Multiply with Partial Products and the Standard Algorithm | Grade 5 Math | LevelKite

You already know how to multiply with tens and hundreds, as in $4 \times 30 = 120$ and $20 \times 30 = 600$ . In this lesson, you will use those facts to multiply any two big numbers.

The answer to a multiplication is called the product. A big product is easier to find when you break the job into small ones.

You can think of $38 \times 24$ as $24$ groups of $38$ . Count those groups in two parts: $20$ groups of $38$ , then $4$ more groups of $38$ . Each group of $38$ has two parts as well: $30$ and $8$ .

That is why you can split each number into its place values, such as tens and ones: $38 = 30 + 8$ and $24 = 20 + 4$ . When you multiply one part by another part, the answer is a partial product. Partial means it is only a part of the whole answer. Add all the partial products and you get the product.

Worked example

Multiply with partial products

Find $38 \times 24$ by splitting both numbers into place values and adding all the partial products.

Split each number into tens and ones. 38 is 3 tens and 8 ones. 24 is 2 tens and 4 ones. Splitting does not change a number, so the product stays the same.
$38 = 30 + 8 \qquad 24 = 20 + 4$
Find the 20 groups of 38. Each group has a 30 and an 8, so multiply both parts by 20. You know 3 times 2 is 6, so 30 times 20 is 600. You know 8 times 2 is 16, so 8 times 20 is 160.
$30 \times 20 = 600 \qquad 8 \times 20 = 160$
Find the 4 groups of 38. Multiply both parts by 4. You know 3 times 4 is 12, so 30 times 4 is 120. And 8 times 4 is 32.
$30 \times 4 = 120 \qquad 8 \times 4 = 32$
Add all four partial products. The 20 groups and the 4 groups make all 24 groups of 38, so their total is the product. Add in steps: 600 and 160 make 760, then 120 more makes 880, then 32 more makes 912.
$600 + 160 + 120 + 32 = 912$

Answer

The product $38 \times 24$ is $912$ .

A grid of the four partial products: every part of 38 times every part of 24

Part of 38	Times 20	Times 4
$30$	$30 \times 20 = 600$	$30 \times 4 = 120$
$8$	$8 \times 20 = 160$	$8 \times 4 = 32$
Column total	$600 + 160 = 760$	$120 + 32 = 152$

Key concept

Every part times every part

To multiply two numbers, split them into their place values. Every part of one number must multiply every part of the other. Each of those answers is a partial product. Add all the partial products to get the product.

You can write this work in more than one way. A shorter way joins several partial products on one line. The product does not change, because the same parts are still multiplied.

Four partial products are a lot to write. Two 3-digit numbers would have nine! The standard algorithm is a shorter way to write the same work. An algorithm is a set of steps that you follow in order.

Write one number above the other. The bottom number is the multiplier, the number you multiply by. You split only the multiplier, and you write one line for each of its digits. Each line is a bigger partial product: the value of one digit of the multiplier times the whole top number. In $38 \times 24$ , the $4$ makes the line $4 \times 38$ . The $2$ stands for $20$ , so it makes the line $20 \times 38$ .

Look back at the column totals in the grid, $152$ and $760$ . Those are the two lines. Each line joins the small partial products from one column. Here is the finished work, with a label that names each line:

\begin{array}{rl} 38 \\ \times\ 24 \\ \hline 152 & \leftarrow 4 \times 38 \\ 760 & \leftarrow 20 \times 38 \\ \hline 912 \end{array}

Worked example

The standard algorithm, one line at a time

Use the standard algorithm to find $38 \times 24$ again. Build each line one digit at a time, and name the partial product each line stands for.

