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Lesson 3 of 5

We Distribute, Yet Things Multiply · Lesson 3 of 5

Mind the Mistake, Mend the Mistake

“Identify, explain and correct algebraic errors through twelve checked simplifications.”

Learning Objectives

• Check an algebraic simplification one equality at a time. • Identify and explain errors in distribution, signs, powers and like terms. • Correct an expression while preserving its value. • Recognise valid working instead of assuming every example is wrong. • Use substitution to detect errors and algebra to justify general results.

Two students can start with the same expression and finish with different answers. To decide whose work is correct, it helps to inspect the steps rather than to trust the shorter-looking answer. An expression may become tidier while changing its meaning. For example, 5w² + 6w looks longer than 11w², but the two expressions usually give different values. Your task is to explain exactly what changed and whether that change was allowed.

In this lesson, you will examine twelve pieces of working. Some contain mistakes and some are correct. Read each original expression, try simplifying it yourself, and then compare your route with the proposed steps. Keeping correct work is just as important as repairing incorrect work.

An equals sign makes a claim: the expressions on both sides have the same value. Find the first line where that claim fails. Correcting only the final answer can hide the misunderstanding that caused the error, so explain the relevant rule in words as well as symbols.

How to inspect a solution

Read the operations and brackets first. Mark the signed terms. Check that every required product appears. Combine only matching variable parts. Finally, substitute a convenient value to look for a mismatch. A mismatch disproves an identity; a matching numerical check supports your arithmetic but does not prove the identity for all values.

A product mistaken for a sum

Problem
Proposed working: −3p(−5p + 2q) = −3p + 5p − 2q = p − 2q. Is it valid?

  1. 1.The first equality is already wrong. The expression means multiply −3p by the entire bracket. It does not mean add or subtract the written pieces.
  2. 2.Distribute: (−3p)(−5p) + (−3p)(2q). Two negative factors give +15p²; a negative and a positive factor give −6pq.
  3. 3.The correct expansion is 15p² − 6pq. The terms have different variable parts, so they cannot be combined.
  4. 4.There is also an arithmetic error in the proposed middle expression: −3p + 5p would be 2p, not p. Repairing that later error would still leave the original distribution error.
  5. 5.Check p = 1, q = 1. The original gives −3(−5 + 2) = 9. The correct expansion gives 15 − 6 = 9, while p − 2q gives −1.
Constants missed during distribution

Problem
Proposed working: 2(x − 1) + 3(x + 4) = 2x − 1 + 3x + 4 = 5x + 3. Find the error.

  1. 1.In the first expansion, 2 has multiplied x but not −1, and 3 has multiplied x but not 4. An outside factor multiplies every term.
  2. 2.The correct products are 2x − 2 + 3x + 12.
  3. 3.Combine the x terms and the constants separately: 5x + 10.
  4. 4.At x = 0, the original is 2(−1) + 3(4) = 10. The proposed 5x + 3 gives 3, so it cannot be an identity.
Addition confused with a repeated factor

Problem
Proposed working: y + 2(y + 2) = (y + 2)² = y² + 4y + 4. What went wrong?

  1. 1.The first expression is a sum: y plus twice the bracket y + 2. A square would require the product (y + 2)(y + 2).
  2. 2.Distribute only the 2: y + 2y + 4.
  3. 3.Combine the like terms to obtain 3y + 4. The expansion of the squared bracket is correct in itself, but that bracket was not the original expression.
  4. 4.At y = 1, the original gives 1 + 2(3) = 7, whereas the proposed square is 9. A correct rule used on a different expression does not repair the first step.
Read the operation before expandingy + 2(y + 2)y plus twice a bracket → 3y + 4(y + 2)(y + 2)a bracket times itself → y² + 4y + 4
Addition outside a bracket compared with multiplying the bracket by itself
A missing middle term

Problem
Proposed working: (5m + 6n)² = 25m² + 36n². Is anything missing?

