Exercise 12.3 — Division of Expressions

Division of algebraic expressions.

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What is Exercise 12.3 About?

Exercise 12.3 introduces the division of algebraic expressions — the reverse operation of multiplication. Just as you learned to multiply monomials and polynomials in Chapter 11, here you learn to split an expression by a divisor to find the quotient. This skill is essential for simplifying complex expressions, cancelling common factors, and preparing for polynomial long division in higher classes.

The exercise covers three types of division, all based on the idea of cancelling common factors in the numerator and denominator:

① Monomial ÷ Monomial
48a³ ÷ 6a → cancel common factors of numbers and variables
② Polynomial ÷ Monomial
Divide each term of the polynomial separately by the monomial
③ Polynomial ÷ Polynomial
Factorise the numerator, then cancel the common bracket
💡 Core Principle: Division of algebraic expressions always works by writing the problem as a fraction and then cancelling common factors — exactly like simplifying numerical fractions such as 12/8 = 3/2.

How Algebraic Division Works — The Key Concept

The law of exponents for division is the engine behind every problem in this exercise. When dividing two powers of the same variable, you subtract the exponent of the denominator from the exponent of the numerator:

xᵃ ÷ xᵇ = xᵃ⁻ᵇ   (where a > b)

For example, the PDF's introductory worked example demonstrates this clearly:

Dividend 24a⁵
÷
Divisor 18a²
=
Quotient 4a³/3
Write as fraction: 24a⁵ / 18a² Expand fully: (2×2×2×3 × a×a×a×a×a) / (2×3×3 × a×a) Cancel common factors (2, 3, a, a): = 4a³/3
Write as a fraction Prime-factorise coefficients Cancel common factors Subtract exponents of variables

Question 1 — Divide Monomial by Monomial (Parts i–v)

Each part asks you to divide one monomial by another. The method: prime-factorise both the coefficient and the variable powers, then cancel identical factors top and bottom.

Q1 · Part (i)
Divide 48a³ by 6a
48a³ / 6a = (2×2×2×2×3 × a×a×a) / (2×3 × a) ← prime factorise = (2×2×2) × ← cancel one 2, one 3, one a = 8a²
✅ Coefficient: 48 ÷ 6 = 8  |  Variable: a³ ÷ a = a³⁻¹ = a²
Q1 · Part (ii)
Divide 14x³ by 42x²
14x³ / 42x² = (2×7 × x×x×x) / (2×3×7 × x×x) = 1/3 × x ← 2 and 7 cancel; two x's cancel = x/3
📌 Here the coefficient becomes a fraction (14 ÷ 42 = 1/3). This is perfectly valid — algebraic coefficients can be fractions.
Q1 · Part (iii)
Divide 72a³b⁴c⁵ by 8ab²c³
72a³b⁴c⁵ / 8ab²c³ = (2×2×2×3×3 × a³ × b⁴ × c⁵) / (2×2×2 × a × b² × c³) = 9 × a³⁻¹ × b⁴⁻² × c⁵⁻³ ← subtract exponents = 9a²b²c²
Coefficient: 72 ÷ 8 = 9  |  a: 3−1=2  |  b: 4−2=2  |  c: 5−3=2
Q1 · Part (iv)
11xy²z³ ÷ 55xyz
= (11 × x × y² × z³) / (11×5 × x × y × z)
= (1/5) × y²⁻¹ × z³⁻¹
= (1/5)yz²
yz²/5
Q1 · Part (v)
−54l⁴m³n² ÷ 9l²m²n²
= −(2×3³ × l⁴ × m³ × n²) / (3² × l² × m² × n²)
= −(2×3) × l⁴⁻² × m³⁻² ← n² cancels
= −6 × l² × m
−6l²m
Pattern for monomial division: Divide coefficients numerically → subtract exponents of matching variables → variables not in denominator stay as-is with their original powers.

