5th Grade Multiplying Multi-Digit Whole Numbers Lesson: Notes and Examples (Free PDF)

Q: Does the order of the factors matter?

No. 23\times45 and 45\times23 both equal 1,035. Many students put the number with fewer digits on the bottom because it gives fewer rows to add.

5th grade multi digit multiplication lesson – Maths PDF

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This 5th grade multi digit multiplication lesson covers multiplying multi-digit whole numbers (Common Core 5.NBT.B.5), with a free PDF to print.

Key takeaway: This lesson helps fifth grade students multiply multi-digit whole numbers using the standard algorithm.

Topic: Multiplying Multi-Digit Whole Numbers (5th grade)
Content: One-page lesson with a definition, a worked example and key facts
Standards: 5.NBT.B.5
Answers: Explanations included
Format: Free printable PDF, US Letter, black and white

To multiply multi-digit whole numbers, use the standard algorithm. Multiply the top number by each digit of the bottom number, writing a placeholder zero for each place after the ones. These results are partial products. Add the partial products to find the answer, and estimate first so you can check that the answer is reasonable.

This lesson helps fifth grade students multiply multi-digit whole numbers using the standard algorithm. It builds on area models and partial products, so each row of the algorithm has a clear meaning.

Students learn to multiply by each digit of the bottom number, to write a placeholder zero when they multiply by tens and hundreds, and to add the partial products. They also estimate to check, handle zeros inside a factor, and solve multi-step stories. Also, plan about 25 minutes. Then, work each example with an area model beside the algorithm.

Grade 5 · Common Core State Standards: 5.NBT.B.5

Key facts in this 5th grade multi digit multiplication lesson

Key facts

A partial product comes from multiplying by one digit of the bottom number.
Next, write a placeholder zero when you multiply by tens, hundreds and so on.
Add all partial products to get the final product.
Then, estimate by rounding each factor to its greatest place before you multiply.
An area model splits each factor by place value and shows the same partial products.

What is the standard algorithm for multiplication?

The standard algorithm is a short way to multiply big numbers. Multiply the top number by each digit of the bottom number. Each result is a partial product. Then add the partial products to get the answer.

Key words

Factor: a number you multiply.
Partial product: the result of multiplying by one digit of the bottom number.
Placeholder: a zero that keeps digits in the right place.
Product: the final answer to a multiplication.

How do you multiply by a two-digit number?

Multiply \(412\times38\).

Estimate first: \(400\times40=16{,}000\).
Multiply by the 8 ones: \(412\times8=3{,}296\).
Write a 0 in the ones place as a placeholder. Then multiply by the 3 tens: \(412\times30=12{,}360\).
Add the partial products: \(3{,}296+12{,}360=15{,}656\).
Compare with the estimate. 15,656 is close to 16,000, so the answer makes sense.

The product is 15,656.

Two partial products add to 15,656.

Why do we write a zero in the second row?

The 3 in 38 stands for 3 tens, or 30. So the second row is \(412\times30\), not \(412\times3\). The zero is a placeholder. It keeps every digit in its correct place.

The placeholder zero shifts the row one place left.

How does an area model match the algorithm?

Split each factor by place value: \(412=400+10+2\) and \(38=30+8\). Draw a rectangle with six parts. Each small rectangle is one partial product. When you add all six, you get the same total, 15,656. The area model shows why the algorithm works.

Add all six: 15,656.

How do you estimate a product?

Round each factor to its greatest place. Then multiply the rounded numbers. Here 412 rounds to 400 and 38 rounds to 40. The estimate is \(400\times40=16{,}000\). If your exact answer is far from the estimate, check your work for a missing placeholder zero or a carrying mistake. For example, if a friend says \(412\times38=1{,}566\), you can see right away that this is far too small, because the estimate is about 16,000.

The estimate is 16,000.

5th grade multi digit multiplication lesson: questions and answers

What is a partial product?

A partial product is the answer you get when you multiply the top number by one digit of the bottom number. In \(412\times38\), the partial products are \(412\times8=3{,}296\) and \(412\times30=12{,}360\). You add them at the end.

Why do I write a zero in the second row?

The second row multiplies by tens, so the answer must be 10 times larger. The zero holds the ones place so the other digits shift one place left. Without it, you get an answer that is far too small.

How do I estimate a product?

Round each factor to its greatest place value, then multiply the rounded numbers. For \(58\times31\) use \(60\times30=1{,}800\). The exact answer should be close to your estimate.

Does the order of the factors matter?

No. \(23\times45\) and \(45\times23\) both equal 1,035. Many students put the number with fewer digits on the bottom because it gives fewer rows to add.

How can I check my answer?

First compare it with an estimate. Then multiply again with the factors switched, or draw an area model. If both methods give the same product, your answer is very likely right.

Common mistakes with this 5th grade multi digit multiplication lesson

Students skip the placeholder zero in the second row. The second row multiplies by tens, so it needs a zero in the ones place.
Students forget to add a carried digit. Write each carry above the next place and add it in.
Students lose a zero inside a factor. A zero in the middle still holds a place, so write it in the partial products.

How to study this chapter

Estimate by rounding to the greatest place before you multiply. Finally, the estimate tells you how many digits the product should have.
Line up the digits in columns on grid paper. Also, neat columns prevent misplaced digits. Repeat this habit on every problem for the best results.
Parents can ask the child to explain why a zero is written in the second row, using place value.

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