Question

Factorise the following algebraic expression:
${{x}^{4}}+2{{x}^{2}}+9$
[a] $\left( {{x}^{2}}-2x+3 \right)\left( {{x}^{2}}+2x+3 \right)$
[b] $\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}-3 \right)$
[c] The expression cannot be factorised.
[d] $\left( x+3 \right)\left( x-2 \right)$

Answer 1

Hint:Add and subtract $6{{x}^{2}}$ to the expression. Use the formula ${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$ followed by the use of the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. Hence factorise the given expression.

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
$\begin{aligned} & \left( a+b \right)\left( a-b \right)={{a}^{2}}+ab-ab-{{b}^{2}} \\\ & ={{a}^{2}}-{{b}^{2}} \\\ \end{aligned}$
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
Step 1: Add and subtract $6{{x}^{2}}$
We have ${{x}^{4}}+2{{x}^{2}}+9={{x}^{4}}+6{{x}^{2}}+9-6{{x}^{2}}+2{{x}^{2}}$
Step2: Use ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Put $a={{x}^{2}}$ and $b=3$, we get
${{\left( {{x}^{2}}+3 \right)}^{2}}={{x}^{4}}+6{{x}^{2}}+9$
Hence, we have
${{x}^{4}}+2{{x}^{2}}+9={{\left( {{x}^{2}}+3 \right)}^{2}}-4{{x}^{2}}$
Now we will write the above expression in the form of ${{a}^{2}}-{{b}^{2}}$
We have
${{\left( {{x}^{2}}+3 \right)}^{2}}-4{{x}^{2}}={{\left( {{x}^{2}}+3 \right)}^{2}}-{{\left( 2x \right)}^{2}}$
Hence, we have
${{x}^{4}}+2{{x}^{2}}+9={{\left( {{x}^{2}}+3 \right)}^{2}}-{{\left( 2x \right)}^{2}}$
Step 3: Factorise the expression using $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$
We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. Using the above formula, we get
${{\left( {{x}^{2}}+3 \right)}^{2}}-4{{x}^{2}}=\left( {{x}^{2}}+3-2x \right)\left( {{x}^{2}}+3+2x \right)=\left( {{x}^{2}}-2x+3 \right)\left( {{x}^{2}}+2x+3 \right)$ which is in factorised form.
Hence, we have
${{x}^{4}}+2{{x}^{2}}+9=\left( {{x}^{2}}-2x+3 \right)\left( {{x}^{2}}+2x+3 \right)$
Hence option [a] is correct.

Note: [1] In step 2 we need to combine ${{x}^{4}}+6{{x}^{2}}+9$ to get ${{\left( {{x}^{2}}+3 \right)}^{2}}$ and not ${{x}^{4}}-6{{x}^{2}}+9$ to get ${{\left( {{x}^{2}}-3 \right)}^{2}}$ 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.