Grade 4 · 4.NBT.B.5 · 3.MD.C.7

Partial products

Six rows of 34 seats is not a fact you know by heart. How do you break a big multiplication into small ones you do know?

You need

The area modelGrade 3 Expanded formGrade 2

Unlocks

The standard multiplication algorithmGrade 5

Where you are 3 in Multi-digit multiplication

Number9

Arithmetic5

Multiplication and division16

Multi-digit multiplication3

Multiplying by tens and hundreds

Partial products

The standard multiplication algorithm

When you reach for it

The front section of a school auditorium has 6 rows, with 34 seats in each row. How many seats is that? The answer is 6 \times 34. That is not a fact you know by heart, since the facts you know stop at 10 \times 10.

So break the big multiplication into small ones you do know. Split a factor by place value, the way Expanded form writes it: 34 = 30 + 4. Multiply each part. Then add the answers. Each small answer is called a partial product, because it is one part of the whole product.

The words: in 6 \times 34 = 204, the numbers being multiplied, 6 and 34, are the factors, and the answer, 204, is the product.

Two ideas you have met before make this work.

The area model: a rectangle whose sides are the two factors has the product as its area. Cutting the rectangle into pieces does not change its total area.
The The distributive property: multiplying a sum is the same as multiplying each part and adding, a \times (b + c) = a \times b + a \times c.

The area model is the picture. The distributive property is the rule it pictures. Every time this lesson splits a factor, it is using the distributive property, and it will say so.

Reach for partial products whenever a factor has more than one digit.

The procedure

This lesson takes four days, one part a day.

Part 1: two digits times one digit, with an area model

Draw the auditorium section as a rectangle, 6 tall and 34 wide. Now cut it with one line, where 34 splits into 30 and 4.

Figure 16 \times 34 as a rectangle, cut where 34 splits into 30 + 4. Not drawn to scale.

The left piece is 6 by 30. Its area is 6 \times 30. Six groups of 3 tens are 18 tens, which is 180 (see Multiplying by tens and hundreds). The right piece is 6 by 4, and 6 \times 4 = 24. The whole rectangle is the two pieces together:

\begin{aligned} 6 \times 34 &= 6 \times (30 + 4) \\ &= 6 \times 30 + 6 \times 4 \\ &= 180 + 24 \\ &= 204 \end{aligned}

The second line is the distributive property. It says the two small rectangles hold exactly as many seats as the big one. So the section has 204 seats.

Why split by place value, and not some other way? You could split 34 as 20 + 14, and the distributive property would still give the right answer: 120 + 84 = 204. But 6 \times 14 is not a fact you know by heart. Splitting by place value makes every piece a basic fact, perhaps shifted over by a place. That is the whole point.

Order does not matter either. The commutative property says 34 \times 6 = 6 \times 34. In the picture, that is the same rectangle turned on its side.

A quick look at the size helps too. The piece 6 \times 30 = 180 is most of the rectangle, so the product should be a bit more than 180. And 204 is.

Part 2: three and four digits times one digit

Gautam’s orchard packs 3{,}214 apples each day. How many apples does it pack in 6 days?

The idea does not change when the number gets longer. Write 3{,}214 in expanded form: 3{,}000 + 200 + 10 + 4. The distributive property works with any number of parts, so you multiply each part by 6 and add:

\begin{aligned} 6 \times 3{,}214 &= 6 \times 3{,}000 + 6 \times 200 \\ &\quad + 6 \times 10 + 6 \times 4 \\ &= 18{,}000 + 1{,}200 + 60 + 24 \\ &= 19{,}284 \end{aligned}

Each piece is a fact and a shift. For example, 6 \times 200 is 6 groups of 2 hundreds, which is 12 hundreds, or 1{,}200.

An area model would have four pieces side by side. Drawing four pieces gets slow, so most people record the partial products in a column instead. Write the factors stacked by place value. Under a line, write each partial product, lined up by place value too. Then add.

Figure 2The four partial products of 3{,}214 \times 6, lined up by place value and added.

Starting from the ones is a habit, not a rule: you may add the pieces in any order, because addition is commutative and associative. Lining up by place value is the rule, because only digits of the same place can be added. Here the ones column holds 4. The tens column holds 2 + 6 = 8. The hundreds column holds 2. The thousands column holds 1 + 8 = 9. The ten-thousands column holds 1. The orchard packs 19{,}284 apples.

