}+6n+9 $.
So, the correct answer is “ $ {{n}^{2}}+6n+9 $ ”.

Note: we can also apply the identity of $ {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} $ for the simplification of $ {{\left( n+3 \right)}^{2}} $ where we take $ a=n,b=3 $ .
Therefore, $ {{\left( n+3 \right)}^{2}}={{n}^{2}}+2\times 3\times n+{{3}^{2}}={{n}^{2}}+6n+9 $