The Distributive Property

Core Concepts

Learn how to break apart numbers using the distributive property to make multiplication much easier.

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The Distributive Property

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What is the rule for the Distributive Property?
The rule is \(a \times (b + c) = (a \times b) + (a \times c)\). This means you can multiply a number by each part of a sum and then add the results.
How do you break apart the number \(47\) using place value to prepare for the distributive property?
You write it in expanded form: \(40 + 7\).
Using the distributive property, rewrite the problem \(6 \times 58\).
You would rewrite it as \(6 \times (50 + 8)\), which then becomes \((6 \times 50) + (6 \times 8)\).
Solve \(6 \times 58\) using the partial products from the distributive property: \((6 \times 50) + (6 \times 8)\).
$$ (6 \times 50) + (6 \times 8) = 300 + 48 = 348 $$
How do you break apart a three-digit number like \(243\) for the distributive property?
You write it in expanded form based on place value: \(200 + 40 + 3\).
Rewrite the problem \(5 \times 243\) using the distributive property.
Rewrite it as \(5 \times (200 + 40 + 3)\), which equals \((5 \times 200) + (5 \times 40) + (5 \times 3)\).
Find the partial products for \(5 \times 243\) using the distributive property.
The partial products are:\ \(5 \times 200 = 1000\)\ \(5 \times 40 = 200\)\ \(5 \times 3 = 15\)
What is the final product of \(5 \times 243\) after finding the partial products (\(1000\), \(200\), \(15\))?
$$ 1000 + 200 + 15 = 1215 $$
Why is the distributive property a key concept for multi-digit multiplication?
It is the mathematical reason why methods like the area model and the standard algorithm work. It allows us to break a large, difficult problem into several smaller, easier problems.
Show how \(9 \times 99\) can be solved easily with subtraction and the distributive property.
Think of \(99\) as \((100 - 1)\). Then, \(9 \times (100 - 1) = (9 \times 100) - (9 \times 1) = 900 - 9 = 891\).
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