Question

The factors of ${{a}^{2}}-{{b}^{2}}-4{{c}^{2}}+4{{d}^{2}}-4\left( ad-bc \right)$
 $\left( a+2d+b+2c \right)\left( a-2d-b+2c \right)$
$\left( a-2d+b-2c \right)\left( a+2d-b+2c \right)$
$\left( a-2d+b-2c \right)\left( a-2d-b+2c \right)$
$\left( a-2d-b-2c \right)\left( a+2d+b+2c \right)$

Answer 1

Hint: We start solving this problem first by considering the given expression ${{a}^{2}}-{{b}^{2}}-4{{c}^{2}}+4{{d}^{2}}-4\left( ad-bc \right)$ . Then we start simplifying the given expression by rearranging the terms in the expression. We use the formula that the square of difference of two numbers is the sum of squares of the two numbers and the negative of twice the product of the two numbers, that is, ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ . And we also consider the formula that the difference of squares of two numbers is the product of sum of those two numbers and difference of the two numbers, that is, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . Then we finally get the factors of the given expression.

Complete step by step answer:
Let us consider the given expression ${{a}^{2}}-{{b}^{2}}-4{{c}^{2}}+4{{d}^{2}}-4\left( ad-bc \right)$.
We were asked to find the factors of the above expression.
Let us briefly discuss factors now.
Multiplying some numbers gives us a product. Those numbers which are multiplied to get the product are called factors of that product. Simply, a divisor of an integer, also called a factor, is an integer that is multiplied by some integer to produce.
Mathematically, if a number $y$ is a divisible number $x$ , then $x$ is a factor of $y$.
We were given the expression, ${{a}^{2}}-{{b}^{2}}-4{{c}^{2}}+4{{d}^{2}}-4\left( ad-bc \right)$
Now, let us simplify this expression, we get,
$\begin{align}
  & {{a}^{2}}-{{b}^{2}}-4{{c}^{2}}+4{{d}^{2}}-4\left( ad-bc \right) \\
 & ={{a}^{2}}-{{b}^{2}}-4{{c}^{2}}+4{{d}^{2}}-4ad+4bc \\
\end{align}$
Let us rearrange the terms in the above obtained expression.
We get,
$\begin{align}
  & {{a}^{2}}-{{b}^{2}}-4{{c}^{2}}+4{{d}^{2}}-4ad+4bc \\
 & ={{a}^{2}}+4{{d}^{2}}-4ad-{{b}^{2}}-4{{c}^{2}}+4bc \\
 & =\left( {{a}^{2}}+4{{d}^{2}}-4ad \right)+\left( -{{b}^{2}}-4{{c}^{2}}+4bc \right) \\
 & =\left( {{a}^{2}}+4{{d}^{2}}-4ad \right)-\left( {{b}^{2}}+4{{c}^{2}}-4bc \right) \\
\end{align}$
Let us consider the formula, ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
By using the above formula, we say that,
${{a}^{2}}+4{{d}^{2}}-4ad={{\left( a-2d \right)}^{2}}$
${{b}^{2}}+4{{c}^{2}}-4bc={{\left( b-2c \right)}^{2}}$
So, we get,
${{a}^{2}}-{{b}^{2}}-4{{c}^{2}}+4{{d}^{2}}-4\left( ad-bc \right)={{\left( a-2d \right)}^{2}}-{{\left( b-2c \right)}^{2}}$
Let us consider the formula, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
By using the above formula, we get,
$\begin{align}
  & {{a}^{2}}-{{b}^{2}}-4{{c}^{2}}+4{{d}^{2}}-4\left( ad-bc \right)={{\left( a-2d \right)}^{2}}-{{\left( b-2c \right)}^{2}} \\
 & \Rightarrow {{a}^{2}}-{{b}^{2}}-4{{c}^{2}}+4{{d}^{2}}-4\left( ad-bc \right)=\left( \left( a-2d \right)+\left( b-2c \right) \right)\left( \left( a-2d \right)-\left( b-2c \right) \right) \\
 & \Rightarrow {{a}^{2}}-{{b}^{2}}-4{{c}^{2}}+4{{d}^{2}}-4\left( ad-bc \right)=\left( a-2d+b-2c \right)\left( a-2d-b+2c \right) \\
\end{align}$
Therefore, we get, ${{a}^{2}}-{{b}^{2}}-4{{c}^{2}}+4{{d}^{2}}-4\left( ad-bc \right)=\left( a-2d+b-2c \right)\left( a-2d-b+2c \right)$
Hence, the answer is $\left( a-2d+b-2c \right)\left( a-2d-b+2c \right)$.

Note:
The possibilities for making mistakes in this type of problems are, one may make a mistake by considering the formula of difference of squares of two numbers as square of difference of the two numbers, that is, ${{a}^{2}}-{{b}^{2}}={{\left( a-b \right)}^{2}}$ .