NCERT Solutions for Class 8 Maths Chapter 14 - Factorisation

Exercise (14.1)

1. Find the common factors of the terms

i. $12x,36$

Ans:  Write the factors of each term separately:

$12x=2\times 2\times 3\times x$

$36=2\times 2\times 3\times 3$

The factors that appear in both the lists are the common factors. 

Hence, the common factors are $2,2,3$. 

Multiply the common factors, $2\times 2\times 3=12$

ii.  $2y,22xy$

Ans: Write the factors of each term separately:

$2y=2\times y$

$22xy=2\times 11\times x\times y$

The factors that appear in both the lists are the common factors. 

Hence, the common factors are $2,y$. 

Multiply the common factors, $2\times y=2y$. 

iii. $14pq,28{{p}^{2}}{{q}^{2}}$

Ans: Write the factors of each term separately:

$14pq=2\times 7\times p\times q$

$28{{p}^{2}}{{q}^{2}}=2\times 2\times 7\times p\times p\times q\times q$

The factors that appear in both the lists are the common factors. 

Hence, the common factors are$2,7,p,q$. 

Multiply the common factors, $2\times 7\times p\times q=14pq$. 

iv. \[2x,3{{x}^{2}},4\]

Ans: Write the factors of each term separately:

$2x=2\times x$

$3{{x}^{2}}=3\times x\times x$

$4=2\times 2$

The factors that appear in both the lists are the common factors. 

Hence, the common factor is $1$. 

v. $6abc,24a{{b}^{2}},12{{a}^{2}}b$

Ans: Write the factors of each term separately:

$6abc=2\times 3\times a\times b\times c$

$24a{{b}^{2}}=2\times 2\times 2\times 3\times a\times b\times b$

$12{{a}^{2}}b=2\times 2\times 3\times a\times a\times b$

The factors that appear in both the lists are the common factors. 

Hence, the common factors are$2,3,a,b$. 

Multiply the common factors, $2\times 3\times a\times b=6ab$. 

vi. $16{{x}^{3}},-4{{x}^{2}},32x$

Ans: Write the factors of each term separately:

$16{{x}^{3}}=2\times 2\times 2\times 2\times x\times x\times x$

$-4{{x}^{2}}=-1\times 2\times 2\times x\times x$

$32x=2\times 2\times 2\times 2\times 2\times x$

The factors that appear in both the lists are the common factors. 

Hence, the common factors are$2,2,x$ 

Multiply the common factors, $2\times 2\times x=4x$

vii. $10pq,20qr,30rp$

Ans: Write the factors of each term separately:

$10pq=2\times 5\times p\times q$

$20qr=2\times 2\times 5\times q\times r$

$30rp=2\times 3\times 5\times r\times p$

The factors that appear in both the lists are the common factors. 

Hence, the common factors are $2,5$.

Multiply the common factors, $2\times 5=10$. 

viii. $3{{x}^{2}}{{y}^{3}},10{{x}^{3}}{{y}^{2}},6{{x}^{2}}{{y}^{2}}z$

Ans: Write the factors of each term separately:

$3{{x}^{2}}{{y}^{3}}=3\times x\times x\times y\times y\times y$

$10{{x}^{3}}{{y}^{2}}=2\times 5\times x\times x\times x\times y\times y$

$6{{x}^{2}}{{y}^{2}}z=2\times 3\times x\times x\times y\times y\times z$

The factors that appear in both the lists are the common factors. 

Hence, the common factors are $2\times 7\times p\times q=14pq$. 

2. Factorise the following expressions

i. $7x-42$

Ans: Take out the common factors from all the terms to factorize. 

$\text{ }7x=7\times x$

$42=2\times 3\times 7$

The common factor is 7

$\therefore 7x-42=(7\times x)-(2\times 3\times 7)=7(x-6)$

ii. $6p-12q$

Ans: Take out the common factors from all the terms to factorize. 

$6p=2\times 3\times p$

$12q=2\times 2\times 3\times q$

The common factors are $2$ and $3$.

$6p-12q=(2\times 3\times p)-(2\times 2\times 3\times q)$

$=2\times 3[p-(2\times q)]$

$=6(p-2q)$

iii. $7{{a}^{2}}+14a$

Ans: Take out the common factors from all the terms to factorize. 

$7{{a}^{2}}=7\times a\times a$

$14a=2\times 7\times a$

The common factors are $7$ and \[a\]

$\therefore 7{{a}^{2}}+14a=(7\times a\times a)+(2\times 7\times a)$

$=7\times a[a+2]$

$=7a(a+2)$

iv. $-16z+20{{z}^{3}}$

Ans: Take out the common factors from all the terms to factorize. 

$\text{ }16z=2\times 2\times 2\times 2\times z$

$20{{z}^{3}}=2\times 2\times 5\times z\times z\times z$

The common factors are $2,2$, and $z$ .

$-16z+20{{z}^{3}}=-(2\times 2\times 2\times 2\times z)+(2\times 2\times 5\times z\times z\times z)$

$=(2\times 2\times z)[-(2\times 2)+(5\times z\times z)]$

$=4z\left( -4+5{{z}^{2}} \right)$

v. $20{{l}^{2}}m+30\text{alm}$

Ans: Take out the common factors from all the terms to factorize. 

$20{{l}^{2}}m=2\times 2\times 5\times 1\times 1\times m\text{ }$

$\text{30alm }=2\times 3\times 5\times a\times 1\times m$

The common factors are \[2,5,1,\] and $m$.

$\therefore 20{{l}^{2}}m+30alm=(2\times 2\times 5\times 1\times 1\times m)+(2\times 3\times 5\times a\times 1\times m)$

$=(2\times 5\times 1\times m)[(2\times l)+(3\times a)]$

$=10lm(2l+3a)$

vi. $5{{x}^{2}}y-15x{{y}^{2}}$

Ans: Take out the common factors from all the terms to factorize. 

$5{{x}^{2}}y=5\times x\times x\times y$

$15x{{y}^{2}}=3\times 5\times x\times y\times y$

The common factors are$5,x$, and$y$.

$5{{x}^{2}}y-15x{{y}^{2}}=(5\times x\times x\times y)-(3\times 5\times x\times y\times y)$

$=5\times x\times y[x-(3\times y)]$

$=5xy(x-3y)$

vii. $10{{a}^{2}}-15{{b}^{2}}+20{{c}^{2}}$

Ans: Take out the common factors from all the terms to factorize. 

