Exercise 14.2 Factorization- NCERT Solutions Class 8

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Chapter 14 Ex.14.2 Question 1

Factorize the following expressions.

(i) \(\begin{align}  {a^2} + 8a + 16\end{align}\)

(ii) \(\begin{align}  {p^2} - 10p + 25\end{align}\)

(iii) \(\begin{align}  25{m^2} + 30m + 9\end{align}\)

(iv) \(\begin{align} 49{y^2}  + 84yz + 36{z^2}\end{align}\)

(v) \(\begin{align}  4{x^2} - 8x + 4\end{align}\)

(vi) \(\begin{align}  121{b^2} - 88bc + 16{c^2}\end{align}\)

(vii) \( {{(l + m)}^2} - 4lm \)

(Hint: Expand \({{\left( {l + m} \right)}^2}\) first) 

(viii) \(\begin{align}  {a^4} + 2{a^2}{b^2} + {b^4} \end{align}\)

Solution

Video Solution

What is known:

Algebraic expression.

What is unknown:

Factorisation of the algebraic expression.

Reasoning: Use identity:

\[\begin{align}  & {{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}} \\  & {{\left( x-y \right)}^{2}}={{x}^{2}}-2xy+{{y}^{2}} \\ \end{align}\]

Steps:

\(\begin{align}({\rm{i}}) \,{a^2} + 8a + 16 &=  \!\!{{(a)}^2} + 2 \times \!a \times 4 +\! {{(4)}^2} \\&= {{(a + 4)}^2} \end{align}\)

Using identity\(\begin{align}{(x + y)^2} = {x^2} + 2xy + {y^2}\end{align}\),considering \(x = a\)  and \(y = 4\) 

\(\begin{align}({\rm{ii}}) \, & {p^2} - 10p + 25 \\ &=  {(p)}^2 - 2 \times p  \times 5 + {(5)}^2 \\&= {(p - 5)}^2 \end{align}\)

Using identity\(\begin{align}(a - b)^2  = {a^2} - 2ab + {b^2}\end{align}\),considering \(a = p\)  and \(y = 5 \)

\(\begin{align}({\rm{iii}} ) \, & 25{m^2} + 30m + 9 \\&=  {{(5m)}^2} + 2 \times 5m \times 3 + {{(3)}^2} \\&= {{(5m + 3)}^2} \end{align}\)

Using identity\(\begin{align}(a +b)^2  = {a^2} + 2ab + {b^2}\end{align}\),considering \(a = 5m\)  and  \(b = 3\) 

\(\begin{align}({\rm{iv}}) \, & 49{y^2} + 84yz +36{z^2}\\&= {{(7y)}^2} + 2 \times (7y) \times (6z) + {{(6z)}^2}  \\ &= {{(7y + 6z)}^2}\end{align}\)

Using identity\(\begin{align}(a +b)^2  = {a^2} + 2ab + {b^2}\end{align}\),considering \(a = 7y\)  and \(b = 6\)

\(\begin{align}({\rm{v}}) \, & 4{x^2} - 8x + 4 \\ &= {{(2x)}^2} - 2(2x)(2)  + {{(2)}^2}  \\&= {{(2x - 2)}^2} \\&= {{[(2)(x - 1)]}^2} \\ &= 4{{(x - 1)}^2}\end{align}\)

Using identity\(\begin{align}(a -b)^2  = {a^2} - 2ab + {b^2}\end{align}\),considering \(a = 2x\)  and \(y = 2\)

\(\begin{align}({\rm{vi}})\, &121{b^2} - 88bc + 16{c^2} \\ &={{(11b)}^2} - 2(11b)(4c)  + {{(4c)}^2} \\ &= {(11b - 4c)}^2\end{align}\)

Using identity\(\begin{align}(a -b)^2  = {a^2} - 2ab + {b^2}\end{align}\),considering \(a = 11b\)  and \(b = 4c\)

