Ex.14.2 Q2 Factorization - NCERT Maths Class 8

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Question

Factorize

(i)\(\begin{align} 4{p^2} - 9{q^2}\end{align}\)

(ii)\(\begin{align} 63{a^2} - 112{b^2}\end{align}\)

(iii)\(\begin{align} 49{x^2} - 36\end{align}\)

(iv)\(\begin{align} 16{x^5} - 144{x^3}\end{align}\)

(v)\(\begin{align} {{(l + m)}^2} - 16\end{align}\)

(vi)\(\begin{align} 9{x^2}{y^2} - 16\end{align}\)

(vii)\(\begin{align} \left( {{x^2} - 2xy + {y^2}} \right) - {z^2}\end{align}\)

(viii)\(\begin{align}  25{a^2} - 4{b^2} + 28bc - 49{c^2} \end{align}\)

Text Solution

What is known:

Algebraic expression.

What is unknown:

Factorisation of the algebraic expression.

Reasoning: Use identity:

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

Steps:

\(\begin{align}{\rm{(i)}} \quad 4{p^2} - 9{q^2} = {{(2p)}^2} - {{(3q)}^2}\\= (2p + 3q)(2p - 3q)\end{align}\)

\[\begin{align}\text{Using }{a^2} - {b^2} &= (a - b)(a + b),\\\text{considering}\;a &= 2p\;{\rm{and}} \;b = 3q\end{align}\]

\(\begin{align}\text { (ii) } \quad & 63 a^{2}-112 b^{2} \\ &=7\left(9 a^{2}-16 b^{2}\right) \\ &=7\left[(3 a)^{2}-(4 b)^{2}\right] \\ &=7[(3 a+4 b)(3 a-4 b)] \end{align}\)

\[\begin{bmatrix} \text{Using identity } {x^2} - {y^2} \\= (x - y)(x + y),\\\!\text{considering }x = 3a \\ {\rm{and}}\;y = 4b\end{bmatrix}\]

\(\begin{align}({\rm{iii}}) \quad &49{x^2} - 36 \\&= {{(7x)}^2} - {{(6)}^2}\\ &= (7x - 6)(7x + 6)\end{align}\)

\[ \begin{bmatrix} \text{Using identity}\;{a^2} - {b^2} \\ = (a - b)(a + b),\\\!\text{considering}\;a = 7x\; \\ {\rm{and}} \;b = 6.\end{bmatrix}\]

\(\begin{align}{\rm{(iv)}} \quad &16{x^5} - 144{x^3} \\ &= 16{x^3}\left( {{x^2} - 9} \right)\\&= 16{x^3}\left[ {{{(x)}^2} - {{(3)}^2}} \right]\\&= 16{x^3}\left[ {(x - 3)(x + 3) }\right]\end{align}\)\[\begin{bmatrix} \text{Using identity }{a^2} - {b^2}\\ = (a - b)(a + b)\\ \text{Considering }a = x\\ {\rm{and}}\;b = 3.\end{bmatrix} \]

\(\begin{align}({\rm{v}} ) \quad & (l+m)^{2}-(l-m)^{2} \\ \\ &= \begin{Bmatrix} [(l+m)-(l-m)] \\ [(l+m)+(l-m)] \end{Bmatrix} \\ &= \begin{Bmatrix} (l+m-l+m) \\ (l+m+l-m) \end{Bmatrix} \\ &=2 m \times 2 l \\ &=4 m l \\ &=4 l m\end{align}\)

\[\begin{bmatrix} \text{Using identity }{a^2} - {b^2} \\ = (a - b)(a + b)\\ \text{Considering }a = (l+m)\\ {\rm{and}}\;b = (l-m). \end{bmatrix}\]

\(\begin{align}({\rm{vi}}) \quad & 9{x^2}{y^2} - 16 \\ &= {(3xy)}^2 - {(4)}^2\\&= (3xy - 4)(3xy + 4)\end{align}\)

\[\begin{bmatrix} \text { Using the identity } a^{2}-b^{2}\\ =(a-b)(a+b), \\ \text { considering } a=3 x y \\ \text { and } b=4\end{bmatrix}\]

\(\begin{align} \left( {\rm{vii}} \right)\;\;\;&\left( {{x^2} - 2xy + {y^2}} \right) - {z^2} \\ &= \left( {x - y} \right)^2 - \left( z \right)^2 \\ & \quad \begin{bmatrix} {\rm{Using}\,\, \text{identity}}\;{\left( {a - b} \right)^2} \\ = {a^2} - 2ab + {b^2}\\ {\rm{for}}\; \left( {x - y} \right)^2 \\ = {x^2} - 2xy + {y^2} \end{bmatrix} \\ &=\left ( x- y-z \right)\left( x-y+z \right) \\ &  \quad  \begin{bmatrix} \text{Using identity}\ {a}^{2}-{b}^{2} \\=\left( a-b \right)\left( a+b \right) \\ \text{considering}\ a=x-y\ \\ \text{and}\ b=z.\end{bmatrix} \end{align}\)

\(\begin{align}{\rm{(viii)}}\;\;\;\,\;& 25{a^2} - 4{b^2} + 28bc - 49{c^2} \\ & = \begin{Bmatrix} 25{a^2} - \\ \left( \! {4{b^2} -\! 28bc +\! 49{c^2}} \right) \end{Bmatrix} \\ & = \!\!\begin{Bmatrix} {(5a)^2} - \\ \begin{bmatrix} (2b)^2 \!- 2 \times \! 2b \times \! 7c \\ + (7c)^2\end{bmatrix} \end{Bmatrix} \\ & \; \begin{bmatrix}{\rm{Using}}\;\;{\rm{identity }}{(x - y)^2} \\= {x^2} - 2xy + {y^2}\\\text{considering}\;x = 2b\; \\ \text{and}\;y = 7c. \end{bmatrix} \\ & = {(5a)^2} - {(2b - 7c)^2} \\ & \; \begin{bmatrix} {\rm{Using} \,\,\text{identity }}{x^2} - {y^2} \\ = (x - y)(x + y)\\\text{considering}\;x = 5a\; \\ \text{and}\;y = 2b - 7c. \end{bmatrix}\\ & = \begin{Bmatrix} [5a + (2b - 7c)] \\ [5a - (2b - 7c)] \end{Bmatrix} \\ &= \begin{Bmatrix} (5a + 2b - 7c) \\ (5a - 2b + 7c) \end{Bmatrix} \end{align}\)

  
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