$10{{a}^{2}}=2\times 5\times a\times a$

$15{{b}^{2}}=3\times 5\times b\times b$

$20{{c}^{2}}=2\times 2\times 5\times c\times c$

The common factor is $5$

$10{{a}^{2}}-15{{b}^{2}}+20{{c}^{2}}=(2\times 5\times a\times a)-(3\times 5\times b\times b)+(2\times 2\times 5\times c\times c)$

$=5[(2\times a\times a)-(3\times b\times b)+(2\times 2\times c\times c)]$

$=5\left( 2{{a}^{2}}-3{{b}^{2}}+4{{c}^{2}} \right)$

viii. $-4{{a}^{2}}+4ab-4ca$

Ans: Take out the common factors from all the terms to factorize. 

$4{{a}^{2}}=2\times 2\times a\times a$

$4ab=2\times 2\times a\times b$

$4ca=2\times 2\times c\times a$

The common factors are $2,2$, and $a$

$-4{{a}^{2}}+4ab-4ca=-(2\times 2\times a\times a)+(2\times 2\times a\times b)-(2\times 2\times c\times a)$

$=2\times 2\times a[-(a)+b-c]$

$=4a(-a+b-c)\text{ }$

ix. ${{x}^{2}}yz+x{{y}^{2}}z+xy{{z}^{2}}$

Ans: Take out the common factors from all the terms to factorize. 

${{x}^{2}}yz=x\times x\times y\times z$

$x{{y}^{2}}z=x\times y\times y\times z$

$xy{{z}^{2}}=x\times y\times z\times z$

The common factors are$x,y$, and \[z\]

${{x}^{2}}yz+x{{y}^{2}}z+xy{{z}^{2}}=(x\times x\times y\times z)+(x\times y\times y\times z)+(x\times y\times z\times z)$

$=x\times y\times z[x+y+z]$

$=xyz(x+y+z)$

x. $a{{x}^{2}}y+bx{{y}^{2}}+cxyz$

Ans: Take out the common factors from all the terms to factorize. 

$a{{x}^{2}}y=a\times x\times x\times y$

$bx{{y}^{2}}=b\times x\times y\times y$

$\operatorname{cxyz}=c\times x\times y\times z$

The common factors are$x$and $y$.

$a{{x}^{2}}y+bx{{y}^{2}}+cxyz=(a\times x\times x\times y)+(b\times x\times y\times y)+(c\times x\times y\times z)$

$=(x\times y)[(a\times x)+(b\times y)+(c\times z)]$

$=xy(ax+by+cz)$

3.  Factorise

i. ${{x}^{2}}+xy+8x+8y$

Ans: 

Write each term in terms of its factors and take out the common factors. 

${{x}^{2}}+xy+8x+8y=x\times x+x\times y+8\times x+8\times y$

$=x(x+y)+8(x+y)$

$=(x+y)(x+8)$

ii. $15xy-6x+5y-2$

Ans: 

Write each term in terms of its factors and take out the common factors. 

$15xy-6x+5y-2=3\times 5\times x\times y-3\times 2\times x+5\times y-2$

$=3x(5y-2)+1(5y-2)$

$=(5y-2)(3x+1)$

iii. $ax+bx-ay-by$

Ans: 

Write each term in terms of its factors and take out the common factors. 

$ax+bx-ay-by=a\times x+b\times x-a\times y-b\times y$

$=x(a+b)-y(a+b)$

$=(a+b)(x-y)$

iv. $15pq+15+9q+25p$

Ans: 

Write each term in terms of its factors and take out the common factors. 

$15pq+15+9q+25p=15pq+9q+25p+15$

$=3\times 5\times p\times q+3\times 3\times q+5\times 5\times p+3\times 5$

$ =3\text{ }q\left( 5\text{ }p+3 \right)+5\left( 5\text{ }p+3 \right) $

$  =\left( 5\text{ }p+3 \right)\left( 3\text{ }q+5 \right) $

v.  $z-7+7xy-xyz$

Ans: 

Write each term in terms of its factors and take out the common factors. 

$z-7+7xy-xyz=z-x\times y\times z-7+7\times x\times y$

\[=z\left( 1-x\text{ }y \right)-7\left( 1-x\text{ }y \right)\]

$=(1-xy)(z-7)$

Exercise (14.2)

1. Factorise the following expressions.

i. ${{a}^{2}}+8a+16$

Ans: Use the formula \[\left[ {{(x+y)}^{2}}={{x}^{2}}+2xy+{{y}^{2}} \right]\] to factorize.

 \[{{a}^{2}}+8a+16={{(a)}^{2}}+2\times a\times 4+{{(4)}^{2}}\]

\[={{(a+4)}^{2}}\]

ii. ${{p}^{2}}-10p+25$

Ans: Use the formula \[\left[ {{(x+y)}^{2}}={{x}^{2}}+2xy+{{y}^{2}} \right]\] to factorize.

${{p}^{2}}-10p+25={{(p)}^{2}}-2\times p\times 5+{{(5)}^{2}}$

$={{(p-5)}^{2}}$

iii. $25{{m}^{2}}+30m+9$

Ans: Use the formula \[\left[ {{(x+y)}^{2}}={{x}^{2}}+2xy+{{y}^{2}} \right]\] to factorize.

$25{{m}^{2}}+30m+9={{(5m)}^{2}}+2\times 5m\times 3+{{(3)}^{2}}$

$={{(5m+3)}^{2}}$

iv. $49{{y}^{2}}+84yz+36{{z}^{2}}$

Ans: Use the formula \[\left[ {{(x+y)}^{2}}={{x}^{2}}+2xy+{{y}^{2}} \right]\] to factorize.

$49{{y}^{2}}+84yz+36{{z}^{2}}={{(7y)}^{2}}+2\times (7y)\times (6z)+{{(6z)}^{2}}$

$={{(11b-4c)}^{2}}$

v. $4{{x}^{2}}-8x+4$

Ans: Use the formula \[\left[ {{(x+y)}^{2}}={{x}^{2}}+2xy+{{y}^{2}} \right]\] to factorize.

$4{{x}^{2}}-8x+4={{(2x)}^{2}}-2(2x)(2)+{{(2)}^{2}}$

$={{(2x-2)}^{2}}$

$={{[(2)(x-1)]}^{2}}$

$=4{{(x-1)}^{2}}$

vi. $121{{b}^{2}}-88bc+16{{c}^{2}}$

Ans: Use the formula \[\left[ {{(x+y)}^{2}}={{x}^{2}}+2xy+{{y}^{2}} \right]\] to factorize.

$121{{b}^{2}}-88bc+16{{c}^{2}}={{(11b)}^{2}}-2(11b)(4c)+{{(4c)}^{2}}$

$={{(11b-4c)}^{2}}$

vii. ${{(l+m)}^{2}}-4lm$ (Hint: Expand ${{(l+m)}^{2}}$first)

Ans: Use identity $\left[ {{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \right]$to factorize.