\(\begin{align}({\rm{vii}})\,& {{(l + m)}^2} - 4lm \\ &=  {l^2} + 2lm +  {m^2} - 4lm  \\&= {l^2} - 2lm + {m^2}\\&= {(l - m)}^2 \end{align}\)

Using identity\(\begin{align}(a -b)^2  = {a^2} - 2ab + {b^2}\end{align}\),considering \(a = l \) and \(b = m\)

\(\begin{align}{\rm{(viii) }} \, & {a^4} + 2{a^2}{b^2} + {b^4} \\ &=  {{\left( {{a^2}} \right)}^2} + 2\left( {{a^2}} \right)\left( {{b^2}} \right)  + {{\left( {{b^2}} \right)}^2}  \\&= {{\left( {{a^2} + {b^2}} \right)}^2} \end{align}\)

Using identity\(\begin{align}(x + y)^2 = {x^2} + 2xy + {y^2}\end{align}\),considering \(x = {a^2} \) and \(y = {b^2}\)

Chapter 14 Ex.14.2 Question 2

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

Solution

Video 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:

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

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

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

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

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

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

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

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

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

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

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

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

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

Chapter 14 Ex.14.2 Question 3

Factorise the expressions

(i) \(\begin{align}a{x^2} + bx\end{align}\)

(ii) \(\begin{align}7{p^2} + 21{q^2}\end{align}\)

(iii) \(\begin{align}2{x^3} + 2x{y^2} + 2x{z^2}\end{align}\)

(iv) \(\begin{align}a{m^2} + b{m^2} + b{n^2} + a{n^2}\end{align}\)

(v) \(\begin{align}(lm + l) + m + 1\end{align}\)

(vi) \(\begin{align}y(y + z) + 9(y + z)\end{align}\)

(vii) \(\begin{align}5{y^2} - 20y - 8z + 2yz\end{align}\)

(viii) \(\begin{align}10ab + 4a + 5b + 2\end{align}\)

(ix) \(\begin{align}6xy - 4y + 6 - 9x\end{align}\)

Solution

Video Solution

What is known:

Algebraic expression.

What is unknown:

Factorisation of given algebraic expression.

Reasoning: For part (\(\rm{i}\)), (\(\rm{ii}\)), (\(\rm{iii}\)) and (\(\rm{vi}\)) - First we will find factors of each terms then find out which factors are common in each term and take out that common factor from expression.

For part (\(\rm iv\)), (\(\rm v\)), (\(\rm vii\)), (\(\rm viii\)), (\(\rm ix\)) - There are \(4\) terms in each expression. First we will make pair of two terms from which we can take out common factors and convert the expression of \(4\) terms into \(2\) terms expression then take out common factors from remaining \(2\) terms.

Steps:

\( \begin{align} {\rm{(i)}}\quad &a{x^2} + bx \\ & = a \times x \times x + b \times x \\ & = x(ax + b) \end{align} \)

\( \begin{align}  \left( {{\rm{ii}}} \right)\quad & 7{p^2} + 21{q^2} \\ &= 7 \times p \times p + 3 \times 7 \times q \times q \\ & = 7\left( {{p^2} + 3{q^2}} \right) \end{align} \)

\( \begin{align} \left( {{\rm{iii}}} \right) \quad & 2{x^3} + 2x{y^2} + 2x{z^2} \\ & = 2x\left( {{x^2} + {y^2} + {z^2}} \right) \end{align} \)

\(\begin{align}({\rm{iv}})\quad &a{m^2} + b{m^2} + b{n^2} + a{n^2} \\ &= a{m^2} + b{m^2} + a{n^2} + b{n^2}\\&= {m^2}(a + b) + {n^2}(a + b)\\&= (a + b)\left( {{m^2} + {n^2}} \right) \end{align}\)