Check the size. The biggest piece is 6 \times 3{,}000 = 18{,}000, and the other pieces are much smaller. So the product is a bit more than 18{,}000, and 19{,}284 is.

A factor may have a zero digit, as in 7 \times 3{,}059. Then the hundreds piece is 7 \times 0 = 0. You may write it or skip it. Either way, the other pieces still land in their right columns (Example 4).

Part 3: two digits times two digits

The main floor of the auditorium has 36 rows with 47 seats in each row. That is 47 \times 36 seats, and now both factors have two digits.

Split both factors: 47 = 40 + 7 and 36 = 30 + 6. Use the distributive property twice. First split 36:

47 \times 36 = 47 \times 30 + 47 \times 6

Then split 47 inside each part. The distributive property again:

\begin{aligned} 47 \times 30 &= 40 \times 30 + 7 \times 30 \\ 47 \times 6 &= 40 \times 6 + 7 \times 6 \end{aligned}

That is four partial products. The area model shows them as four pieces of one rectangle. Each part of one factor meets each part of the other, so 2 parts times 2 parts gives 2 \times 2 = 4 pieces.

Figure 3The four-part area model of 47 \times 36. Every part of one factor meets every part of the other. Not drawn to scale.

Find each piece:

40 \times 30 = 1{,}200, since 4 \times 3 = 12 and tens times tens gives hundreds: 12 hundreds.
7 \times 30 = 210, which is 21 tens.
40 \times 6 = 240, which is 24 tens.
7 \times 6 = 42.

Add the four pieces:

1{,}200 + 210 + 240 + 42 = 1{,}692

The main floor has 1{,}692 seats. The same pieces fit in a column, one line each:

Figure 4The same four pieces of 47 \times 36 recorded in a column.

Look at the right edge of the area model. The top row, for 30, adds to 1{,}200 + 210 = 1{,}410, which is 47 \times 30. The bottom row, for 6, adds to 240 + 42 = 282, which is 47 \times 6. Those two row totals are what the standard algorithm writes later, one line for each digit of the bottom factor.

Part 4: lattice multiplication, and checking with an estimate

Lattice multiplication is another way to record the same partial products. It looks like a trick at first. It is not. It is the area model with the zeros left off and the place values kept by diagonal lines.

To multiply 47 \times 36, draw a grid with one column for each digit of 47 and one row for each digit of 36. Write 4 and 7 above the columns and 3 and 6 beside the rows. Cut every cell with a diagonal from its top right corner to its bottom left corner.

In each cell, write the product of its column digit and its row digit, as two digits: the tens digit above the diagonal and the ones digit below it. A product less than 10 gets a 0 above the diagonal, so 2 \times 3 is written 06.

Figure 547 \times 36 in a lattice. Each diagonal band is one place value, so adding along a band adds digits of the same place.

Now add along each diagonal band, starting at the bottom right:

The bottom right band holds only 2. That is the ones digit.
The next band holds 4 + 4 + 1 = 9. That is the tens digit.
The next band holds 2 + 2 + 2 = 6. That is the hundreds digit.
The top left band holds 1. That is the thousands digit.

Read the digits from the top left, down the left side and along the bottom: 1{,}692. That is the same answer as Part 3.

Why do the bands match place values? Look at the cell for 4 and 3. It holds 12, but it really stands for 40 \times 30 = 1{,}200. Every step to the left in the grid moves to a digit worth ten times as much, and so does every step up (see Multiplying and dividing by ten, one hundred and one thousand). The diagonals are drawn so that one step left or one step up moves a digit into the next band. So the 1 of that 12 lands in the thousands band and the 2 in the hundreds band, which is 1{,}200. In the same way, the other cells stand for 210, 240 and 42. The lattice adds the four partial products of Part 3, place by place.

If a band adds to 10 or more, write its ones digit and add the 1 (or more) to the next band up. That is regrouping, ten of one place for one of the next (Example 7).

Check with an estimate. Whichever record you use, a quick estimate catches a slip of a whole place. Round each factor (see Estimating by rounding): 47 is close to 50 and 36 is close to 40, and 50 \times 40 = 2{,}000. Both factors were rounded up, so the real product should be a little less than 2{,}000. The answer 1{,}692 fits. An answer like 16{,}920 or 169 would not, and the estimate would send you back to look.

Worked examples

Example 1

Rows of apple trees

Rawiri plants 8 rows of apple trees with 28 trees in each row. How many trees does he plant?

