Question

How do you multiply $\left( 2n+2 \right)\left( 6n+1 \right)$ ?

Answer 1

Hint: We will multiply the given 2 terms by simple algebraic multiplication. We know that we can write a ( b + c ) is equal to ab + ac and we know the formula ${{a}^{x}}{{a}^{y}}$ is equal to ${{a}^{x+y}}$ .We can use the 2 formula to multiply ( 2n +2 ) with ( 6n +1).

Complete step-by-step answer:
We have to multiply $\left( 2n+2 \right)\left( 6n+1 \right)$ . we know the algebraic formula a ( b + c ) is equal to ab + ac
So we can write $\left( 2n+2 \right)\left( 6n+1 \right)$ is equal to $2n\left( 6n+1 \right)+2\left( 6n+1 \right)$ .
We can again apply the formula to solve $2n\left( 6n+1 \right)+2\left( 6n+1 \right)$. The value of 2n( 6n +1) is equal to $2n\times 6n+2n\times 1$ . Multiplication of 2n and 6n is equal to $12{{n}^{2}}$ and 2n and 1 is 2n
So we can write is equal to 2n( 6n +1) is equal to $12{{n}^{2}}+2n$ similarly we can write $2\left( 6n+1 \right)$ as 12n + 2
So $2n\left( 6n+1 \right)+2\left( 6n+1 \right)=12{{n}^{2}}+2n+12n+2$
The sum of 2n and 12n is equal to 14n
So we can write $2n\left( 6n+1 \right)+2\left( 6n+1 \right)=12{{n}^{2}}+14n+2$
$\Rightarrow \left( 2n+2 \right)\left( 6n+1 \right)=12{{n}^{2}}+14n+2$

Note: In the above question we have multiplied 2 terms. When we multiply more than 2 terms , first we multiply first 2 terms then we multiply the result with the third term. This process goes on to the last term. Always remember the formula a( b +c ) is equal to ab + ac but the reverse is not true that means $a+\left( bc \right)$ is not equal to $\left( a+b \right)\left( a+c \right)$