NCERT Solutions Class 8 Maths Chapter 12 Factorisation Exercise 12.3

NCERT Solutions Class 8 Maths Chapter 12 Factorisation Exercise 12.3 focuses on dividing algebraic expressions. In this exercise, students learn how to divide a given algebraic expression by a monomial or a binomial. It also explains how factorisation helps in simplifying such divisions quickly and correctly. This is an important skill for solving more advanced algebraic problems in higher classes.

These NCERT Solutions are created in line with the NCERT syllabus and provide step-by-step explanations for each question. The solutions are written in simple language, with clear methods and examples to help students understand easily. By practicing NCERT Solutions from Exercise 12.3, students can build a strong foundation in algebra and improve their confidence in solving expressions involving division.

1.0Download NCERT Solutions Class 8 Maths Chapter 12 Factorisation Exercise 12.3: Free PDF

Download the free NCERT Solutions Class 8 Maths Chapter 12 Factorisation Exercise 12.3 PDF with clear, step-by-step answers to help you understand regrouping and solve problems easily.

2.0Key Concepts in Exercise 12.3 of Class 8 Maths Chapter 12

Exercise 12.3 of Chapter 12 – Factorisation focuses on the factorisation of algebraic expressions using standard algebraic identities and special product forms. This exercise strengthens students’ ability to recognise patterns and apply formulas effectively.

• Factorisation Using Identities:
  Students apply known identities directly to factorise expressions. Key identities used include:
• $$\begin{gathered} a^2-b^2=(a-b)(a+b) \\ {a}^2+2ab+b^2=(a+b{)}^2 \\ {a}^2-2ab+b^2=(a-b{)}^2 \end{gathered}$$
• Recognising Square and Difference Patterns:
  Students learn to quickly identify expressions that are perfect squares or differences of squares and factorise them without splitting the middle term.
• Efficient Factorisation Techniques:
  Students are encouraged to match expressions with identity forms to simplify them in fewer steps, improving efficiency and accuracy.
• Application of Multiple Identities:
  In more complex expressions, students may need to apply more than one identity or combine identity-based factorisation with grouping.

3.0NCERT Class 8 Maths Chapter 12: Other Exercises

4.0NCERT Class 8 Maths Chapter 12 Exercise 12.3: Detailed Solutions

1. 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) $12a^8{b}^8÷\left(-6a^6{b}^4\right)$ $6\times 2\times a\times a\times a\times a\times a\times a\times a\times a\times b\times b$ $$\begin{gathered} =\frac{\times b\times b\times b\times b\times b\times b}{-1\times 6\times a\times a\times a\times a\times a\times a\times b\times b\times b\times b} \\ =\frac{2\times a\times a\times b\times b\times b\times b}{-1}=-2a^2{b}^4 \end{gathered}$$
2. 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) $8\left(x^3{y}^2{z}^2+x^2{y}^3{z}^2+x^2{y}^2{z}^3\right)÷4x^2{y}^2{z}^2$ $$\begin{gathered} =\frac{8\left(x^3{y}^2{z}^2+x^2{y}^3{z}^2+x^2{y}^2{z}^3\right)}{4x^2{y}^2{z}^2} \\ =\frac{8\times x^2{y}^2{z}^2(x+y+z)}{4x^2{y}^2{z}^2}=2(x+y+z) \end{gathered}$$ (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) $\left(p^3{q}^6-p^6{q}^3\right)÷p^3{q}^3=\frac{p^3{q}^6-p^6{q}^3}{p^3{q}^3}$ $=\frac{p^3{q}^3\left(q^3-p^3\right)}{p^3{q}^3}=q^3-p^3$
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) $=\frac{96\mathrm{abc}(3\mathrm{a}-12)(5\mathrm{b}-30)}{144(\mathrm{a}-4)(\mathrm{b}-6)}$ $=\frac{48\times 2\times \mathrm{abc}\times 3\times (\mathrm{a}-4)\times 5\times (\mathrm{b}-6)}{48\times 3\times (\mathrm{a}-4)\times (\mathrm{b}-6)}$ $=\frac{2\times \mathrm{abc}\times 5}{1}=10\mathrm{abc}$
4. 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) $x(x+1)(x+2)(x+3)÷x(x+1)$ $=\frac{x(x+1)(x+2)(x+3)}{x(x+1)}=\frac{(x+2)(x+3)}{1}$ $=(x+2)(x+3)$
5. 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) $39y^3\left(50y^2-98\right)÷26y^2(5y+7)$ $\Rightarrow 39{\mathrm{y}}^3\left(50{\mathrm{y}}^2-98\right)=3\times 13\times \mathrm{y}\times \mathrm{y}\times$ $y\times 2\left[\left(25y^2-49\right)\right]$ $=3\times 13\times 2\times \mathrm{y}\times \mathrm{y}\times \mathrm{y}\times \left[(5\mathrm{y}{)}^2-(7{)}^2\right]$ $=3\times 13\times 2\times \mathrm{y}\times \mathrm{y}\times \mathrm{y}\times (5\mathrm{y}-7)(5\mathrm{y}+7)$ $\Rightarrow 26{\mathrm{y}}^2(5\mathrm{y}+7)=2\times 13\times \mathrm{y}\times \mathrm{y}\times (5\mathrm{y}+7)$ $\Rightarrow 39y^3\left(50y^2-98\right)÷26y^2(5y+7)$ $=$ $\frac{3\times 13\times 2\times y\times y\times y\times (5y-7)(5y+7)}{2\times 13\times y\times y\times (5y+7)}$ $=3y(5y-7)$

5.0Key Features and Benefits of Class 8 Maths Chapter 12 Exercise 12.3

• Factorisation by Regrouping Terms: Exercise 12.3 teaches students how to factorise expressions by rearranging and grouping terms in a smart way.
• Encourages Logical Thinking: This method helps students think creatively to find patterns and group terms for easier factorisation.
• Step-by-Step Learning: The exercise includes simple to slightly challenging problems that help build confidence as students practice.
• Useful for Solving Equations: Learning to factorise by regrouping is an important skill that helps in solving algebraic equations quickly.