\(\begin{align}({\rm{v}})\quad & \left( {lm + l} \right) + m + 1 \\ &= lm + m + l + 1\\&= m\left( {l + 1} \right) + 1\left( {l + 1} \right)\\&= \left( {l + 1} \right)\left( {m + 1} \right)\end{align}\)

\( \begin{align} \left( {{\rm{vi}}} \right)\quad &y{\rm{(}}y + z{\rm{)}} + {\rm{9(}}y + z{\rm{)}} \\ & = {\rm{(}}y + z{\rm{)(}}y + {\rm{9)}} \end{align} \)

\(\begin{align}{\rm{(vii)}}\quad & 5{y^2} - 20y - 8z + 2yz \\ &= 5{y^2} - 20y + 2yz - 8z\\&= 5y(y - 4) + 2z(y - 4)\\&= (y - 4)(5y + 2z)\end{align}\)

\(\begin{align}{\rm{(viii)}}\quad  &10ab + 4a + 5b + 2 \\ &= 10ab + 5b + 4a + 2\\&= 5b(2a + 1) + 2(2a + 1)\\&= (2a + 1)(5b + 2)\end{align}\)

\(\begin{align}({\rm{ix)}}\quad & 6xy - 4y + 6 - 9x \\ &= 6xy - 9x - 4y + 6\\&= 3x(2y - 3) - 2(2y - 3)\\ &= (2y - 3)(3x - 2)\end{align}\)

Chapter 14 Ex.14.2 Question 4

 Factorise

(i) \(\begin{align}{a^4} - {b^4}\end{align}\)

(ii) \(\begin{align}{p^4} - 81\end{align}\)

(iii) \(\begin{align}{x^4} - {{(y + z)}^4}\end{align}\)

(vi) \(\begin{align}{x^4} - {{(x - z)}^4}\end{align}\)

(v) \(\begin{align} {a^4} - 2{a^2}{b^2} + {b^4} \end{align}\)

Solution

Video Solution

What is known:

Algebraic expression.

What is unknown:

Factorisation of the algebraic expression.

Reasoning: Use identity:

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

Steps:

\(\begin{align}{\rm{(i)}}\quad & {a^4} - {b^4} \\ &= {{\left( {{a^2}} \right)}^2} - {{\left( {{b^2}} \right)}^2}\\&= \left( {{a^2} - {b^2}} \right)\left( {{a^2} + {b^2}} \right)\\&= (a - b)(a + b)\left( {{a^2} + {b^2}} \right)\end{align}\)

\(\begin{align}{\rm{(ii)}}\quad & {p^4} - 81 \\ &= {{\left( {{p^2}} \right)}^2} - {{(9)}^2}\\&= \left( {{p^2} - 9} \right)\left( {{p^2} + 9} \right)\\&= \left[ {{{(p)}^2} - {{(3)}^2}} \right]\left( {{p^2} + 9} \right)\\&= (p - 3)(p + 3)\left( {{p^2} + 9} \right)\end{align}\)

\(\begin{align}\left( {{\rm{iii}}} \right) \quad & {x^4} - {{(y + z)}^4} \\ &= {{\left( {{x^2}} \right)}^2} - {{\left[ {{{(y + z)}^2}} \right]}^2}\\&= \begin{Bmatrix} \left[ {{x^2} - {{(y + z)}^2}} \right] \\ \left[ {{x^2} + {{(y + z)}^2}} \right] \end{Bmatrix} \\ &= \begin{Bmatrix}  [x - (y + z)] \\ [x + (y + z)] \\ \left[ {{x^2} + {{(y + z)}^2}} \right] \end{Bmatrix} \\&= \begin{Bmatrix} (x - y - z) \\ (x + y + z)\\ \left[ {{x^2} + {{(y + z)}^2}} \right] \end{Bmatrix} \end{align}\)

