# Exercise 14.2 Factorization- NCERT Solutions Class 8

## 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

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

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

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

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

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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