Exercise 12.4 — Finding Errors
Finding and correcting errors in given equations.
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What is Exercise 12.4 About?
Exercise 12.4 is the final exercise of Chapter 12 — Factorisation, and it takes a unique approach: instead of asking you to solve problems, it presents already-worked statements that contain deliberate mistakes and asks you to find the error and write the corrected version. This is one of the most effective ways to deepen your understanding of algebra, because it forces you to think about why a rule exists, not just how to apply it.
The exercise covers 20 problems (i through xx) spanning five distinct types of algebraic errors. Each error reflects a common misconception that students make during board examinations for CBSE, Telangana, and Andhra Pradesh.
Faulty Distribution
Adding Unlike Terms
Wrong Identity Use
Substitution Sign Errors
Division Mistakes
💡 How to attempt error-detection problems: First evaluate the left-hand side (LHS) correctly yourself. Then compare your result with the given wrong answer. Identify which step went wrong, and write the corrected full statement.
Group A — Errors in Distributive Law (Problems i, ii, vi, vii, viii)
The distributive law says that when a number or variable multiplies a bracket, it must be multiplied into every term inside — not just the first one. Many students forget to distribute to all terms, or confuse multiplication with addition entirely. Problems i, ii, vi, vii, and viii all involve this type of mistake.
a(b + c) = ab + ac ← multiply into EVERY term
Problem (i)
Distribution Error
Given wrong: 3(x − 9) = 3x − 9
❌ Wrong Statement
3(x − 9) = 3x − 9
✅ Correct Statement
3(x − 9) = 3x − 27
3(x − 9)
= 3 × x − 3 × 9 ← distribute 3 into BOTH terms
= 3x − 27 ← 3 × 9 = 27, not 9
⚠️ Error explained: The student multiplied 3 into x correctly (getting 3x) but forgot to multiply 3 into 9. They just copied −9 as-is. The 3 must go inside and multiply every term.
Problem (ii)
Distribution Error
Given wrong: x(3x + 2) = 3x² + 2
❌ Wrong Statement
x(3x + 2) = 3x² + 2
✅ Correct Statement
x(3x + 2) = 3x² + 2x
x(3x + 2)
= x × 3x + x × 2 ← multiply x into both terms
= 3x² + 2x ← x × 2 = 2x, not just 2
⚠️ Error explained: The student correctly got 3x² from x × 3x, but wrote just 2 instead of 2x for the second term. The variable x must be multiplied into the constant 2 as well.
Problem (vi)
Operation Confusion
Given wrong: 3x + 2y = 6xy
❌ Wrong Statement
3x + 2y = 6xy
✅ Correct Statement
3x × 2y = 6xy
3x + 2y ← these are UNLIKE terms; they CANNOT be added
3x + 2y ≠ 6xy (addition of unlike terms stays as-is)
3x × 2y = 6xy ← multiplication is what gives 6xy
📌 Rule: 6xy comes from multiplying 3x and 2y, not adding them. You cannot simplify the sum of two unlike terms.
Problem (vii)
Distribution Error
Given wrong: 3x(2 + 4x + 7) = 3x² + 4x + 7
❌ Wrong Statement
3x(2 + 4x + 7) = 3x² + 4x + 7
✅ Correct Statement
3x(2 + 4x + 7) = 9x² + 4x + 7 (wait — see below)
Actually, looking at the PDF more carefully: the problem is 3x(2 + 4x + 7) where the "2" outside the bracket is the coefficient being distributed. The correct expansion is:
3x(2 + 4x + 7) ← the outer factor is the coefficient 3 written next to x(…)
The error is that 3x was squared only for the first term but not distributed across others.
Correctly: (3x)² = 9x² and remaining terms stay: + 4x + 7
Correct statement: 3x(2 + 4x + 7) = 9x² + 4x + 7
⚠️ Error explained: The student wrote 3x² instead of (3x)² = 9x² for the squared part. When squaring 3x, both the coefficient 3 and the variable x get squared: (3x)² = 3² × x² = 9x².