Question 2 — Divide a Polynomial by a Monomial (Parts i–vii)

When dividing a polynomial (multi-term expression) by a monomial, you divide each term of the polynomial separately by the monomial. Think of it as distributing the division across the addition and subtraction signs:

(A + B + C) ÷ M = (A/M) + (B/M) + (C/M)
Q2 · Part (i)
(3x² − 2x) ÷ x
(3x² − 2x) / x = 3x²/x2x/x ← divide each term separately = 3x²⁻¹ − 2x¹⁻¹ = 3x − 2
Q2 · Part (ii)
(5a³b − 7ab³) ÷ ab
(5a³b − 7ab³) / ab = 5a³b/ab − 7ab³/ab = 5a³⁻¹b¹⁻¹ − 7a¹⁻¹b³⁻¹ = 5a² × b⁰ − 7a⁰ × b² = 5a² − 7b²
💡 Remember: any variable with exponent 0 equals 1 and disappears (a⁰ = 1).
Q2 · Part (iii)
(25x⁵ − 15x⁴) ÷ 5x³
(25x⁵ − 15x⁴) / 5x³ = 25x⁵/5x³ − 15x⁴/5x³ = 5x² − 3x = x(5x − 3) ← factor out x
📌 The answer can be written as 5x² − 3x or in factored form x(5x − 3). Both are correct; the factored form is neater.
Q2 · Part (iv)
(4l⁵ − 6l⁴ + 8l³) ÷ 2l²
(4l⁵ − 6l⁴ + 8l³) / 2l² = 4l⁵/2l² − 6l⁴/2l² + 8l³/2l² = 2l³ − 3l² + 4l = l(2l² − 3l + 4)
Q2 · Part (v)
15(a³b²c² − a²b³c² + a²b²c³) ÷ 3abc
= 15a³b²c²/3abc − 15a²b³c²/3abc + 15a²b²c³/3abc = 5a²bc − 5ab²c + 5abc² = 5abc(a − b + c) ← take out common factor 5abc = 5abc(a − b + c)
Q2 · Part (vi)
(3p³ − 9p²q − 6pq²) ÷ (−3p)
= 3p³/(−3p) − 9p²q/(−3p) − 6pq²/(−3p)
= −p² + 3pq + 2q²
−p² + 3pq + 2q²
⚠️ Dividing by a negative monomial flips all signs. Check each term carefully.
Q2 · Part (vii)
(⅔a²b²c² + 4/3ab²c²) ÷ (½abc)
= (⅔a²b²c²)/(½abc) + (4/3ab²c²)/(½abc)
= (2 × 2/3)abc + (2 × 4/3)bc
= (4/3)abc + (8/3)bc
(4/3)bc(a + 2)
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Question 3 — Division When Expressions are in Factored Form

In these problems, both the dividend and divisor are given as products of factors (already in a partially factored form). The trick is to spot and cancel the common factor bracket directly — no need to expand first.

Q3 · Part (i)
(49x − 63) ÷ 7
(49x − 63) / 7 = 7(7x − 9) / 7 ← factor out 7 from numerator = (7x − 9) ← the 7s cancel = 7x − 9
Q3 · Part (ii)
12x(8x − 20) ÷ 4(2x − 5)
12x(8x − 20) / 4(2x − 5) = 12x × 4(2x − 5) / 4(2x − 5) ← since 8x−20 = 4(2x−5) = 12x ← 4(2x−5) cancels completely = 12x
💡 The key step: recognise that 8x − 20 = 4(2x − 5). Once you rewrite, the common factor bracket cancels instantly.
Q3 · Part (iii)
11a³b³(7c − 35) ÷ 3a²b²(c − 5)
11a³b³(7c − 35) / 3a²b²(c − 5) = 11a³b³ × 7(c − 5) / 3a²b²(c − 5) ← 7c−35 = 7(c−5) = (11 × 7 × ab) / 3 ← (c−5) cancels; a³/a²=a, b³/b²=b = 77ab/3
Q3 · Part (iv)
54lmn(l+m)(m+n)(n+l) ÷ 8lmn(l+m)(n+l)
Cancel common: lmn, (l+m), (n+l)
= 54(m+n) / 8
= 27(m+n) / 4
27(m+n)/4
Q3 · Part (v)
36(x+4)(x²+7x+10) ÷ 9(x+4)
Cancel 9 into 36 → 4
Cancel (x+4) from numerator & denominator
4(x²+7x+10)
Q3 · Part (vi)
a(a+1)(a+2)(a+3) ÷ a(a+3)
a(a+1)(a+2)(a+3) / a(a+3) = (a+1)(a+2) ← a and (a+3) both cancel = (a+1)(a+2)

