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:

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

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

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

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

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

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

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

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

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

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

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

\(\begin{align}{\left( {{\rm{vii}}} \right)\;\;\;\;\;\left( {{x^2} - 2xy + {y^2}} \right) - {z^2} = {{\left( {x - y} \right)}^2} - {{\left( z \right)}^2}{\rm{ }} }\;\;[ {\rm{Using}\,\, \text{identity}}\;{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\\{\rm{for}}\;{\left( {x - y} \right)^2} = {x^2} - 2xy + {y^2} ]\end{align}\)

\[\begin{align}\left ( x-\text{ }y-z \right)\left( x-y+z \right)[  \text{Using identity}\ {a}^{2}-{b}^{2}=\left( a-b \right)\left( a+b \right) \\   \ \text{considering}\ a=x-y\ \text{and}\ b=z.\end{align}\]

\(\begin{align}{\rm{(viii)}}\;\;\;\,\;25{a^2} - 4{b^2} + 28bc - 49{c^2}& = 25{a^2} - \left( {4{b^2} - 28bc + 49{c^2}} \right)\\ & = {(5a)^2} - \left[ {{{(2b)}^2} - 2 \times 2b \times 7c + {{(7c)}^2}} \right]\\\end{align}\)

\[\begin{align}[{\rm{Using}}\;\;{\rm{identity }}{(x - y)^{\rm{2}}} &= {x^2} - 2xy + {y^2}\\\text{considering}\;x &= 2b\;\text{and}\;y = 7c. ] \\&  = {(5a)^2} - {(2b - 7c)^2}\end{align}\]

\[ \begin{align}{\rm{Using} \,\,\text{identity }}{x^2} - {y^2} &= (x - y)(x + y)\\\text{considering}\;x = 5a\;\text{and}\;y &= 2b - 7c.\\ &  = [5a + (2b - 7c)][5a - (2b - 7c)]\\ &= (5a + 2b - 7c)(5a - 2b + 7c)\end{align}\]

  
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