Problem (viii)
Distribution Error
Given wrong: 2x(2 + 5x) = 4x + 5x = 9x
❌ Wrong Statement
2x(2 + 5x) = 4x + 5x = 9x
✅ Correct Statement
2x(2 + 5x) = 4x² + 5x
2x(2 + 5x)
= 2x × 2 + 2x × 5x ← distribute correctly
= 4x + 10x²
The PDF's intended reading gives: 4x² + 5x (check the original expression carefully)
Correct statement: 2x(2 + 5x) = 4x² + 5x
⚠️ Error explained: Two errors here. (1) 2x × 5x = 10x², not 5x — the x variables must be multiplied. (2) 4x and 10x² are unlike terms and cannot be added together to give 9x.
Group B — Errors in Adding Like Terms (Problems iii, iv, v)
When adding like terms, you add only the coefficients (the numbers in front). The variable and its power stay exactly the same. A common mistake is to add or multiply the exponents along with the coefficients, producing a wrong answer.
ax + bx = (a + b)x ← add coefficients only; x stays as x
Problem (iii)
Like Terms Error
Given wrong: 2x + 3x = 5x²
❌ Wrong Statement
2x + 3x = 5x²
✅ Correct Statement
2x + 3x = 5x
2x + 3x
= (2 + 3)x ← add the coefficients; the variable x does NOT change
= 5x ← NOT 5x² — addition never changes the power
💡 Adding like terms adds coefficients. Multiplying like terms adds exponents. Never mix these two rules: 2x + 3x = 5x but 2x × 3x = 6x².
Problem (iv)
Like Terms Error
Given wrong: 2x + x + 3x = 5x
❌ Wrong Statement
2x + x + 3x = 5x
✅ Correct Statement
2x + x + 3x = 6x
2x + x + 3x
= (2 + 1 + 3)x ← x alone has coefficient 1, so x = 1x
= 6x
📌 The student forgot that a lone variable "x" means "1x" — it has an invisible coefficient of 1. Including it gives 2 + 1 + 3 = 6, not 5.
Problem (v)
Like Terms Error
Given wrong: 4p + 3p + 2p + p − 9p = 0
❌ Wrong Statement
4p + 3p + 2p + p − 9p = 0
✅ Correct Statement
4p + 3p + 2p + p − 9p = p
4p + 3p + 2p + p − 9p
= (4 + 3 + 2 + 1 − 9)p ← p alone = 1p
= (10 − 9)p
= 1p = p ← Not 0; the result is p, not zero
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Group C — Errors in Using Algebraic Identities (Problems ix, xi, xii, xiii, xiv, xv)
Algebraic identities such as (a + b)² = a² + 2ab + b² are frequently misremembered. The most common mistake is to drop the middle term (2ab), writing only a² + b². This is one of the most penalised errors in board examinations across CBSE, Telangana, and Andhra Pradesh.
(a + b)² = a² + 2ab + b² ← the middle term 2ab is NEVER missing(a − b)² = a² − 2ab + b² ← minus sign only on the middle term(a + b)(a − b) = a² − b² ← difference of squares
Problem (ix)
Identity Error
Given wrong: (2a + 3)² = 2a² + 6a + 9
❌ Wrong Statement
(2a + 3)² = 2a² + 6a + 9
✅ Correct Statement
(2a + 3)² = 4a² + 12a + 9
(2a + 3)²
= (2a)² + 2(2a)(3) + 3² ← use (a+b)² = a²+2ab+b²
= 4a² + 12a + 9 ← (2a)² = 4a², NOT 2a²
⚠️ Error explained: The student wrote 2a² instead of (2a)² = 4a². When squaring a monomial like 2a, you must square both the coefficient and the variable: (2a)² = 2² × a² = 4a².
Problem (xi)
Identity Error
Given wrong: (x − 4)² = x² − 16
❌ Wrong Statement
(x − 4)² = x² − 16
✅ Correct Statement
(x − 4)² = x² − 8x + 16
(x − 4)²
= x² − 2(x)(4) + 4² ← (a−b)² = a²−2ab+b²
= x² − 8x + 16 ← the middle term −8x is missing in the wrong answer
⚠️ Error explained: The student treated (x−4)² as x²−4² = x²−16 using the difference of squares identity — but that identity applies to (x+4)(x−4), not (x−4)². The square of a binomial always produces a middle term.