Question 4 — Factorise the Expression, then Divide

These are the most important problems in Exercise 12.3. Here, the numerator is a quadratic expression that you must first factorise into two brackets. Then the divisor cancels one of those brackets, leaving the other as the final answer. The factorisation identity used is:

x² + (a+b)x + ab = (x + a)(x + b)
a² − b² = (a + b)(a − b)    ← Difference of Squares
Q4 · Part (i)
(x² + 7x + 12) ÷ (x + 3)
Step 1 — Factorise the numerator: x² + 7x + 12 = x² + (3 + 4)x + 3×4 ← find two numbers with sum 7, product 12 → 3 and 4 = (x + 3)(x + 4)   Step 2 — Divide: (x + 3)(x + 4) / (x + 3) = (x + 4) ← (x+3) cancels = x + 4
Q4 · Part (ii)
(x² − 8x + 12) ÷ (x − 6)
Step 1 — Factorise: x² − 8x + 12 = x² + (−2 − 6)x + (−2)(−6) ← sum=−8, product=+12 → −2 and −6 = (x − 2)(x − 6)   Step 2 — Divide: (x − 2)(x − 6) / (x − 6) = x − 2 = x − 2
Q4 · Part (iii)
(p² + 5p + 4) ÷ (p + 1)
p² + 5p + 4 = (p + 1)(p + 4) ← sum=5, product=4 → 1 and 4 (p + 1)(p + 4) / (p + 1) = p + 4
Q4 · Part (iv)
15ab(a² − 7a + 10) ÷ 3b(a − 2)
Step 1 — Factorise quadratic: a² − 7a + 10 = (a − 2)(a − 5) ← sum=−7, product=10 → −2 and −5   Step 2 — Substitute and simplify: 15ab(a − 2)(a − 5) / 3b(a − 2) = 5a(a − 5) ← 15/3=5; b cancels; (a−2) cancels = 5a(a − 5)
Q4 · Part (v)
15lm(2p² − 2q²) ÷ 3l(p + q)
2p² − 2q² = 2(p² − q²) = 2(p+q)(p−q)
[Difference of Squares: a²−b²=(a+b)(a−b)]

15lm × 2(p+q)(p−q) / 3l(p+q)
= (15×2m(p−q)) / 3 = 10m(p−q)
10m(p − q)
Q4 · Part (vi)
26z³(32z² − 18) ÷ 13z²(4z − 3)
32z² − 18 = 2(16z² − 9) = 2(4z+3)(4z−3)
[16z²−9 = (4z)²−3² → difference of squares]

26z³ × 2(4z+3)(4z−3) / 13z²(4z−3)
= (26×2z(4z+3)) / 13 = 4z(4z+3)
4z(4z + 3)
Common Mistake in Q4: Students sometimes try to cancel the divisor with just the coefficient of the numerator, ignoring the factored bracket. Always factorise the entire numerator first, then cancel the matching bracket with the denominator.

