Question

Factorise the given algebraic expression
${{x}^{4}}+4{{y}^{4}}$

Answer 1

Hint: Add and subtract $4{{x}^{2}}{{y}^{2}}$ to the expression. Use the formula ${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$ followed by the use of the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$

Complete 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.
The expression $\left( a+b \right)\left( a-b \right)$is said to be factorised form of ${{a}^{2}}-{{b}^{2}}$. When factorising an expression, we make use of algebraic identities like ${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$,${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$,${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$, etc. In this question, we will make use of the identities ${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$ and ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to factorise the given expression
We will first complete the square.
Step 1: Express in form \[{{a}^{2}}+{{b}^{2}}\]
We have
${{x}^{4}}+4{{y}^{4}}={{\left( {{x}^{2}} \right)}^{2}}+{{\left( 2{{y}^{2}} \right)}^{2}}$
Step 2: Add and subtract 2ab
Here $a={{x}^{2}}$ and $b=2{{y}^{2}}$.
So, we have $2ab=2{{x}^{2}}\left( 2{{y}^{2}} \right)=4{{x}^{2}}{{y}^{2}}$
Adding and subtracting $4{{x}^{2}}{{y}^{2}}$ we get
$\begin{align}
  & {{x}^{4}}+4{{y}^{4}}={{\left( {{x}^{2}} \right)}^{2}}+{{\left( 2{{y}^{2}} \right)}^{2}}+4{{x}^{2}}{{y}^{2}}-4{{x}^{2}}{{y}^{2}} \\
 & ={{\left( {{x}^{2}}+2{{y}^{2}} \right)}^{2}}-4{{x}^{2}}{{y}^{2}} \\
\end{align}$
Now we will write the above expression in the form of ${{a}^{2}}-{{b}^{2}}$
We have
${{\left( {{x}^{2}}+2{{y}^{2}} \right)}^{2}}-4{{x}^{2}}{{y}^{2}}={{\left( {{x}^{2}}+2{{y}^{2}} \right)}^{2}}-{{\left( 2xy \right)}^{2}}$
Hence we have
${{x}^{4}}+4{{y}^{4}}={{\left( {{x}^{2}}+2{{y}^{2}} \right)}^{2}}-{{\left( 2xy \right)}^{2}}$
We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. Using the above formula we get
${{x}^{4}}+4{{y}^{4}}=\left( {{x}^{2}}+2{{y}^{2}}-2xy \right)\left( {{x}^{2}}+2{{y}^{2}}+2xy \right)$ which is in factorised form.

Note: In step 2 we need to combine ${{\left( {{x}^{2}} \right)}^{2}}+{{\left( 2{{y}^{2}} \right)}^{2}}+4{{x}^{2}}{{y}^{2}}$ to get ${{\left( {{x}^{2}}+2{{y}^{2}} \right)}^{2}}$ and not ${{\left( {{x}^{2}} \right)}^{2}}+{{\left( 2{{y}^{2}} \right)}^{2}}-4{{x}^{2}}{{y}^{2}}$ to get ${{\left( {{x}^{2}}-2{{y}^{2}} \right)}^{2}}$ because the former leads to an expression of the form ${{a}^{2}}-{{b}^{2}}$ whereas the latter leads to an expression of the form ${{a}^{2}}+{{b}^{2}}$. ${{a}^{2}}-{{b}^{2}}$ can be factorised whereas ${{a}^{2}}+{{b}^{2}}$ cannot.