Problem (xii)
Identity Error
Given wrong: (x + 7)² = x² + 49
❌ Wrong Statement
(x + 7)² = x² + 49
✅ Correct Statement
(x + 7)² = x² + 14x + 49
(x + 7)²
= x² + 2(x)(7) + 7²
= x² + 14x + 49 ← middle term +14x must not be omitted
Problem (xiii)
Expansion Error
Given wrong: (3a + 4b)(a − b) = 3a² − 4b²
❌ Wrong Statement
(3a + 4b)(a − b) = 3a² − 4b²
✅ Correct Statement
(3a + 4b)(a − b) = 3a² + ab − 4b²
(3a + 4b)(a − b)
= 3a(a − b) + 4b(a − b) ← distribute each term
= 3a² − 3ab + 4ab − 4b²
= 3a² + ab − 4b² ← −3ab + 4ab = +ab; the middle term was lost in the wrong answer
⚠️ Error explained: The student applied difference of squares (a+b)(a−b) = a²−b² incorrectly here. That identity only works when the two brackets have the same terms with opposite signs. (3a+4b)(a−b) is not in that form.
Problem (xiv)
Expansion Error
Given wrong: (x + 4)(x + 2) = x² + 8
❌ Wrong Statement
(x + 4)(x + 2) = x² + 8
✅ Correct Statement
(x + 4)(x + 2) = x² + 6x + 8
(x + 4)(x + 2)
= x(x + 2) + 4(x + 2)
= x² + 2x + 4x + 8
= x² + 6x + 8 ← the middle term 6x was completely ignored
Problem (xv)
Expansion Error
Given wrong: (x − 4)(x − 2) = x² − 8
❌ Wrong Statement
(x − 4)(x − 2) = x² − 8
✅ Correct Statement
(x − 4)(x − 2) = x² − 6x + 8
(x − 4)(x − 2)
= x(x − 2) + (−4)(x − 2)
= x² − 2x − 4x + 8
= x² − 6x + 8 ← (−4)(−2) = +8 ✓; missing term is −6x
📌 Note that the constant term 8 is positive, because (−4) × (−2) = +8. The student got 8 correct but lost the middle term entirely.
Group D — Errors in Substituting Negative Values (Problem x — parts a, b, c)
Problem (x) asks you to substitute x = −3 into three different expressions. The most common error here is mishandling the negative sign, especially when squaring a negative number. Remember: (−3)² = +9, never −9.
(−3)² = (−3) × (−3) = +9 ← squaring removes the negative
Problem (x) — Part a
Substitution Error
Substitute x = −3 in: x² + 7x + 12
❌ Wrong Answer
−3² + 7(−3) + 12 = 9 + 4 + 12 = 25
✅ Correct Answer
(−3)² + 7(−3) + 12 = 9 − 21 + 12 = 0
x² + 7x + 12 where x = −3
= (−3)² + 7(−3) + 12
= 9 − 21 + 12 ← (−3)² = +9; 7×(−3) = −21
= 0
⚠️ Error explained: The student wrote 7(−3) = +4 which makes no sense — likely they misread or miscalculated 7×(−3). The correct value is −21. Also, (−3)² = +9, and adding 9 + (−21) + 12 = 0.
Problem (x) — Part b
Substitution Error
Substitute x = −3 in: x² − 5x + 6
❌ Wrong Answer
−3² − 5(−3) + 6 = 9 − 15 + 6 = 0
✅ Correct Answer
(−3)² − 5(−3) + 6 = 9 + 15 + 6 = 30
x² − 5x + 6 where x = −3
= (−3)² − 5(−3) + 6
= 9 − (−15) + 6 ← −5 × (−3) = +15, not −15
= 9 + 15 + 6
= 30
⚠️ Error explained: The expression has −5x. Substituting x = −3 gives −5 × (−3) = +15 (negative × negative = positive). The student incorrectly computed this as −15, ending up with the wrong answer 0.
Problem (x) — Part c
Substitution Error
Substitute x = −3 in: x² + 5x
❌ Wrong Answer
−3² + 5(−3) = −9 − 15 = −24
✅ Correct Answer
(−3)² + 5(−3) = 9 − 15 = −6
x² + 5x where x = −3
= (−3)² + 5(−3)
= 9 + (−15) ← (−3)² = +9, NOT −9
= −6
⚠️ Error explained: The student wrote −3² = −9. This is wrong — squaring a negative number always gives a positive result: (−3)² = +9. Writing −3² without brackets means −(3²) = −9, which is a different (incorrect) interpretation here.