${{(l+m)}^{2}}-4lm={{l}^{2}}+2lm+{{m}^{2}}-4lm$

$={{l}^{2}}-2lm+{{m}^{2}}$

$={{(l-m)}^{2}}$

viii. ${{a}^{4}}+2{{a}^{2}}{{b}^{2}}+{{b}^{4}}$

Ans: Use identity $\left[ {{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \right]$to factorize. 

${{a}^{4}}+2{{a}^{2}}{{b}^{2}}+{{b}^{4}}={{\left( {{a}^{2}} \right)}^{2}}+2\left( {{a}^{2}} \right)\left( {{b}^{2}} \right)+{{\left( {{b}^{2}} \right)}^{2}}$

$={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}$

2. Factorise

i. $4{{p}^{2}}-9{{q}^{2}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing. 

$4{{p}^{2}}-9{{q}^{2}}={{(2p)}^{2}}-{{(3q)}^{2}}$ 

$=(2p+3q)(2p-3q)$

ii. $63{{a}^{2}}-112{{b}^{2}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing. 

$63{{a}^{2}}-112{{b}^{2}}=7\left( 9{{a}^{2}}-16{{b}^{2}} \right)$

$=7\left[ {{(3a)}^{2}}-{{(4b)}^{2}} \right]$

$=7(3a+4b)(3a-4b)$

iii. $49{{x}^{2}}-36$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing. 

$49{{x}^{2}}-36={{(7x)}^{2}}-{{(6)}^{2}}$

$=(7x-6)(7x+6)$

iv. $16{{x}^{5}}-144{{x}^{3}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing. 

$16{{x}^{5}}-144{{x}^{3}}=16{{x}^{3}}\left( {{x}^{2}}-9 \right)$

$=16{{x}^{3}}\left[ {{(x)}^{2}}-{{(3)}^{2}} \right]$

$=16{{x}^{3}}(x-3)(x+3)$

v. ${{(l+m)}^{2}}-{{(l-m)}^{2}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing. 

${{(l+m)}^{2}}-{{(l-m)}^{2}}=[(l+m)-(l-m)][(l+m)+(l-m)]$ \[=\left( l+m-1+m \right)\left( I+m+l-m \right)\]

$=2m\times 2l$

$=4\text{ml}$

\[=4lm\]

vi. \[9{{x}^{2}}{{y}^{2}}-16\]

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing. 

$9{{x}^{2}}{{y}^{2}}-16={{(3xy)}^{2}}-{{(4)}^{2}}$

$=(3xy-4)(3xy+4)$

vii. $\left( {{x}^{2}}-2xy+{{y}^{2}} \right)-{{z}^{2}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing. 

$\left( {{x}^{2}}-2xy+{{y}^{2}} \right)-{{z}^{2}}={{(x-y)}^{2}}-{{(z)}^{2}}$

$=(x-y-z)(x-y+z)$

viii. $25{{a}^{2}}-4{{b}^{2}}+28bc-49{{c}^{2}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing. 

$25{{a}^{2}}-4{{b}^{2}}+28bc-49{{c}^{2}}=25{{a}^{2}}-\left( 4{{b}^{2}}-28bc+49{{c}^{2}} \right)$   $={{(5a)}^{2}}-\left[ {{(2b)}^{2}}-2\times 2b\times 7c+{{(7c)}^{2}} \right]$

$={{(5a)}^{2}}-\left[ {{(2b-7c)}^{2}} \right]$

\[=\left[ 5\text{ }a+\left( 2\text{ }b-7\text{ }c \right) \right]\left[ 5\text{ }a-\left( 2\text{ }b-7\text{ }c \right) \right]\]

\[=\left( 5\text{ }a+2\text{ }b-7\text{ }c \right)\left( 5\text{ }a-2\text{ }b+7\text{ }c \right)\]

3. Factorise the expressions

i. $a{{x}^{2}}+bx$

Ans: Take the common factors out to factorize. 

$a{{x}^{2}}+bx=a\times x\times x+b\times x$

$=x(ax+b)$

ii. $7{{p}^{2}}+21{{q}^{2}}$

Ans: Take the common factors out to factorize. 

$7{{p}^{2}}+21{{q}^{2}}=7\times p\times p+3\times 7\times q\times q$

$=7\left( {{p}^{2}}+3{{q}^{2}} \right)$

iii. $2{{x}^{3}}+2x{{y}^{2}}+2x{{z}^{2}}$

Ans: Take the common factors out to factorize. 

$2{{x}^{3}}+2x{{y}^{2}}+2x{{z}^{2}}=2x\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)$

iv. $a{{m}^{2}}+b{{m}^{2}}+b{{n}^{2}}+a{{n}^{2}}$

Ans: Take the common factors out to factorize. 

$a{{m}^{2}}+b{{m}^{2}}+b{{n}^{2}}+a{{n}^{2}}=a{{m}^{2}}+b{{m}^{2}}+a{{n}^{2}}+b{{n}^{2}}$

$={{m}^{2}}(a+b)+{{n}^{2}}(a+b)$

$=(a+b)\left( {{m}^{2}}+{{n}^{2}} \right)$

v. $(lm+l)+m+l$

Ans: Take the common factors out to factorize. 

$(lm+l)+m+1=1m+m+1+1$

$=m(l+1)+1(l+1)$

$=(l+1)(m+1)$

vi. $y(y+z)+9(y+z)$

Ans: Take the common factors out to factorize. 

$y(y+z)+9(y+z)=(y+z)(y+9)$

vii. $5{{y}^{2}}-20y-8z+2yz$

Ans: Take the common factors out to factorize. 

$5{{y}^{2}}-20y-8z+2yz=5{{y}^{2}}-20y+2yz-8z$

\[=5y(y-4)+2z(y-4)\]

\[=(y-4)(5y+2z)\]

viii. $10ab+4a+5b+2$

Ans: Take the common factors out to factorize. 

$10ab+4a+5b+2=10ab+5b+4a+2$

\[=5\text{ }b\left( 2\text{ }a+1 \right)+2\left( 2\text{ }a+1 \right)\]

\[=\left( 2\text{ }a+1 \right)\left( 5\text{ }b+2 \right)\]

ix. $6xy-4y+6-9x$

Ans: Take the common factors out to factorize. 

 \[6xy-4\text{ }y+6-9x=6xy-9x-4y+6\]

 \[=3x\left( 2y-3 \right)-2\left( 2y-3 \right)\]

\[=\left( 2y-3 \right)\left( 3x-2 \right)\]

4. Factorise

i. ${{a}^{4}}-{{b}^{4}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing. 

