Question

Factorize the following using appropriate identities:
A.$9{{x}^{2}}+6xy+{{y}^{2}}$
B.$4{{y}^{2}}-4y+1$
C.${{x}^{2}}-\dfrac{{{y}^{2}}}{100}$

Answer 1

Hint:Use the algebraic identities ${{\left( a+b \right)}^{2}},{{\left( a-b \right)}^{2}},{{a}^{2}}-{{b}^{2}}$ to factorize the given equations. And try to observe the given relations carefully for solving them. Use the following algebraic identities wherever required:
$\begin{align}
  & {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\
 & {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} \\
 & {{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right) \\
\end{align}$

Complete step-by-step answer:
i) $9{{x}^{2}}+6xy+{{y}^{2}}$
As we know the algebraic identity of ${{\left( a+b \right)}^{2}}$ can be expressed as
${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ …………………… (i)
Now, observe the right hand side of the expression (i) and the given equation i.e. $9{{x}^{2}}+6xy+{{y}^{2}}$.
We can rewrite the terms of the given equation $9{{x}^{2}}+6xy+{{y}^{2}}$ by replacing $9{{x}^{2}}$ by${{\left( 3x \right)}^{2}}$ , ${{y}^{2}}$ term as ${{\left( y \right)}^{2}}$ and 6xy can be rearranged as$\left( 2\times 3x\times y \right)$. So, we can write the given equation as:
$9{{x}^{2}}+6xy+{{y}^{2}}={{\left( 3x \right)}^{2}}+\left( 2\times 3x\times y \right)+{{\left( y \right)}^{2}}$
Now, compare the right hand side of the above expression with the right hand side of the equation (i), so, we get
a = 3x, b = y
So, we can replace the given equation by the identity (i) as
${{\left( 3x \right)}^{2}}+2\times 3x\times y+{{\left( y \right)}^{2}}={{\left( 3x+y \right)}^{2}}$
Or
$9{{x}^{2}}+6xy+{{y}^{2}}={{\left( 3x+y \right)}^{2}}$
Hence, we can give factorization of equation $9{{x}^{2}}+6xy+{{y}^{2}}$ as${{\left( 3x+y \right)}^{2}}$.
$4{{y}^{2}}-4y+1$
We know the algebraic identity of ${{\left( a-b \right)}^{2}}$ can be given as
${{\left( a-b \right)}^{2}}={{a}^{a}}-2ab+{{b}^{2}}$ …………….. (ii)
Now, observe the right hand side of the equation (ii) i.e. ${{a}^{2}}-2ab+{{b}^{2}}$, with the given equation in the problem i.e. $4{{y}^{2}}-4y+1$.
So, we can rewrite the terms of$4{{y}^{2}}-4y+1$, by replacing $4{{y}^{2}}$by${{\left( 2y \right)}^{2}}$, 4y as $\left( 2\times 2y\times 1 \right)$ and term ‘1’ as${{\left( 1 \right)}^{2}}$. So, we get
$4{{y}^{2}}-4y+1={{\left( 2y \right)}^{2}}-2\times 2y\times 1+{{\left( 1 \right)}^{2}}$
Now, compare the right hand side of the above expression with the right hand side of equation (ii). So, we get
a = 2y and b = 1. Hence, we get
$4{{y}^{2}}-4y+1={{\left( 2y \right)}^{2}}-2\times 2y\times 1+{{\left( 1 \right)}^{2}}={{\left( 2y-1 \right)}^{2}}$
Or
$4{{y}^{2}}-4y+1={{\left( 2y-1 \right)}^{2}}$
Hence, we can give factorization of $4{{y}^{2}}-4y+1$ as ${{\left( 2y-1 \right)}^{2}}$
${{x}^{2}}-\dfrac{{{y}^{2}}}{100}$
As we know the algebraic identity of ${{a}^{2}}-{{b}^{2}}$ can be given as
${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ …………. (iii)
Hence, observe the left hand side of the equation (iii) to the given expression i.e.
${{x}^{2}}-\dfrac{{{y}^{2}}}{100}$ .
So, we can rewrite the terms ${{x}^{2}}$ as ${{\left( x \right)}^{2}}$ and $\dfrac{{{y}^{2}}}{100}$ as ${{\left( \dfrac{y}{10} \right)}^{2}}$ in the expression ${{x}^{2}}-\dfrac{{{y}^{2}}}{100}$ . Hence, we get
${{x}^{2}}-\dfrac{{{y}^{2}}}{100}={{\left( x \right)}^{2}}-{{\left( \dfrac{y}{10} \right)}^{2}}$
Now, compare the right hand side of the above expression to the left hand side of the equation (iii). So, we get values of ‘a’ and ‘b’ as
a = x and\[b=\dfrac{y}{10}\].
So, we get
${{x}^{2}}-\dfrac{{{y}^{2}}}{100}={{\left( x \right)}^{2}}-{{\left( \dfrac{y}{10} \right)}^{2}}=\left( x-\dfrac{y}{10} \right)\left( x+\dfrac{y}{10} \right)$
Or
${{x}^{2}}-\dfrac{{{y}^{2}}}{100}=\left( x-\dfrac{y}{10} \right)\left( x+\dfrac{y}{10} \right)$
Hence, the factorization of the given equation ${{x}^{2}}-\dfrac{{{y}^{2}}}{100}$ is given as $\left( x-\dfrac{y}{10} \right)\left( x+\dfrac{y}{10} \right)$

Note: Observing the given equations and relating them with the algebraic identities is the key point of the question.
One may factorize $4{{y}^{2}}-4y+1$ as ${{\left( 1-2y \right)}^{2}}$ as well. As we can write the expression as ${{\left( 1 \right)}^{2}}-2\times 1\times 2y+{{\left( 2y \right)}^{2}}$ and hence, on comparing with${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$,
We can write $4{{y}^{2}}-4y+1$ as${{\left( 1-2y \right)}^{2}}$. So, if someone factorizes it as${{\left( 1-2y \right)}^{2}}$, don't be confused with that part of the solution.