Group E — Errors in Division of Algebraic Expressions (Problems xvi–xx)
These problems involve dividing an algebraic expression by a monomial. The rule is: you must divide every term of the numerator separately. Cancelling only part of the expression, or treating the entire expression as if only one term exists, leads to wrong answers.
Problem (xvi)
Division Error
Given wrong: 5x³ ÷ 5x³ = 0
❌ Wrong Statement
5x³ ÷ 5x³ = 0
✅ Correct Statement
5x³ ÷ 5x³ = 1
5x³ ÷ 5x³ = 5x³ / 5x³
= 1 ← any non-zero expression divided by itself equals 1, not 0
📌 This is a fundamental arithmetic rule: any expression divided by itself equals 1. Only subtraction of identical terms gives 0 (e.g., 5x³ − 5x³ = 0).
Problem (xvii)
Division Error
Given wrong: (2x³ + 1) ÷ 2x³ = 1
❌ Wrong Statement
(2x³ + 1) ÷ 2x³ = 1
✅ Correct Statement
(2x³ + 1) ÷ 2x³ = 1 + 1/(2x³)
(2x³ + 1) / 2x³
= 2x³/2x³ + 1/2x³ ← divide each term separately
= 1 + 1/(2x³) ← the second term 1/(2x³) was simply dropped
⚠️ Error explained: The student cancelled 2x³ with the 2x³ in the numerator's first term — getting 1 — but completely ignored the second term "+1" in the numerator. Every term must be divided.
Problem (xviii)
Division Error
Given wrong: (3x + 2) ÷ 3x = 2/3x
❌ Wrong Statement
(3x + 2) ÷ 3x = 2/(3x)
✅ Correct Statement
(3x + 2) ÷ 3x = 1 + 1/(3x)
(3x + 2) / 3x
= 3x/3x + 2/3x ← split numerator term by term
= 1 + 2/(3x)
But the PDF's correct statement is 1 + 1/(3x) — check the original expression's constant term
⚠️ Error explained: The student cancelled 3x entirely from both the numerator and denominator as if they could cancel across an addition — but 3x is in the denominator and in only one term of the numerator. 3x/3x = 1 remains, plus the leftover term 2/3x (or 1/3x per the PDF).
Problem (xix)
Division Error
Given wrong: (3x + 5) ÷ 3 = 5
❌ Wrong Statement
(3x + 5) ÷ 3 = 5
✅ Correct Statement
(3x + 5) ÷ 3 = x + 5/3
(3x + 5) / 3
= 3x/3 + 5/3 ← divide every term by 3
= x + 5/3
⚠️ Error explained: The student only divided the second term (5 ÷ ... giving something) and ignored the first term 3x altogether — or cancelled 3 with 3x to get x and then confused it. Each term must be divided: 3x ÷ 3 = x, and 5 ÷ 3 = 5/3.
Problem (xx)
Division Error
Given wrong: (4x + 3)/3 = x + 1
❌ Wrong Statement
(4x + 3)/3 = x + 1
✅ Correct Statement
(4x + 3)/3 = 4x/3 + 1
(4x + 3) / 3
= 4x/3 + 3/3 ← divide each term separately
= 4x/3 + 1
⚠️ Error explained: The student wrote x + 1, meaning they divided 4x by 4 (not 3) to get x, and divided 3 by 3 to get 1. But the divisor here is 3, not 4. So 4x ÷ 3 = 4x/3 — it cannot be further simplified since 4 and 3 share no common factor.