${{a}^{4}}-{{b}^{4}}={{\left( {{a}^{2}} \right)}^{2}}-{{\left( {{b}^{2}} \right)}^{2}}$

$=\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{a}^{2}}+{{b}^{2}} \right)$

$=(a-b)(a+b)\left( {{a}^{2}}+{{b}^{2}} \right)$

ii. ${{p}^{4}}-81$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing. 

${{p}^{4}}-81={{\left( {{p}^{2}} \right)}^{2}}-{{(9)}^{2}}$

$=\left( {{p}^{2}}-9 \right)\left( {{p}^{2}}+9 \right)$

$=\left[ {{(p)}^{2}}-{{(3)}^{2}} \right]\left( {{p}^{2}}+9 \right)$

$=(p-3)(p+3)\left( {{p}^{2}}+9 \right)$

iii. ${{x}^{4}}-{{(y+z)}^{4}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing. 

${{x}^{4}}-{{(y+z)}^{4}}={{\left( {{x}^{2}} \right)}^{2}}-{{\left[ {{(y+z)}^{2}} \right]}^{2}}$

$=\left[ {{x}^{2}}-{{(y+z)}^{2}} \right]\left[ {{x}^{2}}+{{(y+z)}^{2}} \right]$

$=[x-(y+z)][x+(y+z)]\left[ {{x}^{2}}+{{(y+z)}^{2}} \right]$

$=(x-y-z)(x+y+z)\left[ {{x}^{2}}+{{(y+z)}^{2}} \right]$

iv. ${{x}^{4}}-{{(x-z)}^{4}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing. 

${{x}^{4}}-{{(x-z)}^{4}}={{\left( {{x}^{2}} \right)}^{2}}-{{\left[ {{(x-z)}^{2}} \right]}^{2}}$

$=\left[ {{x}^{2}}-{{(x-z)}^{2}} \right]\left[ {{x}^{2}}+{{(x-z)}^{2}} \right]$

$=[x-(x-z)][x+(x-z)]\left[ {{x}^{2}}+{{(x-z)}^{2}} \right]$

$=z(2x-z)\left[ {{x}^{2}}+{{x}^{2}}-2xz+{{z}^{2}} \right]$

$=z(2x-z)\left( 2{{x}^{2}}-2xz+{{z}^{2}} \right)$

v. ${{a}^{4}}-2{{a}^{2}}{{b}^{2}}+{{b}^{4}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing.${{a}^{4}}-2{{a}^{2}}{{b}^{2}}+{{b}^{4}}={{\left( {{a}^{2}} \right)}^{2}}-2\left( {{a}^{2}} \right)\left( {{b}^{2}} \right)+{{\left( {{b}^{2}} \right)}^{2}}$

$={{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}$

$={{[(a-b)(a+b)]}^{2}}$

$={{(a-b)}^{2}}{{(a+b)}^{2}}$

5. Factorise the following expressions

i. ${{p}^{2}}+6p+8$

Ans: We can see that, $8=4\times 2$and $4+2=6$

$\therefore {{p}^{2}}+6p+8={{p}^{2}}+2p+4p+8$

\[=p\left( p+2 \right)+4\left( p+2 \right)\]

\[=\left( p+2 \right)\left( p+4 \right)\]

ii. ${{q}^{2}}-10q+21$

Ans: We can see that, $21=(-7)\times (-3)$and $(-7)+(-3)=-10$

$\therefore {{q}^{2}}-10q+21={{q}^{2}}-7q-3q+21$

\[=q\left( q-7 \right)-3\left( q-7 \right)\]

\[=\left( q-7 \right)\left( q-3 \right)\]

iii. ${{p}^{2}}+6p-16$

Ans: We can see that, $16=(-2)\times 8$ and $8+(-2)=6$

${{p}^{2}}+6p-16={{p}^{2}}+8p-2p-16$

\[=p\left( p+8 \right)-2\left( p+8 \right)\]

\[=\left( p+8 \right)\left( p-2 \right)\]

Exercise (14.3)

1. Carry out the following divisions.

i. $28{{x}^{4}}\div 56x$

Ans: Write the numerator and denominator in its factors and divide. 

$28{{x}^{4}}=2\times 2\times 7\times x\times x\times x\times x$

$56x=2\times 2\times 2\times 7\times x$

$28{{x}^{4}}\div 56x=\dfrac{2\times 2\times 7\times x\times x\times x\times x}{2\times 2\times 2\times 7\times x}$

$=\dfrac{{{x}^{3}}}{2}$

$=\dfrac{1}{2}{{x}^{3}}$

ii. $-36{{y}^{3}}\div 9{{y}^{2}}$

Ans: Write the numerator and denominator in its factors and divide. 

$ 36{{y}^{3}}=2\times 2\times 3\times 3\times y\times y\times y$

$9{{y}^{2}}=3\times 3\times y\times y$

$-36{{y}^{3}}\div 9{{y}^{2}}=\dfrac{-2\times 2\times 3\times 3\times y\times y\times y}{3\times 3\times y\times y}$

$=-4y$

iii. $66p{{q}^{2}}{{r}^{3}}\div 11q{{r}^{2}}$

Ans: Write the numerator and denominator in its factors and divide. 

$66p{{q}^{2}}{{r}^{3}}=2\times 3\times 11\times p\times q\times q\times r\times r\times r$

$11q{{r}^{2}}=11\times q\times r\times r$

$66p{{q}^{2}}{{r}^{3}}\div 11q{{r}^{2}}=\dfrac{2\times 3\times 11\times p\times q\times q\times r\times r\times r}{11\times q\times r\times r}=6pqr$ 

iv. $34{{x}^{3}}{{y}^{3}}{{z}^{3}}\div 51x{{y}^{2}}{{z}^{3}}$

Ans: Write the numerator and denominator in its factors and divide. 

$34{{x}^{3}}{{y}^{3}}{{z}^{3}}=2\times 17\times x\times x\times x\times y\times y\times y\times z\times z\times z$

$51x{{y}^{2}}{{z}^{3}}=3\times 17\times x\times y\times y\times z\times z\times z$

$34{{x}^{3}}{{y}^{3}}{{z}^{3}}\div 51x{{y}^{2}}{{z}^{3}}=\dfrac{2\times 17\times x\times x\times x\times y\times y\times y\times z\times z\times z}{3\times 17\times x\times y\times y\times z\times z\times z}$

$=\dfrac{2}{3}{{x}^{2}}y$

v. $12{{a}^{3}}{{b}^{8}}\div \left( -6{{a}^{6}}{{b}^{4}} \right)$

Ans: Write the numerator and denominator in its factors and divide. 

