Solution - Long multiplication
Step-by-step explanation
1. Rewrite the numbers from top to bottom aligned to the right
| Place value | tens | ones | . | tenths | hundredths |
| 3 | . | 1 | 4 | ||
| × | 2 | 1 | |||
Ignore the decimal points and multiply as if these are whole numbers (as if each most right digit is the ones digit):
In this case we removed 2 decimal place(s). So once calculated, the result will be reduced by the factor of 100.
| Place value | thousands | hundreds | tens | ones |
| 3 | 1 | 4 | ||
| × | 2 | 1 | ||
2. Multiply the numbers using long multiplication method
Start by multiplying the ones digit (1) of the multiplier 21 by each digit of the multiplicand 314, from right to left.
Multiply the ones digit (1) of the multiplicator by the number in the ones place value:
1×4=4
Write 4 in the ones place.
| Place value | thousands | hundreds | tens | ones |
| 3 | 1 | 4 | ||
| × | 2 | 1 | ||
| 4 | ||||
Multiply the ones digit (1) of the multiplicator by the number in the tens place value:
1×1=1
Write 1 in the tens place.
| Place value | thousands | hundreds | tens | ones |
| 3 | 1 | 4 | ||
| × | 2 | 1 | ||
| 1 | 4 | |||
Multiply the ones digit (1) of the multiplicator by the number in the hundreds place value:
1×3=3
Write 3 in the hundreds place.
| Place value | thousands | hundreds | tens | ones |
| 3 | 1 | 4 | ||
| × | 2 | 1 | ||
| 3 | 1 | 4 | ||
314 is the first partial product.
Proceed by multiplying the tens digit (2) of the multiplier (21) by each digit of the multiplicand (314), from right to left.
Because digit (2) is in tens place, we shift partial result by 1 place(s) by placing 1 zero(s).
| Place value | thousands | hundreds | tens | ones |
| 3 | 1 | 4 | ||
| × | 2 | 1 | ||
| 3 | 1 | 4 | ||
| 0 |
Multiply the tens digit (2) of the multiplicator by the number in the ones place value:
2×4=8
Write 8 in the tens place.
| Place value | thousands | hundreds | tens | ones |
| 3 | 1 | 4 | ||
| × | 2 | 1 | ||
| 3 | 1 | 4 | ||
| 8 | 0 |
Multiply the tens digit (2) of the multiplicator by the number in the tens place value:
2×1=2
Write 2 in the hundreds place.
| Place value | thousands | hundreds | tens | ones |
| 3 | 1 | 4 | ||
| × | 2 | 1 | ||
| 3 | 1 | 4 | ||
| 2 | 8 | 0 |
Multiply the tens digit (2) of the multiplicator by the number in the hundreds place value:
2×3=6
Write 6 in the thousands place.
| Place value | thousands | hundreds | tens | ones |
| 3 | 1 | 4 | ||
| × | 2 | 1 | ||
| 3 | 1 | 4 | ||
| 6 | 2 | 8 | 0 |
6,280 is the second partial product.
3. Add the partial products
314+6280=6594 long addition steps can be seen here
| Place value | thousands | hundreds | tens | ones |
| 3 | 1 | 4 | ||
| × | 2 | 1 | ||
| 3 | 1 | 4 | ||
| + | 6 | 2 | 8 | 0 |
| 6 | 5 | 9 | 4 |
Because we have 2 digit(s) to the right of the decimal point in the numbers that are being multiplied, we move the decimal point 2 time(s) to the left (reducing the result by the factor of 100) to get the final result:
The solution is: 65.94
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