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Copyright © Swun Math Grade 5 Unit 2 Lesson 6 C TE

MPs Applied MP

* Embedded MP

Conceptual Lesson Grade 5 · Unit 2 · Lesson 6 MC: 5.NBT.5

Today, I will link my understanding of multiplying multi-digit numbers using an area model with the standard algorithm of multiplication.

Vocabulary Teacher Resources

Digit: any single numeral from 0 to 9 Product: the result of multiplying factors together; the solution to a multiplication problem

Algorithm: a set of rules for solving a math problem; a step-by-step procedure to calculate a solution Partial Product:

Considerations: In Grade 4 Unit 2, students were introduced to the concept of multiplying whole numbers in an area model using base ten blocks, then learned to multiply using numbers in an area model. Some students may benefit from a review of these area models in preparation for this lesson.

Although the area model gives a good visual, it is very time consuming. The standard algorithm provides a more efficient way to multiply.

This lesson builds upon prior knowledge and conceptual understanding of multiplication in support of 5.NBT.5, which requires students to fluently multiply multi-digit whole numbers using the standard algorithm.

A common error for students is overlooking the place value of digits, or forgetting to use zeros as place holders. These errors will result in an incorrect partial product and ultimately the wrong answer when using an algorithm.

Steps: 1. Write the greater factor in expanded form above

the area model proportionally. 2. Write the lesser factor in expanded form on the

left side of the area model proportionally. 3. Write each equation in the corresponding place

value section and find each partial product. 4. Add the partial products to solve.

Application of MPs: MP1: How does understanding the area model help

you understand the standard algorithm? Understanding the area model helps me

understand the standard algorithm by .

MP7: Show how to multiply two numbers in multiple ways.

I can show how to multiply two numbers by and .

1 2 3 4 5 6 7 8

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*

Student Journal Pages

131-134

partial product

partial product

partial product

partial product

sum of partial product

42 × 31

2 40 60

+ 1200 1302

3 × 4 = 12

factor factor product


Answer Key: I/M SGP FCU HW

1. 1,302 1. 276; check area models 1. 576; check area models 1. 150

2. 5,775 2. 4,356; check area models 2. 4,853; check area models 2. 73

3. 282

4. 384; check area models

5. 7,157; check area models

Sample Response to MP questions:

MP1: How does understanding the area model help you understand the standard algorithm? Understanding the area model helps me understand the standard algorithm by giving me a

visual model of what happens when I multiply hundreds and tens, hundreds and ones, tens and tens, tens and ones, and ones by ones.

MP7: Show how to multiply two numbers in multiple ways. I can show how to multiply two numbers by using an area model and the standard algorithm.


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*MP1: Make sense of the problem and persevere in solving it!

*MP8: Find a strategy to help solve the problem. Directions: Find the product using the area model and an algorithm.

1. 42 × 31

1. Write the greater factor in expanded form above the area model proportionally.

2. Write the lesser factor in

expanded form on the left side of the area model proportionally.

3. Write each equation in the corresponding place value section and find each partial product.

4. Add the partial products to solve.

Although the area model helps conceptualize multiplication, it is time consuming. It is necessary to find a more efficient way to multiply larger numbers. This is why we use the algorithm.

Continued on the next page

× 40 + 2

30 30 × 40 = 1,200

30 × 2 =

60 +

1 × 40 = 40

1 × 2 = 2 1

× 40 + 2

30 30 × 40 = 1,200

30 × 2 =

60 +

1 × 40 = 40

1 × 2 =

2 1

× 40 + 2

30 30 × 40 = 1,200

30 × 2 =

60 +

1 × 40 = 40

1 × 2 =

2 1

1,200 60 40 + 2 1,302

42 × 31 = 1,302


/

1. Continued from previous page. The standard algorithm is:

Step 1: First line up the two factors one above the other. First, line up the numbers.

Step 2: Multiply each digit from one factor with each digit from the other factor. Then multiply 1 by every digit in 42. Begin by multiplying 1 by 2 which is 2.

Write 2 in the ones column.

Then multiply 1 by 4, remembering that 4 is in the tens place. This is like saying 1 times 4 tens, which is 4 tens or 40.

Write the 4 in the tens column.

Now multiply 30 by each of the digits in the top factor. (Remember the 3 has a value of 30 because the 3 is in the tens place.)

