NCERT Solutions Class 8 Maths Chapter 12 Factorisation

Chapter 12 of Class 8 Maths, Factorization, deals with the factorization of algebraic expressions. The NCERT solutions for Class 8 Maths Chapter 12 Factorization are designed as per the textbook to gain a comprehensive knowledge of the entire syllabus. The simple format of these solutions will allow students to read and explore more about the topic.

This chapter presents several methods and their applications for factoring algebraic expressions. Allen’s NCERT solutions enable students to understand without much hassle how to break expressions into factors and thereby solve complex problems more efficiently.

1.0Class 8 Maths NCERT Solutions Chapter 12 PDF - Free Download

ALLEN offers detailed and step-by-step explanations of all the problems covered in the NCERT textbook class 8 Maths, helping students understand the concepts easily. This chapter includes the laws of exponents, expressing large numbers in standard form, and making easy calculations with exponents. Students can download the solutions from the link given below.

2.0Key Features of NCERT Solutions for Class 8 Maths - Factorisation

• Step-by-Step Explanations: Each solution is meticulously crafted with detailed steps to ensure that students easily grasp the core concepts.
• Enhanced Problem-Solving Skills: A variety of exercises are designed to strengthen analytical thinking and boost confidence.
• Strong Foundation for Competitive Exams: These solutions provide comprehensive support for school exams and lay a strong foundation for competitive tests.

By practising the problems in these NCERT solutions, students can discover interesting ways to approach factorization, developing a thorough understanding of the topic.

3.0NCERT Questions with Solutions for Class 8 Maths Chapter 12 - Detailed Solutions

Exercise: 12.1

• Find the common factors of the given terms : (i) $12x,36$ (ii) $2\mathrm{y},22\mathrm{xy}$ (iii) $14pq,28p^2{q}^2$ (iv) $2\mathrm{x},3{\mathrm{x}}^2,4$ (v) $6abc,24a{b}^2,12a^{2}b$ (vi) $16x^3,-4x^2,32x$ (vii)10pq, 20qr, 30 rp (viii) $3x^2{y}^3,10x^3{y}^2,6x^2{y}^{2}z$ Sol. (i) The numerical coefficients in the given monomials are 12 and 36 . The highest common factor of 12 and 36 is 12 . But there is no common literal appearing in the given monomials 12 x and 36. $\therefore$ The highest common factor $=12$ (ii) $2\mathrm{y}=2\times \mathrm{y}$ $22xy=2\times 11\times x\times y$ The common factors are $2,\mathrm{y}$. And, $2\times y=2y$ (iii) The numerical coefficients of the given monomials are 14 and 28. The highest common factor of 14 and 28 is 14 . The common literals appearing in the given monomials are $p$ and $q$. The smallest power of p and q in the two monomials = 1 The monomial of common literals with smallest powers $=\mathrm{pq}$ $\therefore$ The highest common factor $=14\mathrm{pq}$ (iv) $2\mathrm{x}=2\times \mathrm{x}$ $3{\mathrm{x}}^2=3\times \mathrm{x}\times \mathrm{x}$ $4=2\times 2$ The common factor is 1 . (v) The numerical coefficients of the given monomials are 6, 24 and 12. The common literals appearing in the three monomials are a and b . The smallest power of a in the three monomials =1 The smallest power of $b$ in the three monomials $=1$ The monomial of common literals with smallest powers $=ab$ Hence, the highest common factor $=6\mathrm{ab}$ (vi) $16{\mathrm{x}}^3=2\times 2\times 2\times 2\times \mathrm{x}\times \mathrm{x}\times \mathrm{x}$ $-4{\mathrm{x}}^2=-1\times 2\times 2\times \mathrm{x}\times \mathrm{x}$ $32\mathrm{x}=2\times 2\times 2\times 2\times 2\times x$ The common factors are $2,2,x$. And $2\times 2\times x=4x$ (vii) The numerical coefficients of the given monomials are 10,20 and 30. The highest common factor of 10,20 and 30 is 10 . There is no common literal appearing in the three monomials. Hence, the highest common factor $=10$ (viii) $3x^2{y}^3=3\times x\times x\times y\times y\times y$ $10x^3{y}^2=2\times 5\times x\times x\times x\times y\times y$ $6x^2{y}^{2}z=2\times 3\times x\times x\times y\times y\times z$ The common factors are $\mathrm{x},\mathrm{x},\mathrm{y},\mathrm{y}$. and, $x\times x\times y\times y=x^2{y}^2$
• Factorise the following expressions (i) $7x-42$ (ii) $6p-12q$ (iii) $7a^2+14a$ (iv) $-16z+20z^3$ (v) $20{\ell }^2\mathrm{m}+30\mathrm{a}\ell \mathrm{m}$ (vi) $5x^{2}y-15x{y}^2$ (vii) $10a^2-15b^2+20c^2$ (viii) $-4a^2+4ab-4ca$ (ix) $x^{2}yz+x{y}^{2}z+xy{z}^2$ (x) $a{x}^{2}y+bx{y}^2+cxyz$ Sol. (i) We have, $7\mathrm{x}=7\times \mathrm{x}$ and $42=2\times 3\times 7$ The two terms have 7 as a common factor $7\mathrm{x}-42=(7\times \mathrm{x})-2\times 3\times 7$ $=7\times (x-2\times 3)=7(x-6)$ (ii) $6\mathrm{p}=2\times 3\times \mathrm{p}$ $12q=2\times 2\times 3\times q$ The common factors are 2 and 3 . $\therefore 6\mathrm{p}-12\mathrm{q}=(2\times 3\times \mathrm{p})-(2\times 2\times 3\times \mathrm{q})$ $=2\times 3[p-(2\times q)]$ $=6(p-2q)$ (iii) We have, $7{\mathrm{a}}^2=7\times \mathrm{a}\times \mathrm{a}$ and, $14\mathrm{a}=2\times 7\times \mathrm{a}$ The two terms have 7 and a as common factors $7{\mathrm{a}}^2+14\mathrm{a}=(7\times \mathrm{a}\times \mathrm{a})+(2\times 7\times \mathrm{a})$ $=7\times a\times (a+2)=7a(a+2)$ (iv) $16z=2\times 2\times 2\times 2\times z$ $20{\mathrm{z}}^3=2\times 2\times 5\times \mathrm{z}\times \mathrm{z}\times \mathrm{z}$ The common factors are 2,2 , and z . $\therefore -16z+20z^3=-(2\times 2\times 2\times 2\times z)+$ $(2\times 2\times 5\times \mathrm{z}\times \mathrm{z}\times \mathrm{z})$ $=(2\times 2\times z)[-(2\times 2)+(5\times z\times z)]$ $=4\mathrm{z}\left(-4+5{\mathrm{z}}^2\right)$ (v) We have, $20{\ell }^2\mathrm{m}=2\times 2\times 5\times \ell \times \ell \times \mathrm{m}$ and, $30\mathrm{a}\ell \mathrm{m}=3\times 2\times 5\times \mathrm{a}\times \ell \times \mathrm{m}$ The two terms have $2,5,\ell$ and m as common factors. $\therefore 20{\ell }^2\mathrm{m}+30\mathrm{a}\ell \mathrm{m}=(2\times 2\times 5\times \ell \times \ell$ $\times m)+(3\times 2\times 5\times a\times \ell \times m)$ $=2\times 5\times \ell \times \mathrm{m}\times (2\times \ell +3\times \mathrm{a})$ $=10\ell \mathrm{m}(2\ell +3\mathrm{a})$ (vi) $5x^{2}y=5\times x\times x\times y$ $15x^2=3\times 5\times x\times y\times y$ The common factors are $5,\mathrm{x}$, and y . $\therefore 5x^{2}y-15x{y}^2$ $=(5\times x\times x\times y)-(3\times 5\times x\times y\times y)$ $=5\times x\times y[x-(3\times y)]$ $=5xy(x-3y)$ (vii)We have, $10{\mathrm{a}}^2=2\times 5\times \mathrm{a}\times \mathrm{a}$, $15b^2=3\times 5\times b\times b$ and, $20c^2=2\times 2\times 5\times c\times c$ The three terms have 5 as a common factor $\therefore 10a^2-15b^2+20c^2$ $=(2\times 5\times a\times a)-(3\times 5\times b\times b)$ $+(2\times 2\times 5\times c\times c)$ $=5\times (2\times a\times a-3\times b\times b+4\times c\times c)$ $=5\left(2{\mathrm{a}}^2-3{\mathrm{b}}^2+4{\mathrm{c}}^2\right)$ (viii) We have, $4a^2=2\times 2\times a\times a$, $4\mathrm{ab}=2\times 2\times \mathrm{a}\times \mathrm{b}$ and, $4\mathrm{ca}=2\times 2\times \mathrm{c}\times \mathrm{a}$ The three terms have 2,2 and a as common factors $$\begin{gathered} \therefore -4a^2+4ab-4ca=-(2\times 2\times a\times a) \\ +(2\times 2\times a\times b)-(2\times 2\times c\times a) \\ =2\times 2\times a\times (-a+b-c) \\ =4a(-a+b-c) \end{gathered}$$ (ix) $x^{2}yz=x\times x\times y\times z$ $x{y}^{2}z=x\times y\times y\times z$ $xy{z}^2=x\times y\times z\times z$ The common factors are $\mathrm{x},\mathrm{y}$, and z . $\therefore {\mathrm{x}}^2\mathrm{yz}+{\mathrm{xy}}^2\mathrm{z}+{\mathrm{xyz}}^2=(\mathrm{x}\times \mathrm{x}\times \mathrm{y}\times \mathrm{z})$ $+(x\times y\times y\times z)+(x\times y\times z\times z)$ $=x\times y\times z[x+y+z]$ $=xyz(x+y+z)$ (x) We have, ${\mathrm{ax}}^2\mathrm{y}=\mathrm{a}\times \mathrm{x}\times \mathrm{x}\times \mathrm{y}$ $bx{y}^2=b\times x\times y\times y$ and, $cxyz=c\times x\times y\times z$ The three terms have $x$ and $y$ as common factors $\therefore a^{2}y+bx{y}^2+cxyz=(a\times x\times x\times y)$ $+(b\times x\times y\times y)+(c\times x\times y\times z)$ $=x\times y\times (a\times x+b\times y+c\times z)$ $=xy(ax+by+cz)$
• Factorise : (i) ${\mathrm{x}}^2+\mathrm{xy}+8\mathrm{x}+8\mathrm{y}$ (ii) $15xy-6x+5y-2$ (iii) $ax+bx-ay-by$ (iv) $15pq+15+9q+25p$ (v) $z-7+7xy-xyz$ Sol. (i) $x^2+xy+8x+8y=\left(x^2+xy\right)+(8x+8y)$ $=x(x+y)+8(x+y)$ $=(x+y)(x+8)$ [Taking $(x+y)$ common] (ii) $15xy-6x+5y-2$ $=3\times 5\times \mathrm{x}\times \mathrm{y}-3\times 2\times \mathrm{x}+5\times \mathrm{y}-2$ $=3x(5y-2)+1(5y-2)$ $=(5y-2)(3x+1)$ (iii) $ax+bx-ay-by=(ax+bx)-(ay+by)$ [Grouping the terms] $=(a+b)x-(a+b)y$ $=(a+b)(x-y)$ [Taking $(a+b)$ common] (iv) $15pq+15+9q+25p$ $=15pq+9q+25p+15$ $=3\times 5\times p\times q+3\times 3\times q+5\times 5\times p+3\times 5$ $=3q(5p+3)+5(5p+3)$ $=(5p+3)(3q+5)$ (v) $z-7+7xy-xyz=z-7-xyz+7xy$ $=1(z-7)-xy(z-7)$ $=(\mathrm{z}-7)(1-\mathrm{xy})$ [Taking (z-7) common]

Exercise : 12.2

• Factorise the following expressions : (i) ${\mathrm{a}}^2+8\mathrm{a}+16$ (ii) ${\mathrm{p}}^2-10\mathrm{p}+25$ (iii) $25m^2+30m+9$ (iv) $49y^2+84yz+36z^2$ (v) $4x^2-8x+4$ (vi) $121b^2-88bc+16c^2$ (vii) $(\ell +m{)}^2-4\ell m$ (viii) $a^4+2a^2{b}^2+b^4$ Sol. (i) $a^2+8a+16=a^2+2\times a\times 4+4^2$ $=(a+4{)}^2$ [Using : $a^2+2ab+b^2=(a+b{)}^2]$ $=(a+4)(a+4)$ (ii) ${\mathrm{p}}^2-10\mathrm{p}+25=(\mathrm{p}{)}^2-2\times \mathrm{p}\times 5+(5{)}^2$ $=(p-5{)}^2\left[\because (a-b{)}^2=a^2-2ab+b^2\right]$ (iii) $25{\mathrm{m}}^2+30\mathrm{m}+9=(5\mathrm{m}{)}^2+2\times 5\mathrm{m}\times 3+(3{)}^2$ $=(5m+3{)}^2$ [Using : ${\mathrm{a}}^2+2\mathrm{ab}+{\mathrm{b}}^2=(\mathrm{a}+\mathrm{b}{)}^2$ ] $=(5m+3)(5m+3)$ (iv) $49y^2+84yz+36z^2$ $=(7y{)}^2+2\times (7y)\times (6z)+(6z{)}^2$ $=(7y+6z{)}^2$ $\left[\because (a+b{)}^2=a^2+2ab+b^2\right]$ (v) $4{\mathrm{x}}^2-8\mathrm{x}+4=4\left({\mathrm{x}}^2-2\mathrm{x}+1\right)$ $=4\left(x^2-2\times x\times 1+1^2\right)$ $=4(x-1{)}^2$ [Using : ${\mathrm{a}}^2-2\mathrm{ab}+{\mathrm{b}}^2=(\mathrm{a}-\mathrm{b}{)}^2$ ] $=4(\mathrm{x}-1)(\mathrm{x}-1)$ (vi) $121b^2-88bc+16c^2$ $=(11b{)}^2-2(11b)(4c)+(4c{)}^2$ $=(11b-4c{)}^2$ $\left[\because (a-b{)}^2=a^2-2ab+b^2\right]$ (vii) $(\ell +m{)}^2-4\ell m={\ell }^2+2\ell m+m^2-4\ell m$ $$\begin{gathered} ={\ell }^2-2\ell \mathrm{m}+{\mathrm{m}}^2 \\ =(\ell -\mathrm{m}{)}^2=(\ell -\mathrm{m})(\ell -\mathrm{m}) \end{gathered}$$ (viii) $a^4+2a^2{b}^2+b^4={\left(a^2\right)}^2+2\left(a^2\right)\left(b^2\right)+{\left(b^2\right)}^2$ $={\left(a^2+b^2\right)}^2$ $\left[\because (a+b{)}^2=a^2+2ab+b^2\right]$
• Factorise (i) $4p^2-9q^2$ (ii) $63a^2-112b^2$ (iii) $49x^2-36$ (iv) $16x^5-144x^3$ (v) $(\ell +m{)}^2-(\ell -m{)}^2$ (vi) $9x^2{y}^2-16$ (vii) $\left({\mathrm{x}}^2-2\mathrm{xy}+{\mathrm{y}}^2\right)-{\mathrm{z}}^2$ (viii) $25a^2-4b^2+28bc-49c^2$ Sol. (i) $4p^2-9q^2=(2p{)}^2-(3q{)}^2$ $=(2p+3q)(2p-3q)$ [Using : ${\mathrm{a}}^2-{\mathrm{b}}^2=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})$ ] (ii) $63a^2-112b^2=7\left(9a^2-16b^2\right)$ $=7\left[(3a{)}^2-(4b{)}^2\right]$ $=7(3a+4b)(3a-4b)$ $\left[\because a^2-b^2=(a-b)(a+b)\right]$ (iii) $49x^2-36=(7x{)}^2-(6{)}^2$ $=(7x-6)(7x+6)$ $\left[\because a^2-b^2=(a-b)(a+b)\right]$ (iv) $16x^5-144x^3=16x^3\left(x^2-9\right)$ $=16x^3\left[(x{)}^2-(3{)}^2\right]$ $=16x^3(x-3)(x+3)$ $\left[\because a^2-b^2=(a-b)(a+b)\right]$ (v) $(\ell +m{)}^2-(\ell -m{)}^2$ $=[(\ell +m)+(\ell -m)][(\ell +m)-(\ell -m)]$ $=(2\ell )(2m)=4\ell m\left[\because a^2-b^2=(a-b)(a+b)\right]$ (vi) $9x^2{y}^2-16=(3xy{)}^2-(4{)}^2$ $=(3xy-4)(3xy+4)$ $\left[\because a^2-b^2=(a-b)(a+b)\right]$ (vii) $\left(x^2-2xy+y^2\right)-z^2$ $=(x-y{)}^2-(z{)}^2$ $\left[\because (a-b{)}^2=a^2-2ab+b^2\right]$ $=(x-y-z)(x-y+z)$ $\left[\because a^2-b^2=(a-b)(a+b)\right]$ (viii) $25a^2-4b^2+28bc-49c^2$ $=25{\mathrm{a}}^2-\left(4{\mathrm{b}}^2-28\mathrm{bc}+49{\mathrm{c}}^2\right)$ $=(5a{)}^2-\left[(2b{)}^2-2\times 2b\times 7c+(7c{)}^2\right]$ $=(5a{)}^2-\left[(2b-7c{)}^2\right]$ [Using identity $(a-b{)}^2=a^2-2ab+b^2$ ] $=[5\mathrm{a}+(2\mathrm{b}-7\mathrm{c})][5\mathrm{a}-(2\mathrm{b}-7\mathrm{c})]$ [Using identity ${\mathrm{a}}^2-{\mathrm{b}}^2=(\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})$ ] $=(5a+2b-7c)(5a-2b+7c)$
• Factorise the following expressions (i) $a{x}^2+bx$ (ii) $7p^2+21q^2$ (iii) $2x^3+2x{y}^2+2x{z}^2$ (iv) $a{m}^2+b{m}^2+b{n}^2+a{n}^2$ (v) $(\ell m+\ell )+m+1$ (vi) $y(y+z)+9(y+z)$ (vii) $5y^2-20y-8z+2yz$ (viii) $10ab+4a+5b+2$ (ix) $6xy-4y+6-9x$ Sol. (i) $a{x}^2+bx=x(ax+b)$ (ii) $7{\mathrm{p}}^2+21{\mathrm{q}}^2=7\times \mathrm{p}\times \mathrm{p}+3\times 7\times \mathrm{q}\times \mathrm{q}$ $=7\left(p^2+3q^2\right)$ (iii) $2x^3+2x{y}^2+2x{z}^2=2x\left(x^2+y^2+z^2\right)$ (iv) ${\mathrm{am}}^2+b{m}^2+b{n}^2+a{n}^2$ $$\begin{gathered} =\left(a{m}^2+b{m}^2\right)+\left(b{n}^2+a{n}^2\right) \\ =(a+b)m^2+(b+a)n^2 \\ =(a+b)\left(m^2+n^2\right) \end{gathered}$$ (v) $(\ell m+\ell )+m+1=\ell (m+1)+1(m+1)$ $=(m+1)(\ell +1)$ (vi) $y(y+z)+9(y+z)=(y+z)(y+9)$ (vii) $5y^2-20y-8z+2yz=5y^2-20y+2yz-8z$ $$\begin{gathered} =5y(y-4)+2z(y-4) \\ =(y-4)(5y+2z) \end{gathered}$$ (viii) $10\mathrm{ab}+4\mathrm{a}+5\mathrm{b}+2$ $$\begin{gathered} =(10ab+5b)+(4a+2) \\ =5b(2a+1)+2(2a+1) \\ =(2a+1)(5b+2) \end{gathered}$$ (ix) $6xy-4y+6-9x=6xy-9x-4y+6$ $$\begin{gathered} =3x(2y-3)-2(2y-3) \\ =(2y-3)(3x-2) \end{gathered}$$
• Factorise as far as you can : (i) $a^4-b^4$ (ii) ${\mathrm{p}}^4-81$ (iii) $x^4-(y+z{)}^4$ (iv) $x^4-(x-z{)}^4$ (v) $a^4-2a^2{b}^2+b^4$ Sol. (i) $a^4-b^4={\left(a^2\right)}^2-{\left(b^2\right)}^2$ $$\begin{gathered} =\left(a^2+b^2\right)\left(a^2-b^2\right) \\ =\left(a^2+b^2\right)(a+b)(a-b) \end{gathered}$$ (ii) ${\mathrm{p}}^4-81={\left({\mathrm{p}}^2\right)}^2-(9{)}^2$ $=\left(p^2-9\right)\left(p^2+9\right)$ $=\left[(p{)}^2-(3{)}^2\right]\left(p^2+9\right)$ $=(p-3)(p+3)\left(p^2+9\right)$ (iii) $x^4-(y+z{)}^4={\left(x^2\right)}^2-{\left[(y+z{)}^2\right]}^2$ $=\left[x^2-(y+z{)}^2\right]\left[x^2+(y+z{)}^2\right]$ $=[x-(y+z)][x+(y+z)]\left[x^2+(y+z{)}^2\right]$ $=(x-y-z)(x+y+z)\left[x^2+(y+z{)}^2\right]$ (iv) ${\mathrm{x}}^4-(\mathrm{x}-\mathrm{z}{)}^4={\left({\mathrm{x}}^2\right)}^2-{\left[(\mathrm{x}-\mathrm{z}{)}^2\right]}^2$ $=\left[x^2-(x-z{)}^2\right]\left[x^2+(x-z{)}^2\right]$ $=[\mathrm{x}-(\mathrm{x}-\mathrm{z})][\mathrm{x}+(\mathrm{x}-\mathrm{z})]\left[{\mathrm{x}}^2+(\mathrm{x}-\mathrm{z}{)}^2\right]$ $=\mathrm{z}(2\mathrm{x}-\mathrm{z})\left({\mathrm{x}}^2+{\mathrm{x}}^2-2\mathrm{xz}+{\mathrm{z}}^2\right)$ $=\mathrm{z}(2\mathrm{x}-\mathrm{z})\left(2{\mathrm{x}}^2-2\mathrm{xz}+{\mathrm{z}}^2\right)$ (v) $a^4-2a^2{b}^2+b^4={\left(a^2\right)}^2-2\times a^2\times b^2+{\left(b^2\right)}^2$ $={\left(a^2-b^2\right)}^2=[(a+b)(a-b){]}^2$ $=(a+b{)}^2(a-b{)}^2$ $=(a+b)(a+b)(a-b)(a-b)$
• Factorise the following expressions (i) ${\mathrm{p}}^2+6\mathrm{p}+8$ (ii) $q^2-10q+21$ (iii) $p^2+6p-16$ Sol. (i) $p^2+6p+8=\left(p^2+6p+9\right)-1$ [Using : 8=9-1] $=\left(p^2+2\times p\times 3+3^2\right)-1$ $=(p+3{)}^2-1^2=(p+3+1)(p+3-1)$ $=(p+4)(p+2)$ (ii) $q^2-10q+21$ It can be observed that, $21=(-7)\times (-3)$ and $(-7)+(-3)=-10$ $\therefore q^2-10q+21=q^2-7q-3q+21$ $=q(q-7)-3(q-7)$ $=(q-7)(q-3)$ (iii) $p^2+6p-16=\left(p^2+6p+9\right)-9-16$ $=(p+3{)}^2-5^2=(p+3+5)(p+3-5)$ $=(p+8)(p-2)$

Exercise : 12.3

• Carry out the following division (i) $28x^4÷56x$ (ii) $-36y^3÷9y^2$ (iii) $66{\mathrm{pq}}^2{\mathrm{r}}^3÷11{\mathrm{qr}}^2$ (iv) $34x^3{y}^3{z}^3÷51x^2{z}^3$ (v) $12a^8{b}^8÷\left(-6a^6{b}^4\right)$ Sol. (i) $28{\mathrm{x}}^4÷56\mathrm{x}$ $=\frac{28\times \mathrm{x}\times \mathrm{x}\times \mathrm{x}\times \mathrm{x}}{28\times 2\times \mathrm{x}}=\frac{\mathrm{x}\times \mathrm{x}\times \mathrm{x}}{2}=\frac{{\mathrm{x}}^3}{2}$ (ii) $36y^3=2\times 2\times 3\times 3\times y\times y\times y$ $9y^2=3\times 3\times y\times y$ $-36y^3÷9y^2=\frac{-2\times 2\times 3\times 3\times y\times y\times y}{3\times 3\times y\times y}$ $=-4y$ (iii) $66{\mathrm{pq}}^2{\mathrm{r}}^3÷11{\mathrm{qr}}^2$ $=\frac{6\times 11\times p\times q\times q\times r\times r\times r}{11\times q\times r\times r}$ $=\frac{6\times \mathrm{p}\times \mathrm{q}\times \mathrm{r}}{1}=6\mathrm{pqr}$ (iv) $34x^3{y}^3{z}^3$ $=2\times 17\times \mathrm{x}\times \mathrm{x}\times \mathrm{x}\times \mathrm{y}\times \mathrm{y}\times \mathrm{y}\times \mathrm{z}\times \mathrm{z}\times \mathrm{z}$ $51{\mathrm{xy}}^2{\mathrm{z}}^3=3\times 17\times \mathrm{x}\times \mathrm{y}\times \mathrm{y}\times \mathrm{z}\times \mathrm{z}\times \mathrm{z}$ $34x^3{y}^3{z}^3÷51x^2{z}^3$ $=\frac{2\times 17\times \mathrm{x}\times \mathrm{x}\times \mathrm{x}\times \mathrm{y}\times \mathrm{y}\times \mathrm{y}\times \mathrm{z}\times \mathrm{z}\times \mathrm{z}}{3\times 17\times \mathrm{x}\times \mathrm{y}\times \mathrm{y}\times \mathrm{z}\times \mathrm{z}\times \mathrm{z}}$ $=\frac{2}{3}{x}^{2}y$ (v) $12a^8{b}^8÷\left(-6a^6{b}^4\right)$ $6\times 2\times a\times a\times a\times a\times a\times a\times a\times a\times b\times b$ $$\begin{gathered} =\frac{\times b\times b\times b\times b\times b\times b}{-1\times 6\times a\times a\times a\times a\times a\times a\times b\times b\times b\times b} \\ =\frac{2\times a\times a\times b\times b\times b\times b}{-1}=-2a^2{b}^4 \end{gathered}$$
• Divide the given polynomial by the given monomial. (i) $\left(5x^2-6x\right)÷3x$ (ii) $\left(3y^8-4y^6+5y^4\right)÷y^4$ (iii) $8\left(x^3{y}^2{z}^2+x^2{y}^3{z}^2+x^2{y}^2{z}^3\right)÷4x^2{y}^2{z}^2$ (iv) $\left(x^3+2x^2+3x\right)÷2x$ (v) $\left(p^3{q}^6-p^6{q}^3\right)÷p^3{q}^3$ Sol. (i) $\left(5x^2-6x\right)÷3x=\frac{5x^2-6x}{3x}=\frac{5x^2}{3x}-\frac{6x}{3x}$ $=\frac{5}{3}x-2=\frac{1}{3}(5x-6)$ (ii) $3y^8-4y^6+5y^4=y^4\left(3y^4-4y^2+5\right)$ $\left(3y^8-4y^6+5y^4\right)÷y^4$ $=\frac{y^4\left(3y^4-4y^2+5\right)}{y^4}$ $=3y^4-4y^2+5$ (iii) $8\left(x^3{y}^2{z}^2+x^2{y}^3{z}^2+x^2{y}^2{z}^3\right)÷4x^2{y}^2{z}^2$ (iv) $x^3+2x^2+3x=x\left(x^2+2x+3\right)$ $$\begin{gathered} \left(x^3+2x^2+3x\right)÷2x=\frac{x\left(x^2+2x+3\right)}{2x} \\ =\frac{1}{2}\left(x^2+2x+3\right) \end{gathered}$$ (v) $\left(p^3{q}^6-p^6{q}^3\right)÷p^3{q}^3=\frac{p^3{q}^6-p^6{q}^3}{p^3{q}^3}$ $=\frac{p^3{q}^3\left(q^3-p^3\right)}{p^3{q}^3}=q^3-p^3$
• Work out the following divisions. (i) $(10x-25)÷5$ (ii) $(10x-25)÷(2x-5)$ (iii) $10y(6y+21)÷5(2y+7)$ (iv) $9x^2{y}^2(3z-24)÷27xy(z-8)$ (v) $96abc(3a-12)(5b-30)÷144(a-4)(b-6)$ Sol. (i) $(10\mathrm{x}-25)÷5=\frac{10\mathrm{x}-25}{5}=\frac{5\times (2\mathrm{x}-5)}{5}$ $=2x-5$ (ii) $(10x-25)÷(2x-5)=\frac{2\times 5\times x-5\times 5}{(2x-5)}$ $=\frac{5(2x-5)}{2x-5}=5$ (iii) $10y(6y+21)÷5(2y+7)$ $$\begin{gathered} =\frac{10y(6y+21)}{5(2y+7)} \\ =\frac{5\times 2\times y\times 3\times (2y+7)}{5\times (2y+7)}=\frac{2\times y\times 3}{1}=6y \end{gathered}$$ (iv) $9x^2{y}^2(3z-24)÷27xy(z-8)$ $$\begin{gathered} =\frac{9x^2{y}^2[3\times z-2\times 2\times 2\times 3]}{27xy(z-8)} \\ =\frac{xy\times 3(z-8)}{3(z-8)}=xy \end{gathered}$$ (v) $96abc(3a-12)(5b-30)÷144(a-4)$ (b-6) $=\frac{96\mathrm{abc}(3\mathrm{a}-12)(5\mathrm{b}-30)}{144(\mathrm{a}-4)(\mathrm{b}-6)}$ $=\frac{48\times 2\times \mathrm{abc}\times 3\times (\mathrm{a}-4)\times 5\times (\mathrm{b}-6)}{48\times 3\times (\mathrm{a}-4)\times (\mathrm{b}-6)}$ $=\frac{2\times \mathrm{abc}\times 5}{1}=10\mathrm{abc}$
• Divide as directed. (i) $5(2x+1)(3x+5)÷(2x+1)$ (ii) $26xy(x+5)(y-4)÷13x(y-4)$ (iii) $52\mathrm{pqr}(\mathrm{p}+\mathrm{q})(\mathrm{q}+\mathrm{r})(\mathrm{r}+\mathrm{p})÷104\mathrm{pq}(\mathrm{q}$ $+\mathrm{r})(\mathrm{r}+\mathrm{p})$ (iv) $20(y+4)\left(y^2+5y+3\right)÷5(y+4)$ (v) $x(x+1)(x+2)(x+3)÷x(x+1)$ Sol. (i) $5(2x+1)(3x+5)÷(2x+1)$ $$\begin{gathered} =\frac{5(2x+1)(3x+5)}{(2x+1)}=\frac{5(3x+5)}{1} \\ =5(3x+5) \end{gathered}$$ (ii) $26xy(x+5)(y-4)÷13x(y-4)$ $=\frac{2\times 13\times xy(x+5)(y-4)}{13x(y-4)}=2y(x+5)$ (iii) $52\mathrm{pqr}(\mathrm{p}+\mathrm{q})(\mathrm{q}+\mathrm{r})(\mathrm{r}+\mathrm{p})÷104\mathrm{pq}$ $$\begin{gathered} (\mathrm{q}+\mathrm{r})(\mathrm{r}+\mathrm{p}) \\ =\frac{52\times \mathrm{p}\times \mathrm{q}\times \mathrm{r}\times (\mathrm{p}+\mathrm{q})\times (\mathrm{q}+\mathrm{r})\times (\mathrm{r}+\mathrm{p})}{52\times 2\times \mathrm{p}\times \mathrm{q}\times (\mathrm{q}+\mathrm{r})\times (\mathrm{r}+\mathrm{p})} \\ =\frac{1}{2}\mathrm{r}(\mathrm{p}+\mathrm{q}) \end{gathered}$$ (iv) $20(y+4)\left(y^2+5y+3\right)$ $=2\times 2\times 5\times (y+4)\left(y^2+5y+3\right)$ $$\begin{gathered} 20(y+4)\left(y^2+5y+3\right)÷5(y+4) \\ =\frac{2\times 2\times 5\times (y+4)\times \left(y^2+5y+3\right)}{5\times (y+4)} \\ =4\left(y^2+5y+3\right) \end{gathered}$$ (v) $x(x+1)(x+2)(x+3)÷x(x+1)$ $=\frac{x(x+1)(x+2)(x+3)}{x(x+1)}=\frac{(x+2)(x+3)}{1}$ $=(x+2)(x+3)$
• Factorise the expressions and divide them as directed. (i) $\left(y^2+7y+10\right)÷(y+5)$ (ii) $\left(m^2-14m-32\right)÷(m+2)$ (iii) $\left(5p^2-25p+20\right)÷(p-1)$ (iv) $4yz\left(z^2+6z-16\right)÷2y(z+8)$ (v) $5pq\left(p^2-q^2\right)÷2p(p+q)$ (vi) $12xy\left(9x^2-16y^2\right)÷4xy(3x+4y)$ (vii) $39y^3\left(50y^2-98\right)÷26y^2(5y+7)$ Sol. (i) $\left(y^2+7y+10\right)=y^2+2y+5y+10$ $=y(y+2)+5(y+2)$ $=(y+2)(y+5)$ $\therefore \left(y^2+7y+10\right)÷(y+5)$ $=\frac{(y+2)(y+5)}{y+5}[U\mathrm{sing}(1)]$ $=y+2$ (ii) $m^2-14m-32=m^2+2m-16m-32$ $=m(m+2)-16(m+2)$ $=(m+2)(m-16)$ Now, $\left(m^2-14m-32\right)÷(m+2)$ $=\frac{(m+2)(m-16)}{(m+2)}=m-16$ (iii) $\left(5p^2-25p+20\right)=5\left(p^2-5p+4\right)$ $=5\left(p^2-p-4p+4\right)$ $=5[p(p-1)-4(p-1)]$ $=5(p-1)(p-4)$ $\therefore \left(5p^2-25p+20\right)÷(p-1)$ $=\frac{5p^2-25p+20}{p-1}$ $=\frac{5(p-1)(p-4)}{p-1}[$ Using (1)] $=5(p-4)$ (iv) $4yz\left(z^2+6z-16\right)=4yz\left(z^2+8z-2z-16\right)$ $=4yz[z(z+8)-2(z+8)$ $=4yz(z+8)(z-2)$ $\therefore 4yz\left(z^2+6z-16\right)÷2y(z+8)$ $=\frac{4yz\left(z^2+6z-16\right)}{2y(z+8)}=\frac{4yz(z+8)(z-2)}{2y(z+8)}$ [Using (1)] $=\frac{2\mathrm{z}(\mathrm{z}-2)}{1}=2\mathrm{z}(\mathrm{z}-2)$ (v) $5pq\left(p^2-q^2\right)=5pq(p-q)(p+q)\dots (1)$ $\therefore 5pq\left(p^2-q^2\right)÷2p(p+q)$ $=\frac{5\mathrm{pq}\left({\mathrm{p}}^2-{\mathrm{q}}^2\right)}{2\mathrm{p}(\mathrm{p}+\mathrm{q})}$ $=\frac{5\times \mathrm{p}\times \mathrm{q}\times (\mathrm{p}-\mathrm{q})\times (\mathrm{p}+\mathrm{q})}{2\times \mathrm{p}\times (\mathrm{p}+\mathrm{q})}[$ Using (1) $]$ $=\frac{5}{2}q(p-q)$ (vi) $12xy\left(9x^2-16y^2\right)=12xy[(3x{)}^2-$ (4y) ${}^2$ ] $=12xy(3x-4y)(3x+4y)$ $=12xy\left(9x^2-16y^2\right)÷4xy(3x+4y)$ $=\frac{2\times 2\times 3\times x\times y\times (3x-4y)\times (3x+4y)}{2\times 2\times x\times y\times (3x+4y)}$ $=3(3x-4y)$ (vii) $39y^3\left(50y^2-98\right)÷26y^2(5y+7)$ $\Rightarrow 39{\mathrm{y}}^3\left(50{\mathrm{y}}^2-98\right)=3\times 13\times \mathrm{y}\times \mathrm{y}\times$ $y\times 2\left[\left(25y^2-49\right)\right]$ $=3\times 13\times 2\times \mathrm{y}\times \mathrm{y}\times \mathrm{y}\times \left[(5\mathrm{y}{)}^2-(7{)}^2\right]$ $=3\times 13\times 2\times \mathrm{y}\times \mathrm{y}\times \mathrm{y}\times (5\mathrm{y}-7)(5\mathrm{y}+7)$ $\Rightarrow 26{\mathrm{y}}^2(5\mathrm{y}+7)=2\times 13\times \mathrm{y}\times \mathrm{y}\times (5\mathrm{y}+7)$ $\Rightarrow 39y^3\left(50y^2-98\right)÷26y^2(5y+7)$ $\frac{3\times 13\times 2\times y\times y\times y\times (5y-7)(5y+7)}{2\times 13\times y\times y\times (5y+7)}$ $=3y(5y-7)$