$12{{a}^{8}}{{b}^{8}}=2\times 2\times 3\times {{a}^{8}}\times {{b}^{8}}$

$6{{a}^{6}}{{b}^{4}}=2\times 3\times {{a}^{6}}\times {{b}^{4}}$

$12{{a}^{8}}{{b}^{8}}\div \left( -6{{a}^{6}}{{b}^{4}} \right)=\dfrac{2\times 2\times 3\times {{a}^{8}}\times {{b}^{8}}}{-2\times 3\times {{a}^{6}}\times {{b}^{4}}}$

$=-2{{a}^{2}}{{b}^{4}}$

2. Divide the given polynomial by the given monomial.

i. $\left( 5{{x}^{2}}-6x \right)\div 3x$

Ans: Write the numerator its factors and divide. 

$5{{x}^{2}}-6x=x(5x-6)$ 

$\left. 5{{x}^{2}}-6x \right)\div 3x=\dfrac{x(5x-6)}{3x}$

$=\dfrac{1}{3}(5x-6)$

ii. $\left( 3{{y}^{8}}-4{{y}^{6}}+5{{y}^{4}} \right)\div {{y}^{4}}$

Ans: Write the numerator its factors and divide.

$3{{y}^{8}}-4{{y}^{6}}+5{{y}^{4}}={{y}^{4}}\left( 3{{y}^{4}}-4{{y}^{2}}+5 \right)$

$\left( 3{{y}^{8}}-4{{y}^{6}}+5{{y}^{4}} \right)\div {{y}^{4}}=\dfrac{{{y}^{4}}\left( 3{{y}^{4}}-4{{y}^{2}}+5 \right)}{{{y}^{4}}}$

$=3{{y}^{4}}-4{{y}^{2}}+5$

iii.  $8\left( {{x}^{3}}{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{y}^{3}}{{z}^{2}}+{{x}^{2}}{{y}^{2}}{{z}^{3}} \right)\div 4{{x}^{2}}{{y}^{2}}{{z}^{2}}$

Ans: Write the numerator its factors and divide.  

$8\left( {{x}^{3}}{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{y}^{3}}{{z}^{2}}+{{x}^{2}}{{y}^{2}}{{z}^{3}} \right)=8{{x}^{2}}{{y}^{2}}{{z}^{2}}(x+y+z)$

$8\left( {{x}^{3}}{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{y}^{3}}{{z}^{2}}+{{x}^{2}}{{y}^{2}}{{z}^{3}} \right)\div 4{{x}^{2}}{{y}^{2}}{{z}^{2}}=\dfrac{8{{x}^{2}}{{y}^{2}}{{z}^{2}}(x+y+z)}{4{{x}^{2}}{{y}^{2}}{{z}^{2}}}$

$=2(x+y+z)$

iv. $\left( {{x}^{3}}+2{{x}^{2}}+3x \right)\div 2x$

Ans: Write the numerator its factors and divide.

${{x}^{3}}+2{{x}^{2}}+3x=x\left( {{x}^{2}}+2x+3 \right)$

$\left( {{x}^{3}}+2{{x}^{2}}+3x \right)\div 2x=\dfrac{x\left( {{x}^{2}}+2x+3 \right)}{2x}$

$=\dfrac{1}{2}\left( {{x}^{2}}+2x+3 \right)$

v. $\left( {{p}^{3}}{{q}^{6}}-{{p}^{6}}{{q}^{3}} \right)\div {{p}^{3}}{{q}^{3}}$

Ans: Write the numerator its factors and divide.

) ${{p}^{3}}{{q}^{6}}-{{p}^{6}}{{q}^{3}}={{p}^{3}}{{q}^{3}}\left( {{q}^{3}}-{{p}^{3}} \right)$

$\left( {{p}^{3}}{{q}^{6}}-{{p}^{6}}{{q}^{3}} \right)\div {{p}^{3}}{{q}^{3}}=\dfrac{{{p}^{3}}{{q}^{3}}\left( {{q}^{3}}-{{p}^{3}} \right)}{{{p}^{3}}{{q}^{3}}}$

$={{q}^{3}}-{{p}^{3}}$

3. Work out the following divisions.

i. $(10x-25)\div 5$

Ans: Write the numerator and denominator in its factors and divide. 

$(10x-25)\div 5=\dfrac{2\times 5\times x-5\times 5}{5}$

$=\dfrac{5(2x-5)}{5}$

$=2x-5$

ii. $(10x-25)\div (2x-5)$

Ans: Write the numerator and denominator in its factors and divide. 

$(10x-25)\div (2x-5)=\dfrac{2\times 5\times x-5\times 5}{(2x-5)}$

$=\dfrac{5(2x-5)}{2x-5}$

$=5$

iii. $10y(6y+21)\div 5(2y+7)$

Ans: Write the numerator and denominator in its factors and divide. 

$10y(6y+21)\div 5(2y+7)=\dfrac{2\times 5\times y[2\times 3\times y+3\times 7]}{5(2y+7)}$

 $=\dfrac{2\times 5\times y\times 3(2y+7)}{5(2y+7)}$

$=6y$

iv. $9{{x}^{2}}{{y}^{2}}(3z-24)\div 27xy(z-8)$

Ans: Write the numerator and denominator in its factors and divide. 

$9{{x}^{2}}{{y}^{2}}(3z-24)\div 27xy(z-8)=\dfrac{9{{x}^{2}}{{y}^{2}}[3\times z-2\times 2\times 2\times 3]}{27xy(z-8)}$

  $=\dfrac{xy\times 3(z-8)}{3(z-8)}$

$=xy$

v. $96abc(3a-12)(5b-30)\div 144(a-4)(b-6)$

Ans: Write the numerator and denominator in its factors and divide. 

$96abc(3a-12)(5b-30)\div 144(a-4)(b-6)$

$=\dfrac{96abc(3\times a-3\times 4)(5\times b-2\times 3\times 5)}{144(a-4)(b-6)}$

$=\dfrac{2abc\times 3(a-4)\times 5(b-6)}{3(a-4)(b-6)}$

$=10abc$

4. Divide as directed.

i. $5(2x+1)(3x+5)\div (2x+1)$

Ans: Write the numerator and denominator in its factors and divide. 

$5(2x+1)(3x+5)\div (2x+1)=\dfrac{5(2x+1)(3x+1)}{(2x+1)}$

$=5(3x+1)$

ii. $26xy(x+5)(y-4)\div 13x(y-4)$

Ans: Write the numerator and denominator in its factors and divide. 