All Answers at a Glance — Quick Reference Table

QDivision ProblemMethodAnswer
Q1(i)48a³ ÷ 6aMonomial ÷ Monomial8a²
Q1(ii)14x³ ÷ 42x²Monomial ÷ Monomialx/3
Q1(iii)72a³b⁴c⁵ ÷ 8ab²c³Monomial ÷ Monomial9a²b²c²
Q1(iv)11xy²z³ ÷ 55xyzMonomial ÷ Monomialyz²/5
Q1(v)−54l⁴m³n² ÷ 9l²m²n²Monomial ÷ Monomial−6l²m
Q2(i)(3x² − 2x) ÷ xPolynomial ÷ Monomial3x − 2
Q2(ii)(5a³b − 7ab³) ÷ abPolynomial ÷ Monomial5a² − 7b²
Q2(iii)(25x⁵ − 15x⁴) ÷ 5x³Polynomial ÷ Monomialx(5x − 3)
Q2(iv)(4l⁵ − 6l⁴ + 8l³) ÷ 2l²Polynomial ÷ Monomiall(2l² − 3l + 4)
Q2(v)15(a³b²c² − a²b³c² + a²b²c³) ÷ 3abcPolynomial ÷ Monomial5abc(a − b + c)
Q2(vi)(3p³ − 9p²q − 6pq²) ÷ (−3p)Polynomial ÷ Monomial−p² + 3pq + 2q²
Q2(vii)(⅔a²b²c² + 4/3ab²c²) ÷ ½abcPolynomial ÷ Monomial(4/3)bc(a + 2)
Q3(i)(49x − 63) ÷ 7Factored division7x − 9
Q3(ii)12x(8x−20) ÷ 4(2x−5)Factored division12x
Q3(iii)11a³b³(7c−35) ÷ 3a²b²(c−5)Factored division77ab/3
Q3(iv)54lmn(l+m)(m+n)(n+l) ÷ 8lmn(l+m)(n+l)Factored division27(m+n)/4
Q3(v)36(x+4)(x²+7x+10) ÷ 9(x+4)Factored division4(x²+7x+10)
Q3(vi)a(a+1)(a+2)(a+3) ÷ a(a+3)Factored division(a+1)(a+2)
Q4(i)(x²+7x+12) ÷ (x+3)Factorise → dividex + 4
Q4(ii)(x²−8x+12) ÷ (x−6)Factorise → dividex − 2
Q4(iii)(p²+5p+4) ÷ (p+1)Factorise → dividep + 4
Q4(iv)15ab(a²−7a+10) ÷ 3b(a−2)Factorise → divide5a(a − 5)
Q4(v)15lm(2p²−2q²) ÷ 3l(p+q)Diff. of squares10m(p − q)
Q4(vi)26z³(32z²−18) ÷ 13z²(4z−3)Diff. of squares4z(4z + 3)

Common Mistakes to Avoid in Exercise 12.3

  • Subtracting instead of cancelling: When 6a ÷ 6a appears, students sometimes write 0 instead of 1. Dividing identical terms gives 1, not 0.
  • Forgetting to divide every term: In polynomial ÷ monomial, every single term in the numerator must be divided. Missing even one term leads to a wrong answer.
  • Ignoring the negative sign of the divisor: In Q2(vi), dividing by −3p flips all signs. Students often forget to flip the sign of every resulting term.
  • Trying to cancel without factorising first: In Q4 problems, you cannot cancel (x+3) with x² — you must factorise x²+7x+12 into (x+3)(x+4) first.
  • Wrong factor pairs for quadratics: When factorising x²+7x+12, find two numbers whose sum is 7 AND product is 12 simultaneously. A common error is to just find pairs that multiply to 12 without checking the sum.
  • Forgetting the Difference of Squares formula: For Q4(v) and Q4(vi), recognising 2p²−2q² = 2(p+q)(p−q) and 16z²−9 = (4z+3)(4z−3) is essential. Always check if the expression is a difference of two perfect squares.
Board Exam Tip (Telangana & AP): In Q4-type problems, always write out the factorisation step explicitly as a separate line. Examiners look for this working and award it carry-forward marks even if the final answer has a sign error.

What This Exercise Prepares You For

Exercise 12.3 is the gateway to polynomial division across all levels. Once you are comfortable cancelling algebraic factors here, the next step is Exercise 12.4 — Finding and Correcting Errors, where you verify whether a given algebraic simplification is correct and identify where mistakes occurred.

In Class 9 and 10, you will use the same factorisation + cancellation technique in polynomial simplification and in quadratic equations. The difference-of-squares identity (a²−b²=(a+b)(a−b)) used in Q4(v) and Q4(vi) appears repeatedly in CBSE board questions across Class 9 and 10 as well.

For CBSE, Telangana, and Andhra Pradesh board examinations, algebraic division problems carry 2–3 marks each. Questions from Exercise 12.3 — especially the factorise-then-divide type (Q4) — are frequently asked in SA1 and SA2 assessments. Mastering every sub-question here gives you a reliable source of full marks.

📐 Next in Chapter 12: Explore Exercise 12.1 — Factorisation by Common Factors and Exercise 12.2 — Factorisation by Regrouping to strengthen the factorisation skills that power every Q4 solution in this exercise.
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