  1. 1.Write the square as (5m + 6n)(5m + 6n). There are four products, not two.
  2. 2.They are 25m², 30mn, 30mn and 36n².
  3. 3.Combine the two mixed products: (5m + 6n)² = 25m² + 60mn + 36n².
  4. 4.For m = n = 1, the original is 11² = 121. The proposed answer gives 25 + 36 = 61, missing exactly 60.
A negative sign handled correctly

Problem
Proposed working: (−q + 2)² = q² − 4q + 4. Is this correct?

  1. 1.Yes. Multiply (−q + 2)(−q + 2). The first product is (−q)(−q) = q².
  2. 2.The mixed products are −2q and −2q, and the final product is 4.
  3. 3.Combining gives q² − 4q + 4. Alternatively, write the bracket as 2 − q and apply the square-of-a-difference identity.
  4. 4.At q = 3, both sides give 1. This confirms the arithmetic at that value; the valid expansion supplies the general reason.
Distribution applied to a product

Problem
Proposed working: 3a(2b × 3c) = 6ab × 9ac = 54a²bc. Is distribution being used correctly?

  1. 1.No. Inside the bracket, 2b and 3c are multiplied. Distribution splits an addition or subtraction, not the factors of a product.
  2. 2.First compute 2b × 3c = 6bc. Then multiply by 3a to obtain 18abc.
  3. 3.You can also group the factors as 3a × 2b × 3c = (3 × 2 × 3)abc = 18abc. The factor 3a appears once.
  4. 4.The proposed method includes 3a twice, creating an extra factor 3a. At a = b = c = 1, the original is 18 and the proposed answer is 54.
Two different structures

For A(B + C), make two products and add them: AB + AC. For A(B × C), multiply the three factors once: ABC. The operation inside the brackets decides the method.

A fraction distributed correctly

Problem
Proposed working: (1/2)(10s − 6) + 3 = 5s − 3 + 3 = 5s. Is this valid?

  1. 1.Yes. Half of 10s is 5s and half of −6 is −3.
  2. 2.The +3 outside the bracket is then added: 5s − 3 + 3.
  3. 3.The constant terms cancel, giving 5s. Nothing has been omitted from the multiplication.
  4. 4.At s = 2, the original gives half of 14 plus 3, which is 10. The final expression also gives 10.
Unlike powers combined

Problem
Proposed working: 5w² + 6w = 11w². Can these terms be combined?

  1. 1.No. The variable part w² contains two factors of w; the variable part w contains one. They are unlike terms.
  2. 2.The expression is already a sum of unlike terms: 5w² + 6w. You may factor it as w(5w + 6), but that is a different form, not 11w².
  3. 3.At w = 2, the original is 5 × 4 + 6 × 2 = 32. The proposed expression gives 44.
  4. 4.At w = 1 both expressions give 11. This shows why choosing only one convenient check can miss a mistake.
Variable powers changed during addition

Problem
Proposed working: 2a³ + 3a³ + 6a²b + 6ab² = 5a³ + 12a²b². Which part is correct?

  1. 1.Combining 2a³ and 3a³ as 5a³ is correct: their variable parts match.
  2. 2.The terms 6a²b and 6ab² are not like terms. Adding them cannot turn both variables into squares.
  3. 3.The correctly collected expression is 5a³ + 6a²b + 6ab². A factored form for the last two terms is 6ab(a + b).
  4. 4.At a = 1, b = 2, the original equals 5 + 12 + 24 = 41. The proposed final expression equals 5 + 48 = 53.
  5. 5.Adding powers in a product with the same base is a multiplication rule. It is not a rule for adding unlike terms.
All four products included

Problem
Proposed working: (x + 2)(x + 5) = (x + 2)x + (x + 2)5 = x² + 2x + 5x + 10 = x² + 7x + 10. Is this correct?