Solution

The trees form a rectangle 8 rows by 28 trees, so the total is 8 \times 28. Split 28 by place value as 20 + 8. The distributive property lets you multiply each part by 8 and add:

8 \times 28 = 8 \times 20 + 8 \times 8 = 160 + 64 = 224

Here 8 \times 20 is 8 groups of 2 tens, which is 16 tens, or 160. Rawiri plants 224 trees.

Check: 28 is close to 30, and 8 \times 30 = 240. Since 28 is a little less than 30, the product should be a little less than 240. It is.

Example 2

When the pieces cross into hundreds

In Yupanqui’s orchard, each row has 86 pear trees, and there are 7 rows. How many pear trees are there?

Solution

The total is 7 \times 86. Split 86 as 80 + 6 and use the distributive property:

7 \times 86 = 7 \times 80 + 7 \times 6 = 560 + 42

Adding 560 + 42 takes care. The tens give 6 + 4 = 10 tens, and ten tens make one hundred. So the sum is 5 hundreds plus 1 more hundred, 0 tens and 2 ones:

560 + 42 = 602

The orchard has 602 pear trees. The zero in the tens place is real: all the tens were traded for a hundred.

Example 3

Three digits by one

The auditorium has 527 seats. The school puts on 4 shows, and every seat is filled at each show. How many tickets are used?

Solution

The total is 4 \times 527. Expanded form splits 527 as 500 + 20 + 7. By the distributive property, multiply each part by 4:

4 \times 500 = 2{,}000, since 4 groups of 5 hundreds are 20 hundreds.
4 \times 20 = 80.
4 \times 7 = 28.

In a column, lined up by place value:

Partial product	From
28	4 \times 7
80	4 \times 20
2{,}000	4 \times 500
2{,}108	sum

The ones column holds 8. The tens column holds 2 + 8 = 10 tens, which is one hundred and no tens. The hundreds column holds only that 1 traded hundred. The thousands column holds 2. So 4 \times 527 = 2{,}108, and 2{,}108 tickets are used.

Check: 4 \times 500 = 2{,}000 is the biggest piece, and the product is a little more. It is.

Example 4

A zero inside a factor

Rawiri’s orchard has 7 fields with 3{,}059 trees in each field. How many trees are there in all?

Solution

The total is 7 \times 3{,}059. In expanded form, 3{,}059 = 3{,}000 + 0 + 50 + 9, with no hundreds. The distributive property gives four pieces:

7 \times 3{,}000 = 21{,}000, since 7 groups of 3 thousands are 21 thousands.
7 \times 0 = 0, for the hundreds. This piece adds nothing, so you may skip it.
7 \times 50 = 350, which is 35 tens.
7 \times 9 = 63.

Add the pieces that are not zero:

21{,}000 + 350 + 63 = 21{,}413

The orchard has 21{,}413 trees. Skipping the zero piece is safe because each other piece was written in full, so its digits still sit in the right places. The digit 0 in 3{,}059 was what kept the 3 in the thousands place to begin with.

Example 5

Two digits by two

The auditorium balcony has 14 rows with 23 seats in each row. How many seats are in the balcony?

Solution

The total is 23 \times 14. Split both factors: 23 = 20 + 3 and 14 = 10 + 4. The distributive property, used once for each factor, gives one piece for each pair of parts. Draw the area model with 20 and 3 across the top and 10 and 4 down the side:

\times	20	3
10	200	30
4	80	12

Each cell is its column number times its row number. Add all four pieces:

200 + 30 + 80 + 12 = 322

The balcony has 322 seats. Check: 20 \times 14 would be 280, and 23 is a bit more than 20, so the product should be a bit more than 280. It is.

Example 6

Bigger pieces, and the row totals

Yupanqui’s orchard has 54 rows of plum trees with 68 trees in each row. How many plum trees does it have?

Solution

The total is 68 \times 54. Split both factors: 68 = 60 + 8 and 54 = 50 + 4. By the distributive property there are 2 \times 2 = 4 pieces:

\times	60	8
50	3{,}000	400
4	240	32

The big piece is 60 \times 50: 6 \times 5 = 30, and tens times tens gives hundreds, so it is 30 hundreds, 3{,}000. Then 8 \times 50 = 400, 60 \times 4 = 240 and 8 \times 4 = 32.