$26xy(x+5)(y-4)\div 13x(y-4)=\dfrac{2\times 13\times xy(x+5)(y-4)}{13x(y-4)}$

$=2$

iii. $52pqr(p+q)(q+r)(r+p)\div 104pq(q+r)(r+p)$

Ans: Write the numerator and denominator in its factors and divide. 

$52pqr(p+q)(q+r)(r+p)\div 104pq(q+r)(r+p)$

$=\dfrac{2\times 2\times 13\times p\times q\times r\times (p+q)\times (q+r)\times (r+p)}{2\times 2\times 2\times 13\times p\times q\times (q+r)\times (r+p)}$

$=\dfrac{1}{2}r(p+q)$

iv. $20(y+4)\left( {{y}^{2}}+5y+3 \right)\div 5(y+4)$

Ans: Write the numerator and denominator in its factors and divide. 

$20(y+4)\left( {{y}^{2}}+5y+3 \right)=2\times 2\times 5\times (y+4)\left( {{y}^{2}}+5y+3 \right)$

$20(y+4)\left( {{y}^{2}}+5y+3 \right)\div 5(y+4)=\dfrac{2\times 2\times 5\times (y+4)\times \left( {{y}^{2}}+5y+3 \right)}{5\times (y+4)}$                    $=4\left( {{y}^{2}}+5y+3 \right)$ 

v. $x(x+1)(x+2)(x+3)\div x(x+1)$

Ans: Write the numerator and denominator in its factors and divide. 

$x(x+1)(x+2)(x+3)\div x(x+1)=\dfrac{x(x+1)(x+2)(x+3)}{x(x+1)}$

\[=\left( x+2 \right)\left( x+3 \right)\]

5. Factorise the expressions and divide them as directed.

i. $\left( {{y}^{2}}+7y+10 \right)\div (y+5)$

Ans: Factorise the given terms separately.

$\left( {{y}^{2}}+7y+10 \right)={{y}^{2}}+2y+5y+10$

$=y(y+2)+5(y+2)$

$=(y+2)(y+5)$

Divide the two terms. 

$\left( {{y}^{2}}+7y+10 \right)\div (y+5)=\dfrac{(y+5)(y+2)}{(y+5)}$

$=y+2$

ii. $\left( {{m}^{2}}-14m-32 \right)\div (m+2)$

Ans: Factorise the given terms separately.

${{m}^{2}}-14m-32={{m}^{2}}+2m-16m-32$

$=m(m+2)-16(m+2)$

$=(m+2)(m-16)$

Divide the two terms. 

$\left( {{m}^{2}}-14m-32 \right)\div (m+2)=\dfrac{(m+2)(m-16)}{(m+2)}$

$=m-16$

iii. $\left( 5{{p}^{2}}-25p+20 \right)\div (p-1)$

Ans: Factorise the given terms separately.

$5{{p}^{2}}-25p+20=5\left( {{p}^{2}}-5p+4 \right)$

=5[p(p-1)-4(p-1)]

$=5(p-1)(p-4)$

Divide the two terms. 

$\left( 5{{p}^{2}}-25p+20 \right)\div (p-1)=\dfrac{5(p-1)(p-4)}{(p-1)}$

$=5(p-4)$

iv. $4yz\left( {{z}^{2}}+6z-16 \right)\div 2y(z+8)$

Ans: Factorise the given terms separately.

$4yz\left( {{z}^{2}}+6z-16 \right)=4yz\left[ {{z}^{2}}-2z+8z-16 \right]$

\[=4\text{ }y\text{ }z\left[ z\left( z-2 \right)+8\left( z-2 \right) \right]\]

\[=4\text{ }y\text{ }z\left( z-2 \right)\left( z+8 \right)\]

Divide the two terms. 

$4yz\left( {{z}^{2}}+6z-16 \right)\div 2y(z+8)=\dfrac{4yz(z-2)(z+8)}{2y(z+8)}$

$=2z(z-2)$

v. $5pq\left( {{p}^{2}}-{{q}^{2}} \right)\div 2p(p+q)$

Ans: Factorise the given terms separately.

$5pq\left( {{p}^{2}}-{{q}^{2}} \right)=5pq(p-q)(p+q)$

Divide the two terms. 

$5pq\left( {{p}^{2}}-{{q}^{2}} \right)\div 2p(p+q)=\dfrac{5pq(p-q)(p+q)}{2p(p+q)}$

$=\dfrac{5}{2}q(p-q)$

vi. $12xy\left( 9{{x}^{2}}-16{{y}^{2}} \right)\div 4xy(3x+4y)$

Ans: Factorise the given terms separately.

$12xy\left( 9{{x}^{2}}-16{{y}^{2}} \right)=12xy\left[ {{(3x)}^{2}}-{{(4y)}^{2}} \right]$

$=12xy(3x-4y)(3x+4y)$

Divide the two terms. 

$12xy\left( 9{{x}^{2}}-16{{y}^{2}} \right)\div 4xy(3x+4y)$

$=\dfrac{2\times 2\times 3\times x\times y\times (3x-4y)\times (3x+4y)}{2\times 2\times x\times y\times (3x+4y)}$

\[=3\left( 3\text{ }x-4\text{ }y \right)\]

vii. $39{{y}^{3}}\left( 50{{y}^{2}}-98 \right)\div 26{{y}^{2}}(5y+7)$

Ans: Factorise the given terms separately.

$39{{y}^{3}}\left( 50{{y}^{2}}-98 \right)=3\times 13\times y\times y\times y\times 2\left[ \left( 25{{y}^{2}}-49 \right) \right]$

$=3\times 13\times 2\times y\times y\times y\times \left[ {{(5y)}^{2}}-{{(7)}^{2}} \right]$

$=3\times 13\times 2\times y\times y\times y(5y-7)(5y+7)$

$26{{y}^{2}}(5y+7)=2\times 13\times y\times y\times (5y+7)$

Divide the two terms. 

\[39{{y}^{3}}\left( 50{{y}^{2}}-98 \right)\div 26{{y}^{2}}(5y+7)=\dfrac{3\times 13\times 2\times y\times y\times y(5y-7)(5y+7)}{2\times 13\times y\times y\times (5y+7)}\]

\[=3y(5y-7)\]

\[=15{{y}^{2}}-21y\]

Exercise (14.4)

1.   Find and correct the errors in the statement: $4(x-5)=4x-5$

Ans: Compare the L.HS. with the R.H.S.

L.H.S $4(x-5)=4x-20 \ne $R.H.S.

The correct statement is  $4(x-5)=4x-20$

2.   Find and correct the errors in the statement: $x(3x+2)=3x^2 +2$

Ans: Compare the L.HS. with the R.H.S.