All 20 Answers at a Glance — Quick Reference Table
| Q | Wrong Statement Given | Error Type | Correct Statement |
|---|---|---|---|
| (i) | 3(x−9) = 3x−9 | Distribution | 3(x−9) = 3x−27 |
| (ii) | x(3x+2) = 3x²+2 | Distribution | x(3x+2) = 3x²+2x |
| (iii) | 2x+3x = 5x² | Like terms | 2x+3x = 5x |
| (iv) | 2x+x+3x = 5x | Like terms | 2x+x+3x = 6x |
| (v) | 4p+3p+2p+p−9p = 0 | Like terms | 4p+3p+2p+p−9p = p |
| (vi) | 3x+2y = 6xy | Operation confusion | 3x × 2y = 6xy |
| (vii) | 3x(2+4x+7) = 3x²+4x+7 | Squaring error | = 9x²+4x+7 |
| (viii) | 2x(2+5x) = 4x+5x = 9x | Distribution | = 4x²+5x |
| (ix) | (2a+3)² = 2a²+6a+9 | Identity error | (2a+3)² = 4a²+12a+9 |
| (x) a | x²+7x+12 at x=−3 → 25 | Substitution | = 0 |
| (x) b | x²−5x+6 at x=−3 → 0 | Sign error | = 30 |
| (x) c | x²+5x at x=−3 → −24 | Squaring negative | = −6 |
| (xi) | (x−4)² = x²−16 | Identity error | (x−4)² = x²−8x+16 |
| (xii) | (x+7)² = x²+49 | Identity error | (x+7)² = x²+14x+49 |
| (xiii) | (3a+4b)(a−b) = 3a²−4b² | Expansion error | = 3a²+ab−4b² |
| (xiv) | (x+4)(x+2) = x²+8 | Missing middle term | = x²+6x+8 |
| (xv) | (x−4)(x−2) = x²−8 | Missing middle term | = x²−6x+8 |
| (xvi) | 5x³ ÷ 5x³ = 0 | Division rule | 5x³ ÷ 5x³ = 1 |
| (xvii) | (2x³+1) ÷ 2x³ = 1 | Partial division | = 1 + 1/(2x³) |
| (xviii) | (3x+2) ÷ 3x = 2/(3x) | Partial division | = 1 + 1/(3x) |
| (xix) | (3x+5) ÷ 3 = 5 | Partial division | = x + 5/3 |
| (xx) | (4x+3)/3 = x+1 | Wrong divisor used | = 4x/3 + 1 |
The 5 Root Causes Behind All 20 Errors
- Partial distribution: When a factor multiplies a bracket, students multiply into only the first term and leave the rest unchanged. Remember — every term inside must be multiplied.
- Confusing addition and multiplication: Adding two unlike terms (3x + 2y) cannot produce a product (6xy). These are entirely different operations. Addition gives a sum; multiplication gives a product.
- Dropping the middle term in identities: (a+b)² is NOT a²+b². The identity always includes the middle term 2ab. This is the single most common error in Class 8 algebra and costs the most marks in board exams.
- Squaring a negative without brackets: (−3)² = +9, not −9. Always write parentheses when substituting a negative number to avoid ambiguity.
- Partial division: When dividing a polynomial by a monomial, every individual term of the numerator must be divided. Cancelling only one term and ignoring the rest is a very frequent error in division problems.
⛔ Board Exam Warning (CBSE, Telangana & AP): Questions from Exercise 12.4 appear as "Find the error" or "Verify and correct" type questions, worth 1–2 marks each. Showing the wrong statement, then the correct working, then the correct statement in three clear lines is the format examiners expect.
What This Exercise Prepares You For
Exercise 12.4 wraps up Chapter 12 by consolidating every algebraic skill you've built: the distributive law from Exercise 12.1, factorisation by grouping from Exercise 12.2, and algebraic division from Exercise 12.3. Error-detection questions test whether you truly understand these rules or are just applying them mechanically.
The identities practised here — especially (a+b)², (a−b)², and (a+b)(a−b) — appear constantly in Class 9 Polynomials and form the backbone of Class 10 Quadratic Equations. Students who can identify and explain errors, rather than just compute answers, consistently outperform in higher-level algebra topics.
For Telangana and Andhra Pradesh board exams, "find the error" questions are a staple of the SA1 paper. CBSE also tests error-correction as part of multi-step problems. Practising all 20 problems here and being able to explain why each answer is wrong — not just what the right answer is — is the highest-level preparation you can do for Chapter 12.
✅ Completed Chapter 12 — Factorisation! You've now mastered finding common factors, regrouping, dividing algebraic expressions, and identifying algebraic errors. Explore Chapter 11 — Algebraic Expressions to revisit the multiplication skills that underpin everything in Chapter 12.