Question

Factorise the following algebraic expression
$ab\left( {{x}^{2}}+{{y}^{2}} \right)-xy\left( {{a}^{2}}+{{b}^{2}} \right)$

Answer 1

Hint: Expand the expressions $ab\left( {{x}^{2}}+{{y}^{2}} \right)$ and $xy\left( {{a}^{2}}+{{b}^{2}} \right)$. Take $ax$ common from $ab{{x}^{2}}\text{ and}\ {{a}^{2}}xy$ and $by$ common from $ab{{y}^{2}}\text{ and }a{{b}^{2}}y$. Finally, take $ax-by$ as common in the resulting expression and factorise to get the result.

Complete step-by-step answer:
Before solving the question, we need to know the meaning of factorisation. Consider two algebraic expressions ${{a}^{2}}-{{b}^{2}}$ and $\left( a+b \right)\left( a-b \right)$. Let us simplify the latter expression. Applying distributive property we get $\left( a+b \right)\left( a-b \right)=\left( a+b \right)a-\left( a+b \right)b$
Applying distributive property again we get
$\left( a+b \right)\left( a-b \right)={{a}^{2}}+ab-\left[ ab+{{b}^{2}} \right]$
Simplifying, we get
$\begin{align}
  & \left( a+b \right)\left( a-b \right)={{a}^{2}}+ab-ab-{{b}^{2}} \\
 & ={{a}^{2}}-{{b}^{2}} \\
\end{align}$
Hence the two expressions are equal.
Step 1: Expand the expressions
We have
$ab\left( {{x}^{2}}+{{y}^{2}} \right)=ab{{x}^{2}}+ab{{y}^{2}}$ and $xy\left( {{a}^{2}}+{{b}^{2}} \right)=xy{{a}^{2}}+xy{{b}^{2}}$
Hence, we have
$ab\left( {{x}^{2}}+{{y}^{2}} \right)-xy\left( {{a}^{2}}+{{b}^{2}} \right)=ab{{x}^{2}}+ab{{y}^{2}}-xy{{a}^{2}}-xy{{b}^{2}}$
Step 2: Take $ax$ common in the first and the third term and $by$ common in the second and the last term.
We have
$ab{{x}^{2}}-xy{{a}^{2}}=ax\left( bx-ay \right)$ and $ab{{y}^{2}}-xy{{b}^{2}}=by\left( ay-bx \right)=-by\left( bx-ay \right)$
Hence, we have
$ab\left( {{x}^{2}}+{{y}^{2}} \right)-xy\left( {{a}^{2}}+{{b}^{2}} \right)=ax\left( bx-ay \right)-by\left( bx-ay \right)$
Step 3: Taking $bx-ay$ common from the terms, we get
$ab\left( {{x}^{2}}+{{y}^{2}} \right)-xy\left( {{a}^{2}}+{{b}^{2}} \right)=\left( bx-ay \right)\left( ax-by \right)$, which is the required factored from of the given expression

Note: Verification:
We have
$\left( bx-ay \right)\left( ax-by \right)=\left( bx \right)\left( ax \right)-\left( bx \right)\left( by \right)-\left( ay \right)\left( ax \right)+\left( ay \right)\left( by \right)$
Hence, we have
$\left( bx-ay \right)\left( ax-by \right)=ab{{x}^{2}}-{{b}^{2}}xy-{{a}^{2}}xy+ab{{y}^{2}}$
Taking ab common from the first and the last term and xy common form the second and the third term, we get
$\left( bx-ay \right)\left( ax-by \right)=ab\left( {{x}^{2}}+{{y}^{2}} \right)-xy\left( {{a}^{2}}+{{b}^{2}} \right)$
Hence our answer is verified to be correct.
[2] Alternative Solution:
We have
$ab\left( {{x}^{2}}+{{y}^{2}} \right)-xy\left( {{a}^{2}}+{{b}^{2}} \right)={{x}^{2}}ab-xy\left( {{a}^{2}}+{{b}^{2}} \right)+ab{{y}^{2}}$, which is quadratic in x.
We will factorise the expression using the method of completing the square.
Making coefficient of ${{x}^{2}}$ equal to 1
We have $ab\left( {{x}^{2}}+{{y}^{2}} \right)-xy\left( {{a}^{2}}+{{b}^{2}} \right)=ab\left( {{x}^{2}}-\dfrac{xy\left( {{a}^{2}}+{{b}^{2}} \right)}{ab}+{{y}^{2}} \right)$
Adding and subtracting ${{\left( \dfrac{y\left( {{a}^{2}}+{{b}^{2}} \right)}{2ab} \right)}^{2}}$, we get
$\begin{align}
  & ab\left( {{x}^{2}}+{{y}^{2}} \right)-xy\left( {{a}^{2}}+{{b}^{2}} \right)=ab\left( {{x}^{2}}-2\times \left( \dfrac{y\left( {{a}^{2}}+{{b}^{2}} \right)}{2ab} \right)+{{\left( \dfrac{y\left( {{a}^{2}}+{{b}^{2}} \right)}{2ab} \right)}^{2}}-{{\left( \dfrac{y\left( {{a}^{2}}+{{b}^{2}} \right)}{2ab} \right)}^{2}}+{{y}^{2}} \right) \\
 & \Rightarrow ab\left( {{x}^{2}}+{{y}^{2}} \right)-xy\left( {{a}^{2}}+{{b}^{2}} \right)=ab\left( {{\left( x-\dfrac{y\left( {{a}^{2}}+{{b}^{2}} \right)}{2ab} \right)}^{2}}-{{y}^{2}}{{\left( \dfrac{{{a}^{2}}-{{b}^{2}}}{2ab} \right)}^{2}} \right) \\
\end{align}$
Hence, we have
$\begin{align}
  & ab\left( {{x}^{2}}+{{y}^{2}} \right)-xy\left( {{a}^{2}}+{{b}^{2}} \right)=ab\left( x-\dfrac{y\left( {{a}^{2}}+{{b}^{2}} \right)}{2ab}+y\left( \dfrac{{{a}^{2}}-{{b}^{2}}}{2ab} \right) \right)\left( x-\dfrac{y\left( {{a}^{2}}+{{b}^{2}} \right)}{2ab}-y\left( \dfrac{{{a}^{2}}-{{b}^{2}}}{2ab} \right) \right) \\
 & =ab\left( x-\dfrac{ya}{b} \right)\left( x-\dfrac{yb}{a} \right)=\left( bx-ay \right)\left( ax-by \right) \\
\end{align}$