L.H.S $x(3x+2)=3x^2 +2x \ne $R.H.S.

3.  Find and correct the errors in the statement: $2x+3y=5xy$

Ans: Compare the L.HS. with the R.H.S.

L.H.S $=2x+3y=2x+3y\ne $R.H.S.

The correct statement is $2x+3y=2x+3y$

4. Find and correct the errors in the statement: $x+2x+3x=5x$

Ans: Compare the L.HS. with the R.H.S.

L.H.S \[=x+2x+3x=1x+2x+3x=x(1+2+3)=6x\ne RHS\]

The correct statement is $x+2x+3x=6x$

5. Find and correct the errors in the statement: $5y+2y+y-7y=0$

Ans: Compare the L.HS. with the R.H.S.

L.H.S. $=5y+2y+y-7y=8y-7y=y\ne \$R.H.S$

The correct statement is $5y+2y+y-7y=y$

6. Find and correct the errors in the statement: $3x+2x=5{{x}^{2}}$

Ans: Compare the L.HS. with the R.H.S.

L.H.S. $=3x+2x=5x\ne \$R.H.S$

The correct statement is $3x+2x=5x$.

7.  Find and correct the errors in the statement: ${{(2x)}^{2}}+4(2x)+7=2{{x}^{2}}+8x+7$

Ans: Compare the L.H.S with the R.H.S. 

L.H.S$={{(2x)}^{2}}+4(2x)+7=4{{x}^{2}}+8x+7\ne $R.H.S 

The correct statement is ${{(2x)}^{2}}+4(2x)+7=4{{x}^{2}}+8x+7$

8. Find and correct the errors in the statement: ${{(2x)}^{2}}+5x=4x+5x=9x$

Ans: Compare the L.HS. with the R.H.S.

LH.S $={{(2x)}^{2}}+5x=4{{x}^{2}}+5x\ne \$R.H.S.$

The correct statement is ${{(2x)}^{2}}+5x=4{{x}^{2}}+5x$

9. Find and correct the errors in the statement: ${{(3x+2)}^{2}}=3{{x}^{2}}+6x+4$

Ans: Compare the L.HS. with the R.H.S.

Use the property $\left[ {{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \right]$.

$\text{ L}\text{.H}\text{.S}\text{. }={{(3x+2)}^{2}}={{(3x)}^{2}}+2(3x)(2)+{{(2)}^{2}}\text{ =}9{{x}^{2}}+12x+4\ne \text{ R}\text{.H}\text{.S }$

The correct statement is ${{(3x+2)}^{2}}=9{{x}^{2}}+12x+4$.

10. Find and correct the errors in the following mathematical statement. Substituting $x=-3$ in

a. ${{x}^{2}}+5x+4$  gives ${{(-3)}^{2}}+5(-3)+4=9+2+4=15$

Ans: Put the value $x=-3$ and solve. 

${{x}^{2}}+5x+4={{(-3)}^{2}}+5(-3)+4$

$=9-15+4$

$=13-15$

$=-2$

Hence the given statement is wrong. 

The correct statement is: ${{x}^{2}}+5x+4$ gives $-2$

b. ${{x}^{2}}-5x+4$ gives ${{(-3)}^{2}}-5(-3)+4=9-15+4=-2$

Ans: Put the value $x=-3$ and solve.

${{x}^{2}}-5x+4={{(-3)}^{2}}-5(-3)+4$ 

$=9+15+4$

$=28$

Hence the given statement is wrong. 

The correct statement is: ${{x}^{2}}-5x+4$ gives $28$

c. ${{x}^{2}}+5x$  gives  ${{(-3)}^{2}}+5(-3)=-9-15=-24$

Ans: Put the value $x=-3$ and solve. 

${{x}^{2}}+5x={{(-3)}^{2}}+5(-3)$

$=9-15$

$=-6$

Hence the given statement is wrong. 

The correct statement is: ${{x}^{2}}+5x$ gives $-6$. 

11. Find and correct the errors in the statement: ${{(y-3)}^{2}}={{y}^{2}}-9$

Ans: Compare the L.HS. with the R.H.S.

Use the property $\left[ {{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} \right]$

L.H.S $={{(y-3)}^{2}}={{(y)}^{2}}-2(y)(3)+{{(3)}^{2}}={{y}^{2}}-6y+9\ne R.H.S$

The correct statement is ${{(y-3)}^{2}}={{y}^{2}}-6y+9$

12. Find and correct the errors in the statement: ${{(z+5)}^{2}}={{z}^{2}}+25$

Ans: Compare the L.HS. with the R.H.S.

Use the property $\left[ {{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \right]$. 

$\text{ L}\text{.H}\text{.S }={{(z+5)}^{2}}={{(z)}^{2}}+2(z)(5)+{{(5)}^{2}}={{z}^{2}}+10z+25\ne \text{ R}\text{.H}\text{.S }$

The correct statement is ${{(z+5)}^{2}}={{z}^{2}}+10z+25$

13. Find and correct the errors in the statement: $(2a+3b)(a-b)=2{{a}^{2}}-3{{b}^{2}}$

Ans: Compare the L.HS. with the R.H.S. 

L.H.S.

$=(2a+3b)(a-b)$

$=2a\times a+3b\times a-2a\times b-3b\times b$

$=2{{a}^{2}}+3ab-2ab-3{{b}^{2}}$ 

$=2{{a}^{2}}+ab-3{{b}^{2}}\ne \text{ R}\text{.H}\text{.S}\text{. }$

The correct statement is $(2a+3b)(a-b)=2{{a}^{2}}+ab-3{{b}^{2}}$

14. Find and correct the errors in the statement: $(a+4)(a+2)={{a}^{2}}+8$

Ans: Compare the L.HS. with the R.H.S.

$\text{ L}\text{.H}\text{.S}\text{. }=(a+4)(a+2)={{(a)}^{2}}+(4+2)(a)+4\times 2={{a}^{2}}+6a+8\ne \text{ R}\text{.H}\text{.S }$

The correct statement is $(a+4)(a+2)={{a}^{2}}+6a+8$

15. Find and correct the errors in the statement: $(a-4)(a-2)={{a}^{2}}-8$

Ans: Compare the L.HS. with the R.H.S.

$L.H.S.=(a-4)(a-2)={{(a)}^{2}}+[(-4)+(-2)](a)+(-4)(-2)={{a}^{2}}-6a+8\ne \text{ R}\text{.H}\text{.S}\text{. }$

The correct statement is $(a-4)(a-2)={{a}^{2}}-6a+8$

16. Find and correct the errors in the statement: $\dfrac{3{{x}^{2}}}{3{{x}^{2}}}=0$

Ans: Compare the L.HS. with the R.H.S.