\(\begin{align} \left( \rm{iv} \right)\quad & {x^{\rm{4}}} - {{\left( {x - z} \right)}^{\rm{4}}} \\ &= {{\left( {{x^2}} \right)}^2} - {{\left[ {{{\left( {x - z} \right)}^2}} \right]}^2}\\ &=\begin{Bmatrix}  \left[ {{x^2} - {{\left( {x - z} \right)}^{\rm{2}}}} \right] \\ \left[ {{x^{\rm{2}}} + {{\left( {x - z} \right)}^{\rm{2}}}} \right] \end{Bmatrix} \\ &= \begin{Bmatrix} \left[ {x - \left( {x - z} \right)} \right] \\ \left[ {x + \left( {x - z} \right)} \right] \\ \left[ {{x^{\rm{2}}} + {{\left( {x - z} \right)}^{\rm{2}}}} \right] \end{Bmatrix} \\ &= \begin{Bmatrix} z\left( {2x - z} \right) \\ \begin{bmatrix}  x^{\rm{2}} + x^{\rm{2}} \\ - 2xz + {z^{\rm{2}}} \end{bmatrix} \end{Bmatrix} \\&= \begin{Bmatrix} z \left( {{\rm{2}}x - z} \right) \\ \left( {{\rm{2}}{x^{\rm{2}}} - {\rm{2}}xz + {z^{\rm{2}}}} \right)\end{Bmatrix} \end{align}\)

\(\begin{align}({\rm{v}})\quad & {a^4} - 2{a^2}{b^2} + {b^4} \\ &= \begin{Bmatrix} \left( {{a^2}} \right)^2 - 2\left( {{a^2}} \right)\left( {{b^2}} \right) \\ + \left( {{b^2}} \right)^2 \end{Bmatrix} \\&= {{\left( {{a^2} - {b^2}} \right)}^2}\\&= {{\left[ {(a - b)(a + b)} \right]}^2}\\&= {{(a - b)}^2}{{(a + b)}^2}\end{align}\)

Chapter 14 Ex.14.2 Question 5

Factorise the following expressions

(i) \(\begin{align}{p^2} + 6p + 8\end{align}\)

(ii) \(\begin{align}{q^2} - 10q + 21\end{align}\)

(iii) \(\begin{align}{p^2} + 6p - 16\end{align}\)

Solution

Video Solution

What is known:

Algebraic expression.

What is unknown:

Factorisation of the algebraic expression.

Reasoning:

In general, for factorising an algebraic expression of the type \({{x}^{2}}+px+q\) , we find two factors \(a\) and \(b\) of \(q\) (i.e., the constant term) such that \(ab = q\) and \(a + b = p.\)

Steps:

(i) \(\quad {p^2} + 6p + 8\)

It can be observed that , \(8 = 4 \times 2\) and \(4 + 2 = 6\)

\[\begin{align}\therefore {p^2} + 6p + 8 &= {p^2} + 2p + 4p + 8\\&= p(p + 2) + 4(p + 2)\\&= (p + 2)(p + 4)\end{align}\]

(ii)\(\quad {q^2} - 10q + 21\)

It can be observed that, 

\(21 = \left( { - 7} \right) \times \left( { - 3} \right) \; \text{and} \\ \;  \left( { - 7} \right) + \left( { - 3} \right) = - 10\)

\[\begin{align}\therefore \quad & {q^2} - 10q + 21 \\ &= {q^2} - 7q - 3q + 21\\ &= q(q - 7) - 3(q - 7)\\&= (q - 7)(q - 3)\end{align}\]

(iii) \(\quad {p^2} + 6p - 16\)

It can be observed that, \(16{\rm{ }} = {\rm{ }}\left( { - 2} \right){\rm{ }} \times {\rm{ }}8\) and  \(8{\rm{ }} + {\rm{ }}\left( { - 2} \right){\rm{ }} = {\rm{ }}6\)

\[\begin{align}\therefore \quad & {p^2} + 6p - 16 \\ &= {p^2} + 8p - 2p - 16\\&= p(p + 8) - 2(p + 8)\\&= (p + 8)(p - 2)\end{align}\]

  
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