Ex.2.5 Q3 Polynomials Solution - NCERT Maths Class 9

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Question

Factorise the following using appropriate identities:

(i) \(\begin{align}9 x^{2}+6 x y+y^{2}\end{align}\)

(ii) \(\begin{align} 4 y^{2}-4 y+1\end{align}\) 
(iii) \(\begin{align} x^{2}-\frac{y^{2}}{100}\end{align}\)

 

 Video Solution
Polynomials
Ex 2.5 | Question 3

Text Solution

Reasoning:

Identities:

\[\begin{align}& {(a+b)^{2}=a^{2}+2 a b+b^{2}} \\ {} & {(a-b)^{2}=a^{2}-2 a b+b^{2}} \\ {} & {(a+b)(a-b)=a^{2}-b^{2}}\end{align}\]

Steps:

(i) \(\begin{align}9 x^{2}+6 x y+y^{2}=(3 x)^{2}+2(3 x)(y)+(y)^{2}\end{align}\)

Identity:  \(\begin{align}(a+b)^{2}=a^{2}+2 a b+b^{2}\end{align}\)

Here \(\begin{align} a=3 x, b=y \end{align}\) 

Hence   \(\begin{align}9 x^{2}+6 x y+y^{2}=(3 x+y)^{2}\end{align}\)

(ii) \(\begin{align}4 y^{2}-4 y+1=\left(2 y^{2}\right)-2(2 y)(1)+(1)^{2}\end{align}\)

Identity: \(\begin{align}(a-b)^{2}=a^{2}-2 a b+b^{2}\end{align}\)

Here \(\begin{align} a=2 y, b=1\end{align}\) 

Hence \(\begin{align} 4 y^{2}-4 y+1=(2 y-1)^{2}\end{align}\)

(iii) \(\begin{align}x^{2}-\frac{y^{2}}{100}=(x)^{2}-\left(\frac{y}{10}\right)^{2}\end{align}\)

Identity: \(\begin{align}(a+b)(a-b)=a^{2}-b^{2}\end{align}\)

Here \(\begin{align}&{ a=x, b=\frac{y}{10}} \end{align}\) 

Hence \(\begin{align}{ x^{2}-\frac{y^{2}}{100}=\left(x+\frac{y}{10}\right)\left(x-\frac{y}{10}\right)}\end{align}\)