$\text{ L}\text{.H}\text{.S }=\dfrac{3{{x}^{2}}}{3{{x}^{2}}}=\dfrac{3\times x\times x}{3\times x\times x}=1\ne \text{ R}\text{.H}\text{.S}\text{. }$

The correct statement is $\dfrac{3{{x}^{2}}}{3{{x}^{2}}}=1$

17. Find and correct the errors in the statement: $\dfrac{3{{x}^{2}}+1}{3{{x}^{2}}}=1+1=2$

Ans: Compare the L.HS. with the R.H.S.

$\dfrac{3{{x}^{2}}+1}{3{{x}^{2}}}=\dfrac{3{{x}^{2}}}{3{{x}^{2}}}+\dfrac{1}{3{{x}^{2}}}=1+\dfrac{1}{3{{x}^{2}}}\ne \text{ R}\text{.H}\text{.S}\text{. }$

The correct statement is $\dfrac{3{{x}^{2}}+1}{3{{x}^{2}}}=1+\dfrac{1}{3{{x}^{2}}}$

18. Find and correct the errors in the statement: $\dfrac{3x}{3x+2}=\dfrac{1}{2}$

Ans: Compare the L.HS. with the R.H.S.

$\text{ L}.\text{H}.\text{S=}\dfrac{\text{3x}}{\text{3x+2}}\ne \text{ R}\text{.H}\text{.S}\text{. }$

The correct statement is $\dfrac{3x}{3x+2}=\dfrac{3x}{3x+2}$

19. Find and correct the errors in the statement: $\dfrac{3}{4x+3}=\dfrac{1}{4x}$

Ans: Compare the L.HS. with the R.H.S. 

$\text{L}\text{.H}\text{.S}=\dfrac{3}{4x+3}\ne \text{R}\text{.H}.\text{S}$

The correct statement is $\dfrac{3}{4x+3}=\dfrac{3}{4x+3}$

20. Find and correct the errors in the statement: $\dfrac{4x+5}{4x}=5$

Ans: Compare the L.HS. with the R.H.S.

L.H.S. $=\dfrac{4x+5}{4x}=\dfrac{4x}{4x}+\dfrac{5}{4x}=1+\dfrac{5}{4x}\ne R.H.S$

The correct statement is $\dfrac{4x+5}{4x}=1+\dfrac{5}{4x}$

21. Find and correct the errors in the statement: $\dfrac{7x+5}{5}=7x$

Ans: Compare the L.HS. with the R.H.S. 

L.H.S. $=\dfrac{7x+5}{5}=\dfrac{7x}{5}+\dfrac{5}{5}=\dfrac{7x}{5}+1\ne R.H.S.$

The correct statement is $\dfrac{7x+5}{5}=\dfrac{7x}{5}+1$.

NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Free PDF

Students can have the benefit of downloading NCERT Solutions for Class 8 Factorisation in a pdf format for free of cost. Take a physical copy for the uninterrupted study hours of the student as these hard copies can be taken anywhere to practice any time, along with friends also. Also, these can be stored and utilised during the time of examinations and can revise while attending olympiads and competitive exams.

NCERT Solutions for Class 8 Factorisation

14.1 Introduction

The first section of every chapter might be the same in most cases. It is the Introduction part. Here the introduction may remind the knowledge of factors for the students which they have studied in the earlier classes. Students have already learned to find factors for natural numbers and the meaning of factors for algebraic expressions. Then in this chapter, they are going to learn the factorisation of big expressions using different methods in detail.

14.2 What is Factorisation?

The 8th class maths chapter 12 factorisation Pdf explains the meaning of factorisation. The factorisation is a process of representing the given expression in the form of the product of two factors. These factors may be either of the numbers or variables or algebraic expressions themselves. The PDF also explains that it has several systematic approaches to find factors for the given expressions. Let's see the available methods to find the factors.

14.2.1 Method of Common Factors

It is the first method of finding factors for the given expression. According to NCERT Solutions Class 8 Factorisation, the given expression can be expressed in the product of terms. This can be understood clearly by observing two or more examples given in the PDF.

14.2.2 Factorization by Regrouping Terms

The next section of this chapter deals with another method of factorization. Chapter 14 Class 8 explains that while finding the factors of the given expression, we will take an individual term as common and consider it as a common factor. This can be done by forming two groups. For an instance,

6𝑝 − 12𝑞

= 6𝑝 − (6 × 2)𝑞

= 6(𝑝 − 2𝑞)(taking 6 as common)

Exercise 14.1 with three short answer type questions.

14.2.3 Factorization Using Identities

NCERT Class 8 Maths Chapter 14 came to the third method of factorization in this section. Here students need to find out the identical terms and then separate those terms which give the factors directly.

14.2.4 Factors of the form (x+a)(x+b)

This is the 4th method of solving factorization. The Class 8 Maths Solutions Chapter 14 explains that this method was usually used if the expression doesn't have perfect squares. Even though it is suitable for perfect squares, without having perfect squares, also we can use this method to get factors of the given expression. Even though we have different methods to find factors, the solution is one and the same, but the procedures may differ.

Exercise 14.2 has five short answer type questions in which each question has some set of expressions to solve.

14.3 Division of Algebraic Expressions

The Maths Class 8 NCERT Solutions Chapter 14 explains an advanced concept in factorization. As the students have learned addition, subtraction, multiplication of algebraic expressions, now class 8 maths factorization explains how to divide two or more expressions. But it has two sub-concepts. Because the division can be made based on the degree of expression. From the notes of NCERT Solutions Chapter 14 Class 8, we have:

• Division of monomial by another monomial.

• Division of polynomial by monomial.

14.4 Division of Algebraic Expressions Continued

Class 8 Maths Factorisation Solutions has explained that this section is the extension of the above section. Here the division of expressions can be made by two or more polynomials. It explains how to divide the Algebraic Expressions of higher degrees.

Exercise 14.3 contains five long answer type questions each question has a set of sums.

14.5  Can You Find the Error

The chapter of Factorisation questions for Class 8 NCERT has ended with this last section of correcting errors on the expressions. It is also known as verification and making the right-hand side terms and left-hand side terms equal. In this equalisation process, students can correct the errors.

Exercise 14.4 with 20 short answer type questions.

Key Takeaways of NCERT Solutions Class 8 Maths Chapter 14 Free PDF

Students can enjoy various benefits by using Factorisation class 8 NCERT. Some of them are:

• The previous papers can help the students to practice and understand the question pattern, allotment of marks.

• Well-experienced mathematicians prepare the PDFs based on the current syllabus.

• These are available for free in both English and Hindi.