Splitting a Factor but Adding the Part
Ms. Ortega's third graders turned in exit tickets for 7 × 8. Devon wrote 7 × 5 = 35, then 35 + 3 = 38, explaining, 'I split the 8 into 5 and 3 and add.' His earlier ticket shows 6 × 9 = 6 × 5 + 4 = 34. Priya wrote 7 × 4 = 28 and doubled it to 56, but when asked why, she said, 'Because doubling is what you do.' Malik wrote 8 × 5 = 40 and 8 × 2 = 16, then 40 + 16 = 56, and said he 'just knows' 8 × 5. The unit plan names the distributive property as the bridge to Grade 4 multi-digit multiplication. Ms. Ortega has 20 minutes before specials and wants one discussion that targets the most consequential misconception without restarting multiplication instruction from the beginning. Arrays and grid paper are available.
1. Diagnose Devon's error precisely. Why does his method produce an answer that is too low, and what should replace the added 3?
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Devon correctly decomposes 8 into 5 + 3 but then adds the 3 as if it meant 3 ones instead of 7 groups of 3. Multiplication distributes over addition: 7 × 8 = 7 × (5 + 3) = (7 × 5) + (7 × 3) = 35 + 21 = 56. His 35 + 3 = 38 accounts for only one of the three extra groups of 7, so it is 18 short (3 groups of 7 is 21, not 3). The same slip appears in 6 × 9 = 6 × (5 + 4): he needs 30 + (6 × 4) = 54, not 30 + 4. The repair is to keep both factors attached to every part of the split. An array cut into a 7-by-5 rectangle and a 7-by-3 rectangle makes the missing 21 visible, because each region's area is a product, not a bare part.
2. Priya's and Malik's work are both correct. Which one should anchor the 20-minute discussion, and how would you connect it to Devon's paper?
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Anchor on Malik's work, because he has already split one factor into parts and multiplied each part (8 × 5 and 8 × 2), which is exactly the structure Devon needs; the only difference is that Malik multiplied the second part instead of adding it. Sequence: (a) Have Malik build an 8-by-7 array and cut it into 8×5 and 8×2, recording 40 + 16 = 56. (b) Ask Devon to place his 5 + 3 split on the same array and write the product for each piece; he should see 7 × 3 = 21, not 3. (c) Let Priya explain doubling as 7 × 4 plus another 7 × 4 = 28 + 28, then connect it to the same array by cutting at 4. Twenty minutes is enough for these three moves; re-teaching every fact is not needed because the other two students' strategies are already efficient and worth naming.
3. Write one follow-up problem you would give Devon the next day and state the evidence you would accept as showing the misconception is repaired.
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Give 9 × 6 and require two ways: 9 × (5 + 1) = 45 + 9 = 54 and 9 × (3 + 3) = 27 + 27 = 54, each with an array sketch. Evidence of repair: he writes the product of each part (not the part itself) in both decompositions, his two results agree, and he can explain that the second addend is 9 × 1 = 9, not 1. A second item, 8 × 7 with the split 4 + 3, checks that he does not depend on a 5-based split: 8 × 4 = 32 and 8 × 3 = 24, so 32 + 24 = 56. If he still adds the bare part, return to the array and label each region with its product before returning to symbols.