Write 38 above 24, with ones under ones and tens under tens. Split only the multiplier: 24 is 2 tens and 4 ones. So there will be two lines, one for the 4 ones and one for the 2 tens.
$24 = 20 + 4$
Make the ones line, 4 times 38. First, 4 times 8 is 32 ones, which is 3 tens and 2 ones. Write 2 in the ones place. Regroup the 3 tens, which means you save them to add to the tens. Next, 4 times 3 tens is 12 tens. Add the 3 saved tens to get 15 tens, and write 15. The line reads 152.
$4 \times 38 = 152$
Make the tens line, 20 times 38. The 2 stands for 2 tens, and any number of tens has 0 ones. So write 0 in the ones place first. Then find 2 times 38 and write it to the left of that 0. 2 times 8 is 16, so write 6 and regroup 1. 2 times 3 is 6, and 1 more makes 7, so write 7. The line reads 760.
$20 \times 38 = 760$
Add the two lines. The 4 groups of 38 and the 20 groups of 38 make all 24 groups of 38, so their total is the product.
$152 + 760 = 912$

Answer

The product $38 \times 24$ is $912$ , the same as before. The line $152$ stands for the partial product $4 \times 38$ , and the line $760$ stands for the partial product $20 \times 38$ .

A 3-digit multiplier has a hundreds digit too, so it makes a third line. Look at $123 \times 432$ . The multiplier is $432 = 400 + 30 + 2$ .

The $2$ is in the ones place. Its line is $2 \times 123 = 246$ .
The $3$ is in the tens place. Its line is $30 \times 123 = 3{,}690$ . You write one zero first.
The $4$ is in the hundreds place. Its line is $400 \times 123 = 49{,}200$ . You write two zeros first, because any number of hundreds has $0$ tens and $0$ ones.

Add the three lines to get the product: $246 + 3{,}690 + 49{,}200 = 53{,}136$ . Here is the written work:

\begin{array}{rl} 123 \\ \times\ 432 \\ \hline 246 & \leftarrow 2 \times 123 \\ 3{,}690 & \leftarrow 30 \times 123 \\ 49{,}200 & \leftarrow 400 \times 123 \\ \hline 53{,}136 \end{array}

Worked example

A 3-digit multiplier with a 0 digit

Use the standard algorithm to find $142 \times 306$ .

Write 142 above 306 and line up the places. Read the multiplier by place value: 306 is 3 hundreds, 0 tens and 6 ones. Each digit gets a line, so plan three lines.
$306 = 300 + 0 + 6$
Make the ones line, 6 times 142. 6 times 2 is 12, so write 2 and regroup 1. 6 times 4 is 24, and 1 more makes 25, so write 5 and regroup 2. 6 times 1 is 6, and 2 more makes 8, so write 8. The line reads 852.
$6 \times 142 = 852$
Make the tens line. The tens digit is 0, so this line is 0 times 142. Zero groups of 142 is 0, so this line adds nothing to the total. You may write zeros on this line or leave the line out.
$0 \times 142 = 0$
Make the hundreds line, 300 times 142. The 3 stands for 3 hundreds, and any number of hundreds has 0 tens and 0 ones. So write two zeros first. Then find 3 times 142 and write it to the left of those zeros. 3 times 2 is 6, so write 6. 3 times 4 is 12, so write 2 and regroup 1. 3 times 1 is 3, and 1 more makes 4, so write 4. The line reads 42,600.
$300 \times 142 = 42{,}600$
Add the three lines. Together they make all 306 groups of 142, so their total is the product.
$852 + 0 + 42{,}600 = 43{,}452$

Answer

The product $142 \times 306$ is $43{,}452$ . The line $852$ stands for $6 \times 142$ , the tens line is $0$ , and the line $42{,}600$ stands for $300 \times 142$ .

\begin{array}{rl} 142 \\ \times\ 306 \\ \hline 852 & \leftarrow 6 \times 142 \\ 0 & \leftarrow 0 \times 142 \\ 42{,}600 & \leftarrow 300 \times 142 \\ \hline 43{,}452 \end{array}

Here is the idea to keep: every big product is made of partial products added together. You can split both numbers and write every partial product. Or you can use the standard algorithm, where each line is the value of one digit of the multiplier times the whole top number. On a tens line, you write one zero first. On a hundreds line, you write two zeros first. A $0$ digit gives a line of $0$ .

Next, you will use these methods to solve word problems, check that your products make sense, and find and fix multiplication mistakes.