Chapter 12.5 — Exercise 12.4 — Finding Errors
Finding and correcting errors in given equations. This is Lesson 5 of 5 in Chapter 12: Factorisation.
A Chapter's Worth of Mistakes, Collected on Purpose
This exercise flips the usual format around entirely: instead of solving a problem from scratch, the task is to find what's already wrong with someone else's working, and correct it. Every error collected here is a realistic one — the kind that comes from skipping a step or half-remembering a rule, not a random typo — which makes spotting them a genuinely different skill from solving problems cleanly the first time. Twenty-one such errors appear across this exercise, falling into five recognisable categories: incomplete distribution, miscombined like terms, dropped middle terms in a squared or multiplied bracket, mishandled negative signs during substitution, and incomplete division.
One Habit That Catches Almost Every Error Below
Before working through the specific categories of mistake in this exercise, it's worth naming the single check that would have caught nearly every one of them: pick a small number, substitute it into both the claimed answer and a fresh, independent recomputation, and see whether they agree. Take (2a+3)² wrongly claimed to equal 2a²+6a+9. At a=1, the wrong answer gives 2+6+9=17, but (2×1+3)² = 5² = 25 — a clear mismatch, catching the error immediately without needing to spot the specific algebraic mistake first. This substitution check doesn't explain why an answer is wrong, which is why the worked corrections below matter too, but it's the fastest possible way to notice that something is wrong before spending time trying to use a flawed result.
Distributing Only Part of a Multiplication
The most common error in this exercise is multiplying into only one term of a bracket instead of every term:
| Written (wrong) | What went wrong | Correct |
|---|---|---|
| 3(x−9) = 3x−9 | The 3 multiplied x but never reached the −9 | 3(x−9) = 3x−27 |
| x(3x+2) = 3x²+2 | The x multiplied 3x but never reached the +2 | x(3x+2) = 3x²+2x |
In both cases, the fix is the same: every single term inside the bracket needs its own multiplication, not just the first one encountered.
Combining Terms That Aren't Actually Like Terms
A second family of errors treats addition as if it were multiplication, or miscounts a simple running total:
| Written (wrong) | What went wrong | Correct |
|---|---|---|
| 2x+3x = 5x² | Added the exponents as if the terms were being multiplied | 2x+3x = 5x |
| 2x+x+3x = 5x | Added the coefficients wrong (2+1+3=6, not 5) | 2x+x+3x = 6x |
| 4p+3p+2p+p−9p = 0 | Added the coefficients wrong (4+3+2+1−9=1, not 0) | 4p+3p+2p+p−9p = p |
| 3x+2y = 6xy | Treated a sum as if it were a product | 3x×2y = 6xy (the original was never an addition at all) |
The x²+3x confusion in the first row is worth dwelling on, since it's easy to see why it happens: 2x×3x really would give 6x², so it's tempting to assume any combination of 2x and 3x should produce a squared term. But 2x+3x is addition, not multiplication — like terms being added just combine their coefficients (2+3=5), with the variable part, x, staying exactly as it was.
Squaring a Sum or Product Without the Middle Term
The most frequent error in this exercise involves the identities from the previous chapter — specifically, forgetting that (a+b)² needs a middle term at all:
| Written (wrong) | What went wrong | Correct |
|---|---|---|
| (3x)²+4x+7 = 3x²+4x+7 | 3² was never applied — (3x)² needs squaring both the 3 and the x | 9x²+4x+7 |
| (2x)²+5x = 4x+5x = 9x | 4x and 5x were wrongly treated as like terms and combined | 4x²+5x (unlike terms — cannot combine further) |
| (2a+3)² = 2a²+6a+9 | Only 2a was squared, not the whole bracket contents properly | 4a²+12a+9 |
| (x−4)² = x²−16 | The −2ab middle term was dropped entirely | x²−8x+16 |
| (x+7)² = x²+49 | The 2ab middle term was dropped entirely | x²+14x+49 |
| (x+4)(x+2) = x²+8 | Both middle terms (2x and 4x) were dropped | x²+6x+8 |
| (x−4)(x−2) = x²−8 | Both middle terms (−2x and −4x) were dropped | x²−6x+8 |
| (3a+4b)(a−b) = 3a²−4b² | Treated as a difference of squares when it isn't one | 3a²+ab−4b² |
Every row here shares the same root cause: jumping straight from "squared first term" and "squared last term" to a two-term answer, skipping the cross-multiplication that produces the middle term (or terms) entirely. (3a+4b)(a−b) is a useful reminder that this shortcut only ever applies to a genuine (a+b)(a−b) pattern — here the two brackets don't share the same two terms with opposite signs, so there's no shortcut available at all, and every term has to be multiplied out in full.
Substituting a Negative Number and Losing Track of a Sign
Three errors come from substituting x = −3 into an expression and mishandling the negative sign partway through:
| Expression | Wrong answer | Correct working |
|---|---|---|
| x²+7x+12 | 25 | (−3)²+7(−3)+12 = 9−21+12 = 0 |
| x²−5x+6 | 0 | (−3)²−5(−3)+6 = 9+15+6 = 30 |
| x²+5x | −24 | (−3)²+5(−3) = 9−15 = −6 |
The second row shows exactly how this kind of error creeps in: −5×(−3) is +15, not −15 — a double-negative that's easy to mis-simplify back into a single negative if the substitution isn't written out one sign at a time. The safest habit is substituting the value in brackets first, (−3), and resolving every sign explicitly before doing any arithmetic at all, rather than trying to track signs mentally — writing (−3)² as (−3)×(−3) in full, rather than jumping straight to a remembered result, removes almost all of the risk.
Dividing and Forgetting What's Left Over
The final group of errors comes from division, specifically from forgetting that dividing a sum means dividing every term in it separately:
| Written (wrong) | What went wrong | Correct |
|---|---|---|
| 5x³ ÷ 5x³ = 0 | Confused "divides to 1" with "subtracts to 0" | 5x³ ÷ 5x³ = 1 |
| (2x³+1) ÷ 2x³ = 1 | The "+1" term was dropped instead of divided separately | 1 + 1/(2x³) |
| (3x+5) ÷ 3 = 5 | The 3x term vanished instead of dividing to x | x + 5/3 |
| (4x+3)/3 = x+1 | 4x/3 was wrongly reduced all the way down to x | (4x/3) + 1 |
The first row is a genuinely different kind of mistake from the rest of this section — mixing up two entirely separate facts (anything minus itself is 0; anything divided by itself, other than 0, is 1) rather than a dropped term. The other three all share the same fix: (a+b) ÷ c means a/c + b/c, and every one of these wrong answers comes from applying that split to only one of the two terms and letting the other one disappear.
What This Exercise Was Really Testing
None of the twenty-one corrections above needed a new rule that hasn't already appeared somewhere earlier in this chapter or the one before it — combining like terms, distributing a multiplication, applying an identity, substituting carefully, dividing term by term. What changed here was the direction of the work: instead of applying a rule forward to reach an answer, each problem asked whether a rule had already been applied correctly, which is a good final check on whether these methods are genuinely understood or just being followed by rote. The substitution check from the very start of this lesson is worth carrying forward well beyond this one exercise — it costs a single small calculation, and it catches exactly the kind of dropped term or mishandled sign that every one of the twenty-one errors above turned out to involve.