Add by rows. The row for 50 gives 3{,}000 + 400 = 3{,}400, which is 68 \times 50. The row for 4 gives 240 + 32 = 272, which is 68 \times 4. Then

3{,}400 + 272 = 3{,}672

The orchard has 3{,}672 plum trees. Adding by columns instead gives 3{,}240 + 432 = 3{,}672, the same answer, because the same four pieces are being added in a different order.

Example 7

A lattice with regrouping

Rawiri sells 93 tickets to an auditorium concert at 27 dollars each. How much money does he take in? Use a lattice.

Solution

The total is 27 \times 93 dollars. Draw a grid with 2 and 7 above the columns and 9 and 3 beside the rows. Fill each cell with a digit product, tens above the diagonal:

2 \times 9 = 18 and 7 \times 9 = 63 in the top row.
2 \times 3 = 06 and 7 \times 3 = 21 in the bottom row.

Figure 6The lattice for Example 7. Two of the bands add to more than 9, so each passes a 1 to the next band.

Add along the bands from the bottom right:

The ones band holds 1. Write 1.
The tens band holds 2 + 3 + 6 = 11. Write 1, and pass 1 hundred to the next band, since ten tens make one hundred.
The hundreds band holds 6 + 0 + 8 = 14, plus the 1 passed in, which makes 15. Write 5 and pass 1 thousand on.
The thousands band holds 1, plus the 1 passed in, which makes 2. Write 2.

So 27 \times 93 = 2{,}511, and Rawiri takes in 2{,}511 dollars.

The cells are the four partial products with their zeros dropped: 20 \times 90 = 1{,}800, 7 \times 90 = 630, 20 \times 3 = 60 and 7 \times 3 = 21. Adding those gives 1{,}800 + 630 + 60 + 21 = 2{,}511 too. Check with an estimate: 30 \times 90 = 2{,}700, which is close to 2{,}511.

Example 8

An estimate catches a slip

Gautam’s orchard sends out 6 trucks, each holding 1{,}849 apples. Gautam writes the partial products as 6{,}000 + 480 + 240 + 54 = 6{,}774. Is he right? Find the product with a lattice.

Solution

First estimate. The number 1{,}849 is close to 2{,}000, and 6 \times 2{,}000 = 12{,}000. An answer near 6{,}000 is about half of that, which is far too small. Something is wrong.

Find the slip. The pieces should be 6 \times 1{,}000, 6 \times 800, 6 \times 40 and 6 \times 9. The second piece is 6 groups of 8 hundreds, which is 48 hundreds, or 4{,}800. Gautam wrote 480, one place too low.

Now the lattice. The factor 1{,}849 has four digits and 6 has one, so the grid is 4 columns by 1 row. The cells hold 1 \times 6 = 06, 8 \times 6 = 48, 4 \times 6 = 24 and 9 \times 6 = 54.

Figure 7The lattice for Example 8. A one-row lattice is the same as recording four partial products.

Add along the bands from the bottom right:

Ones: 4.
Tens: 5 + 4 = 9.
Hundreds: 2 + 8 = 10. Write 0 and pass 1 on.
Thousands: 4 + 6 + 1 = 11. Write 1 and pass 1 on.
Ten thousands: 0 + 1 = 1.

So 6 \times 1{,}849 = 11{,}094. The corrected partial products agree: 6{,}000 + 4{,}800 + 240 + 54 = 11{,}094. The trucks hold 11{,}094 apples, and that is a little less than the estimate of 12{,}000, as it should be, since 1{,}849 was rounded up.

Boundaries

Every part meets every part. When both factors are split, each part of one must be multiplied by each part of the other. A two-digit by two-digit product has four pieces, not two. Leaving one out leaves part of the rectangle uncounted.
The pieces are added, never multiplied. The distributive property turns one product into a sum of products. The partial products are the parts of that sum.
Each piece keeps its place value. In a column, every partial product is written in full and lined up by place. In a lattice, the place is kept by the band instead, so each cell must hold exactly two digits (a leading 0 when the product is less than 10), and a zero digit in a factor still needs its own column or row, or every band past it shifts.
The area model is a picture of the numbers, not a scale drawing. The pieces need not be drawn to size. What matters is which side length goes with which piece.
The splits must add back to the factor. Any split works under the distributive property, as long as the parts add up to the whole factor. Place-value splits are used because they make each piece a basic fact and a shift.
The whole-number method stops here. These pictures are for whole numbers. Fractions and decimals can be split the same way later, but the lattice as drawn here places digits for whole numbers only.

