Krispy Kreme: Partial Products

How many doughnuts are in the box?

This Krispy Kreme three-act task above–from Graham Fletcher or YummyMath–cries out for partial products.

How would you partition the open array?

But more than once, the partial product strategies and models that I anticipated did not emerge. Not even close. 5 Practices-induced flop sweats. More on that in a future post. First, a progression of partial products across the grades, beginning with the basic multiplication facts:

How many do you see? How do you see them?

Some students will see four rows of seven doughnuts and know that 4 ⨉ 7 = 28. Great. For students who haven’t yet mastered the basic multiplication facts, partial products are helpful. Have students use what they know. For example, they might break apart seven as five and two and then find the sum of two familiar products: 4 ⨉ 7 = 4 ⨉ (5 + 2) = (4 ⨉ 5) + (4 ⨉ 2) = 20 + 8 = 28. Or, they might double a double: 4 ⨉ 7 = (2 ⨉ 2) ⨉ 7 = 2 ⨉ (2 ⨉ 7) = 2 ⨉ 14 = 28. They might do both. They might even break a factor into more than two addends: 4 ⨉ 7 = 4 ⨉ (3 + 3 + 1) = (4 ⨉ 3) + (4 ⨉ 3) + (4 ⨉ 1) = 12 + 12 + 4 = 28. (Admittedly not the most useful relationship to help students derive this fact.) Mastery of the basic multiplication facts aside, playing with partial products–and open arrays–reinforces the big idea that numbers can be broken apart–or decomposed–in flexible ways to make calculations easier.

This idea extends to multiplying two-digit numbers by one-digit numbers:

How many do you see? How do you see them?

Some students will understand that breaking apart by place value makes calculations easier: 5 ⨉ 12 = 5 ⨉ (10 + 2) = (5 ⨉ 10) + (5 ⨉ 2) = 50 + 10 = 60. Others might use doubles and double-doubles. Note that a factor can be broken into addends or smaller factors: 5 ⨉ 12 = 5(3 + 3 + 3 + 3) or 5 ⨉ 12 = 5(3 ⨉ 4). How students choose to express this will provide insight into their thinking.

Again, decomposing numbers in flexible ways extends to larger numbers:

How many do you see? How do you see them?

Breaking apart both factors by place value is a common approach: 25 ⨉ 32 = (20 + 5) ⨉ (30 + 2) = (20 ⨉ 30) + (20 ⨉ 2) + (5 ⨉ 30) + (5 ⨉ 2) = 600 + 40 + 150 + 10 = 800. This approach might be too common if reduced to a procedure (i.e., the box method or FOIL). Again, it’s about flexible ways. Breaking apart just one factor by place value is an efficient mental math strategy: 25 ⨉ 32 = 25 ⨉ (30 + 2) = (25 ⨉ 30) + (25 ⨉ 2) = 750 + 50 = 800. A student who inefficiently decomposes 32 as 10 + 10 + 10 + 2 could be nudged towards 32 as 30 + 2. Or, a factor of 25 might spark thinking about 25 ⨉ 4 = 100, a familiar product.

The different varieties of doughnuts illustrate some helpful ways of partitioning the arrays. But each of these slides draws attention to a specific way of seeing the array. My preference would be to show the slides where all the doughnuts are the same. (Same goes for visual patterns.) Ask students how they see them. If students do not see a helpful way of partitioning the arrays, then corresponding slides with different varieties of doughnuts could be displayed. In a number string, 52 – 40 leads students to think about adjusting 39 in 52 – 39 to make the calculation easier. Similarly, a purposely crafted string of images could lead students to see fives, doubles, or place value–all useful relationships–in an original (glazed) array.

Related: The Math Learning Center’s Partial Product Finder

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