  1. 1.Yes. The first step distributes the second bracket while keeping x + 2 together.
  2. 2.The next step distributes each remaining product, producing all four required terms.
  3. 3.The like terms 2x and 5x combine to 7x, so the final result is correct.
  4. 4.At x = −2, the original is zero, and the final expression gives 4 − 14 + 10 = 0. Negative substitutions can be useful checks as well.
Only the first and last products kept

Problem
Proposed working: (a + 2)(b + 4) = ab + 8. Identify the missing pieces.

  1. 1.The product a × b is present, and the product 2 × 4 is present. But a must also multiply 4, and 2 must multiply b.
  2. 2.Writing all products gives ab + 4a + 2b + 8.
  3. 3.No two of these terms have the same variable part, so the expression cannot be collected further.
  4. 4.At a = 1, b = 1, the original is 3 × 5 = 15. The proposed answer is 9, missing 4a + 2b = 6.
Factoring as distribution in reverse

Problem
Proposed working: ab² + a²b + a²b² = ab(a + b + ab). Is the changed order a problem?

  1. 1.No. Expand the right side: ab × a + ab × b + ab × ab.
  2. 2.The products are a²b + ab² + a²b², which are exactly the original terms in a different order.
  3. 3.Addition permits reordering, so the factorisation is correct. An equally direct form is ab(b + a + ab).
  4. 4.The verification by expansion works even when a or b is zero; no division by a possibly zero quantity is needed.
Sort by the reason, not by appearance

Group the incorrect examples into distribution errors, confusion between operations, and errors involving powers or like terms. Can one example fit more than one group?

Missing distribution appears in 2(x − 1) + 3(x + 4) and (a + 2)(b + 4). Confused operations appear in −3p(−5p + 2q), y + 2(y + 2), and 3a(2b × 3c). Powers and like terms cause errors in the squared sum, 5w² + 6w, and the expression containing a²b and ab². Categories can overlap: omitting the middle terms of a squared sum is also incomplete distribution. Explaining the first broken rule matters more than choosing one category name.

Quiz

Quick check

Which step first goes wrong in 4(x − 2) = 4x − 2 = 2x?

Quick check

Which expression equals 2a(3b × 4c)?

Quick check

Which simplification is correct for every x?

Quick check

Two expressions agree at x = 1. What can you conclude?

Quick check

Which pair can be combined by adding coefficients?

Answer: The first equality: −2 was not multiplied by 4.. The first expansion should be 4x − 8. The later combination is also invalid, but the expression has already changed at the first equality.

Practice Problems

Practice Problems
  1. Correct −2r(3r − 4s) = −6r + 8s and explain the lost factors.
  2. Check 3(x − 2) + 2(x + 5) = 5x + 4. Explain every step.
  3. A student writes (2m − 3n)² = 4m² − 9n². Give the correct expansion and a numerical counterexample.
  4. Explain why 7a²b + 2ab² cannot be written as 9a²b². Give a correct factored form and verify it.
  5. Check this chain: (x − 2)(x + 3) = x² + 3x − 2x − 6 = x² + x − 6. Then explain why substituting x = 0 alone would fail to detect the wrong answer x² − 6.

Multiply −2r by 3r: the coefficient is −6 and r × r is r², giving −6r². Multiply −2r by −4s: the coefficient is +8 and the variable part is rs, giving +8rs. The correct result is −6r² + 8rs. The original working omitted the outside r from both products.

Key Takeaways

Key Takeaways

• Inspect each equals sign and identify the first invalid step. • Multiply every signed term when distributing over addition or subtraction. • Do not distribute over a product by repeating the outside factor. • A squared bracket contains mixed products as well as squared terms. • Combine coefficients only when the complete variable parts match. • A counterexample disproves a general claim; one successful check does not prove an identity. • Some proposed simplifications are correct, including forms with reordered terms or common factors.

Previous · Lesson 2

Special Cases of the Distributive Property

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This Way or That Way, All Ways Lead to the Bay