Where this goes next

Partial products show where every digit of a product comes from. The standard multiplication algorithm keeps exactly the same work but writes less of it: it adds each row of the area model as it goes and stores each traded ten as a small carried digit, so a two-digit bottom factor needs only two lines.

The same splitting runs division backward. In Partial quotients, you divide by taking away easy pieces of the dividend, one at a time.

Much later, in algebra, the same four-part rectangle comes back when multiplying two sums with letters in them.

Exercises

Part 1: two digits times one digit

Draw an area model for 4 \times 57 and find the product.
Find 7 \times 39 with partial products. Name the property you use.
Find 9 \times 83 with partial products.
The auditorium balcony at another school has 8 rows with 47 seats in each row. How many seats does it have?
A rectangle is 5 tall. One cut splits it into two pieces with areas 300 and 35. How long is the top edge of each piece? Which multiplication does the whole rectangle show, and what is the product?

Part 2: three and four digits times one digit

Find 3 \times 742 with partial products, recorded in a column.
Find 4 \times 1{,}523 with partial products.
Find 9 \times 4{,}307. Which piece may you skip, and why?
Rawiri’s orchard has 7 blocks of trees with 286 trees in each block. How many trees is that?
An auditorium has 1{,}264 seats. Every seat is filled at each of 5 shows. How many tickets are used?

Part 3: two digits times two digits

Find 32 \times 17 with a four-part area model.
Find 56 \times 43 with partial products.
Find 79 \times 62 with partial products.
An orchard has 35 rows with 26 trees in each row. How many trees does it have?
An area model has 80 and 4 across the top. Its four pieces are 4{,}000, 200, 240 and 12, where 4{,}000 and 200 share a row. What numbers are down the side, which multiplication does it show, and what is the product?

Part 4: lattice multiplication and estimates

Use a lattice to find 65 \times 27.
Use a lattice to find 94 \times 38.
Use a lattice to find 3 \times 2{,}684.
Yupanqui finds 73 \times 48 and writes 280 + 560 + 120 + 24 = 984. Use an estimate to show the answer cannot be right. Then find the slip and the product.
Each of the 8 rows in Gautam’s orchard gives 1{,}275 apples. Estimate the total first, then find it exactly.

Reasoning

Explain why 47 \times 36 and 3{,}214 \times 6 both have four partial products, even though the numbers look so different.
Rawiri says 25 \times 25 = 20 \times 20 + 5 \times 5 = 425. Explain what he left out, and find the right product.
In the lattice for 65 \times 27, the top left cell holds 12. Which partial product does it stand for, and why does its 1 land in the thousands band?
Gautam finds 7 \times 56 as 7 \times 50 + 7 \times 6. Yupanqui finds it as 7 \times 28 + 7 \times 28. Are both allowed? Find the product both ways and say which split is easier and why.
The biggest piece of the area model for 52 \times 38 is 50 \times 30 = 1{,}500. Explain why the product must be more than 1{,}500, then find it.

