Question

How do you multiply ${{\left( x-4y \right)}^{2}}$?

Answer 1

Hint: The square of $\left( x-4y \right)$ is given. So for the multiplication, multiply $\left( x-4y \right)$ with itself i.e. $\left( x-4y \right)\left( x-4y \right)$. First multiply ‘x’ with $\left( x-4y \right)$ and ‘4y’ with $\left( x-4y \right)$ and then subtract them together to get the required solution.

Complete step-by-step solution:
Given ${{\left( x-4y \right)}^{2}}$
As it is a square, so it can be written as
$\Rightarrow \left( x-4y \right)\left( x-4y \right)$
To multiply these two factors, we have to take ‘x’ from the first factor and multiply with the second factor $\left( x-4y \right)$. Then we have to take ‘4y’ from the first factor and multiply with the second factor $\left( x-4y \right)$. After that subtracting these two terms together we can obtain the solution.
Hence, multiplying ‘x’ with $\left( x-4y \right)$ and ‘4y’ with $\left( x-4y \right)$ and then subtracting together, we get
$\begin{align}
  & \Rightarrow \left[ x\left( x-4y \right) \right]-\left[ 4y\left( x-4y \right) \right] \\
 & \Rightarrow \left( {{x}^{2}}-4xy \right)-\left( 4xy-16{{y}^{2}} \right) \\
\end{align}$
While subtracting the term which is being subtracted i.e. $\left( 4xy-4{{y}^{2}} \right)$, after opening the brackets we have to change the sign of each and every individual terms from positive to negative and vice-versa.
So, changing the sign of 4xy from positive to negative and changing the sign of $16{{y}^{2}}$from negative to positive, we get
$\begin{align}
  & \Rightarrow {{x}^{2}}-4xy-4xy+16{{y}^{2}} \\
 & \Rightarrow {{x}^{2}}-8xy+16{{y}^{2}} \\
\end{align}$
This is the required solution.

Note: We are given with ${{\left( x-4y \right)}^{2}}$. As we know ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab-{{b}^{2}}$, so our expression can be expanded as
$\begin{align}
  & {{\left( x-4y \right)}^{2}} \\
 & \Rightarrow {{\left( x \right)}^{2}}-2\cdot x\cdot 4y+{{\left( 4y \right)}^{2}} \\
 & \Rightarrow {{x}^{2}}-8xy+16{{y}^{2}} \\
\end{align}$
This is the alternative method. In this method also we are getting the same solution. It is the easiest way to expand any square term. But since it is not applicable to normal multiplication, so the method mentioned earlier is the best suited.