When multiplying by a number in the tens place, place a zero in the ones column.

Continue with the multiplication of 3 × 2, which equals 6. Remember the 6 is really in the tens place. Therefore, write down a 6 in the tens place. (3 × 20 = 60)

Next multiply 3 by 4, which is 12. (Remember they are both in the tens place, so 30 × 40 is being multiplied, which equals 1200). Write the 1 in the thousands place and the 2 in the hundreds place..

Step 3: Add the partial products. Now add the two products.

The answer is 1,302.

42 × 31 42

42 × 31 42

1260

1 × 42 is a partial product of 31 and 42.

30 × 42 is a partial product of 31 and 42.

42 × 31 42 + 1260

1302

This is the sum of the two partial products.


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2. 231 × 25

1. Write the greater factor in expanded form above the area model proportionally.

2. Write the lesser factor

in expanded form on the left side of the area model proportionally.

3. Write each equation in the corresponding place value section and find each partial product.

4. Add the partial products to solve.


× 200 30 + 1 +

4,000 1,000 600 150 20 + 5 5,775

231 × 25 = 5,775

× 200 30 + 1 +

20

+ 5

× 200 30 + 1 +

20 20 × 200 = 4,000

30 × 20 =

600

20 × 1 = 20

+ 5 × 200 =

1,000

5 × 30 = 150

5 × 1 =

5 5


/

2. Continued from previous page. The standard algorithm is:

Step 1: First line up the two factors one above the other. First, line up the numbers.

Step 2: Multiply each digit from one factor with each digit from the other factor. Then multiply 5 by every digit in 231. Begin by multiplying 5 by 1 which is 5.

Write 5 in the ones column.

Then multiply 5 by 3, remembering that 3 is in the tens place. This is like saying 5 times 3 tens, which is 15 tens or 150.

Write the 5 at the bottom and carry the 1 to the hundreds place.

Next multiply 5 by 2 which is 10 (remembering the 2 in the top factor is in the hundreds place, thus it is like multiplying 5 by 200 which is 1000.)

Before writing the 10 in the hundreds place, remember the additional 1 in the hundreds place. So 10 + 1 = 11 hundreds. Eleven hundreds is equal to 1,100.

Now multiply 20 by each of the digits in the top factor. (Remember the 2 has a value of 20 because the 2 is in the tens place.)

When multiplying by a number in the tens place, place a zero in the ones column.


231 × 25

5

231 × 25 55

1

231 × 25

1155 0

231 × 25

1155 5 × 231 is a partial product of 25 and 231.


/

2. Continued from previous page. Continue with the multiplication of 2 × 1, which equals 2. Remember the 2 is really in

the tens place. Therefore, write down a 2 in the tens place.

Next multiply 2 by 3, which is 6. (Remember they are both in the tens place, so 20 × 30 is being multiplied, which equals 600). Write the 6 in the hundreds place.

Finally, multiply 2 × 2, which equals 4. (Again, remember the 2 in 25 is in the tens place, while the 2 in 231 is in the hundreds place. Therefore, 20 × 200 equals 4000.) Write the 4 in the thousands place.

Step 3: Add the partial products. Now add the two products.

The answer is 5775.

231 × 25

1155 620

231 × 25

1155 + 4620

5775 This is the sum of the two partial products.

231 × 25

1155 4620 20 × 231 is a partial product of 25 and 231.


/

Directions: Find the product using the area model and an algorithm.

1. What is 12 × 23?

2. What is 132 × 33?


1. What is 24 × 24?

2. What is 211 × 23?



Recap today’s lesson with one or more of the following questions: MP1: How did understanding the area model help you understand the

standard algorithm?

MP7: Show how to multiply two numbers in multiple ways.


Name: ____________________________ Date: ______________________________

Directions: Review evaluating expressions containing parentheses.

1. 7 × (25 — 4) + 3 =

2. (81 ÷ 9) + (8 × 8) =

3. 6 × (74 — 27)



4. What is 16 × 24?

5. What is 421 × 17?

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Vocabulary Teacher Resourcesswunvideos.s3.amazonaws.com/ccss_lessons/G5 U2 L6 TE.pdf · Vocabulary Teacher Resources Digit: ... In Grade 4 Unit 2, ... Although the area model gives