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Introduction to the Distributive Property

The Distributive Property

Arithmetic has a few core rules that make everything else work. One of the most useful is the distributive property. It tells us how multiplication interacts with addition. In simple terms, it lets you 'distribute' a multiplier to each number in a sum.

The distributive property is unique as it connects two different operations, typically multiplication and addition (or subtraction).

Think about it like handing out snacks. If you have 2 bags and each bag contains 3 apples and 4 oranges, how much fruit do you have in total? You could first find the total fruit in one bag (3+4=73+4=7) and then multiply by the number of bags (2×7=142 \times 7 = 14).

Or, you could count all the apples first (2×3=62 \times 3 = 6) and then count all the oranges (2×4=82 \times 4 = 8). Adding them together gives you the same total (6+8=146+8=14). The distributive property says both methods give the same result.

The Rule in Action

Mathematically, the distributive property is written like this. For any numbers aa, bb, and cc:

a(b+c)=ab+aca(b + c) = ab + ac

The term on the outside of the parentheses, aa, is multiplied by each term inside, bb and cc. Then, you add the results.

Let's use the example from the learning objective: 3×(4+5)3 \times (4 + 5). We can solve it two ways.

First, by following the order of operations (parentheses first): 3×(4+5)=3×9=273 \times (4 + 5) = 3 \times 9 = 27

Now, let's use the distributive property. We distribute the 3 to the 4 and to the 5.

3×(4+5)=(3×4)+(3×5)=12+15=27\begin{aligned} 3 \times (4 + 5) &= (3 \times 4) + (3 \times 5) \\ &= 12 + 15 \\ &= 27 \end{aligned}

We get the same answer. This might seem like more work for simple numbers, but it becomes incredibly powerful when we start using variables in algebra.

Visualizing with Area

A great way to see the distributive property is by looking at the area of a rectangle. The area of a rectangle is its length times its width. Consider a rectangle with width aa and a length made of two parts, bb and cc. Its total length is b+cb+c.

The total area of the large rectangle is its width, aa, multiplied by its total length, b+cb+c. So, the area is a(b+c)a(b+c).

You can also find the total area by adding the areas of the two smaller rectangles. The area of the first one is abab, and the area of the second is acac. Their sum is ab+acab + ac.

Since both methods calculate the same total area, we can see that a(b+c)a(b+c) must equal ab+acab+ac. This visual proof makes the property easy to remember.

Quiz Questions 1/5

Which of the following expressions correctly demonstrates the distributive property?

Quiz Questions 2/5

Using the distributive property, the expression 5×(20+4)5 \times (20 + 4) is equivalent to: