NCERT Solutions Class 8 Maths Chapter 12 Factorisation Exercise 12.2

NCERT Solutions Class 8 Maths Chapter 12 Factorisation Exercise 12.2 helps students learn how to factorise algebraic expressions using identities and grouping methods. In this exercise, students practice breaking down more complex expressions by identifying common patterns and applying the right factorisation techniques. It builds on the basic concepts introduced in the previous exercise.

These NCERT Solutions are prepared as per the official NCERT guidelines and explained in an easy step-by-step format. With clear examples and simple language, students can understand how to factorise expressions without confusion. Practicing NCERT Solutions from Exercise 12.2 boosts confidence and improves problem-solving skills in algebra.

1.0Download NCERT Solutions Class 8 Maths Chapter 12 Factorisation Exercise 12.2: Free PDF

Download the free NCERT Solutions Class 8 Maths Chapter 12 Factorisation Exercise 12.2 PDF with simple, step-by-step solutions to help you learn and practice easily for better exam results.

2.0Key Concepts in Exercise 12.2 of Class 8 Maths Chapter 12

Exercise 12.2 of Chapter 12 – Factorisation focuses on factorising algebraic expressions using identities and grouping of terms, helping students move beyond common factors and apply deeper algebraic reasoning.

• Factorisation by Grouping Terms
  Students learn how to group terms in pairs or sets, factor out common terms from each group, and then simplify further.
• Using Algebraic Identities
  Introduction to factorisation using standard identities such as:
• $$\begin{gathered} a^2-b^2=(a-b)(a+b) \\ {a}^2+2ab+b^2=(a+b{)}^2 \\ {a}^2-2ab+b^2=(a-b{)}^2 \end{gathered}$$
• Identifying Special Products
  Students practice recognizing patterns in expressions that match identity forms, enabling quick and accurate factorisation.
• Strategic Rearrangement
  Emphasis is placed on rearranging terms (when necessary) to spot factorisation opportunities more easily.

3.0NCERT Class 8 Maths Chapter 12: Other Exercises

4.0NCERT Class 8 Maths Chapter 12 Exercise 12.2: Detailed Solutions

1. 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) $a^4+2a^2{b}^2+b^4={\left(a^2\right)}^2+2\left(a^2\right)\left(b^2\right)+{\left(b^2\right)}^2$ $={\left(a^2+b^2\right)}^2$ $\left[\because (a+b{)}^2=a^2+2ab+b^2\right]$
2. 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 ${\mathrm{a}}^2-{\mathrm{b}}^2=(\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})$ ] $=(5a+2b-7c)(5a-2b+7c)$
3. 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) $6xy-4y+6-9x=6xy-9x-4y+6$ $$\begin{gathered} =3x(2y-3)-2(2y-3) \\ =(2y-3)(3x-2) \end{gathered}$$
4. 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) $a^4-2a^2{b}^2+b^4={\left(a^2\right)}^2-2\times a^2\times b^2+{\left(b^2\right)}^2$ $={\left(a^2-b^2\right)}^2=[(a+b)(a-b){]}^2$ $=(a+b{)}^2(a-b{)}^2$ $=(a+b)(a+b)(a-b)(a-b)$
5. 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) $p^2+6p-16=\left(p^2+6p+9\right)-9-16$ $=(p+3{)}^2-5^2=(p+3+5)(p+3-5)$ $=(p+8)(p-2)$

5.0Key Features and Benefits of Class 8 Maths Chapter 12 Exercise 12.2

• Learning to Factorise Using Identities: Exercise 12.2 teaches students how to use algebraic identities for factorising expressions.
• Makes Algebra Easier: By applying known identities, students can quickly break down complex expressions into simpler factors.
• Builds Concept Clarity: This exercise helps students understand how and when to apply the correct identity while solving problems.
• Helpful for Future Topics: These skills are useful in solving equations, simplifying expressions, and learning quadratic equations in higher classes.
• Follow NCERT Guidelines: All questions are taken from the official NCERT textbook, which ensures alignment with the school curriculum and exams.