Answers

228. The rectangle is 4 tall, cut into 50 and 7: 4 \times 50 = 200 and 4 \times 7 = 28, so 200 + 28 = 228.
273. By the distributive property, 7 \times 39 = 7 \times 30 + 7 \times 9 = 210 + 63 = 273.
747. 9 \times 83 = 9 \times 80 + 9 \times 3 = 720 + 27 = 747.
376 seats. 8 \times 47 = 8 \times 40 + 8 \times 7 = 320 + 56 = 376.
The pieces are 60 and 7 long, since 5 \times 60 = 300 and 5 \times 7 = 35. The whole top edge is 60 + 7 = 67, so the rectangle shows 5 \times 67 = 335.
2{,}226. Split 742 = 700 + 40 + 2. The pieces are 3 \times 2 = 6, 3 \times 40 = 120 and 3 \times 700 = 2{,}100, and 6 + 120 + 2{,}100 = 2{,}226.
6{,}092. The pieces are 4 \times 1{,}000 = 4{,}000, 4 \times 500 = 2{,}000, 4 \times 20 = 80 and 4 \times 3 = 12, and 4{,}000 + 2{,}000 + 80 + 12 = 6{,}092.
38{,}763. The pieces are 9 \times 4{,}000 = 36{,}000, 9 \times 300 = 2{,}700, 9 \times 0 = 0 and 9 \times 7 = 63. The tens piece is 0, so it may be skipped: 36{,}000 + 2{,}700 + 63 = 38{,}763.
2{,}002 trees. 7 \times 286 = 1{,}400 + 560 + 42 = 2{,}002. The tens give 6 + 4 = 10 tens, one hundred, and the hundreds then give 4 + 5 + 1 = 10 hundreds, one thousand.
6{,}320 tickets. 5 \times 1{,}264 = 5{,}000 + 1{,}000 + 300 + 20 = 6{,}320.
544. With 30 and 2 across the top and 10 and 7 down the side, the pieces are 30 \times 10 = 300, 2 \times 10 = 20, 30 \times 7 = 210 and 2 \times 7 = 14, and 300 + 20 + 210 + 14 = 544.
2{,}408. The pieces are 50 \times 40 = 2{,}000, 6 \times 40 = 240, 50 \times 3 = 150 and 6 \times 3 = 18, and 2{,}000 + 240 + 150 + 18 = 2{,}408.
4{,}898. The pieces are 70 \times 60 = 4{,}200, 9 \times 60 = 540, 70 \times 2 = 140 and 9 \times 2 = 18, and 4{,}200 + 540 + 140 + 18 = 4{,}898.
910 trees. 26 \times 35: 20 \times 30 = 600, 6 \times 30 = 180, 20 \times 5 = 100 and 6 \times 5 = 30, and 600 + 180 + 100 + 30 = 910.
50 and 3, since 80 \times 50 = 4{,}000 and 80 \times 3 = 240. It shows 84 \times 53, and 4{,}000 + 200 + 240 + 12 = 4{,}452.
1{,}755. The cells are 6 \times 2 = 12, 5 \times 2 = 10, 6 \times 7 = 42 and 5 \times 7 = 35. Bands from the bottom right: ones 5; tens 0 + 2 + 3 = 5; hundreds 2 + 1 + 4 = 7; thousands 1.
3{,}572. The cells are 9 \times 3 = 27, 4 \times 3 = 12, 9 \times 8 = 72 and 4 \times 8 = 32. Bands: ones 2; tens 2 + 2 + 3 = 7; hundreds 7 + 1 + 7 = 15, write 5 and pass 1; thousands 2 + 1 = 3.
8{,}052. One row of four cells: 06, 18, 24, 12. Bands: ones 2; tens 4 + 1 = 5; hundreds 8 + 2 = 10, write 0 and pass 1; thousands 6 + 1 + 1 = 8; ten thousands 0.
3{,}504. The estimate 70 \times 50 = 3{,}500 is far above 984. The slip is 70 \times 40, which is 28 hundreds, 2{,}800, not 280. The product is 2{,}800 + 560 + 120 + 24 = 3{,}504, close to the estimate.
About 10{,}400, since 1{,}275 is close to 1{,}300 and 8 \times 1{,}300 = 10{,}400. Exactly, 8 \times 1{,}275 = 8{,}000 + 1{,}600 + 560 + 40 = 10{,}200.
There is one partial product for each pair of parts, one part from each factor. 47 and 36 each have 2 parts, so there are 2 \times 2 = 4 pairs. 3{,}214 has 4 parts and 6 has 1, so there are 4 \times 1 = 4 pairs. The counts match because 2 \times 2 and 4 \times 1 are both 4.
625. He multiplied tens by tens and ones by ones, but each part must meet each part. The missing pieces are 20 \times 5 = 100 and 5 \times 20 = 100. So 25 \times 25 = 400 + 100 + 100 + 25 = 625.
It stands for 60 \times 20 = 1{,}200. The cell sits one step left of the ones column and one step up from the ones row. Each step multiplies by ten, so the cell’s ones digit 2 is in the hundreds band and its tens digit 1 is one band further, in the thousands.
Both are allowed, because the distributive property works for any split whose parts add to the factor, and 50 + 6 = 56 and 28 + 28 = 56. Both give 392: 350 + 42 = 392 and 196 + 196 = 392. The place-value split is easier, because 7 \times 50 and 7 \times 6 are basic facts with a shift, while 7 \times 28 is itself a multi-digit product.
1{,}976. The other three pieces, 2 \times 30, 50 \times 8 and 2 \times 8, all add more area, so the product is more than 1{,}500. The pieces are 1{,}500 + 60 + 400 + 16 = 1{,}976.