Distributive Property

The distributive property is the rule that relates addition and multiplication. Specifically, it states that

• $a(b+c) = ab + ac$
• $(a+b)c = ac + bc .$

It is a useful tool for expanding expressions, evaluating expressions, and simplifying expressions.

For a more advanced treatment of the distributive property, see how it can be applied to multiplying polynomials.

The distributive property helps us make sense of multiplication:

$$3 \times (12) = 3 \times (10+2) = 3\times 10 + 3\times 2 = 30 + 6 = 36.$$

It also shows us how to expand algebraic expressions:

Expand $3(a+4)$.

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We have

$$3(a+4) = 3a + 3\times 4 = 3a + 12.\ _\square$$

Simplify and combine all like terms: $(3a-a^2)4a^3$.

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Applying the distributive rule, we get:

$$\begin{aligned} (3a-a^2)4a^3 &= (3a \times 4a^3) - (a^2 \times 4a^3) \\ &= 12a^4 - 4a^5.\ _\square \end{aligned}$$

Here is an example where the distributive property is applied to a polynomial with two variables, $x$ and $y$.

Simplify and combine all like terms: $4(x+y) + x(x+2)$.

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Applying the distributive rule, we get:

$$\begin{aligned} 4(x+y) + x(x+2) &= 4x + 4y + x^2 + 2x \\ &= x^2 + 6x + 4y.\ _\square \end{aligned}$$

In the previous section, we only had $1$ pair of parentheses to distribute across. When applied to polynomial expressions, we can generalize this rule to the statement "multiply every term by every other". For example, when multiplying $(a+b)(c+d)$, we distribute by multiplying $(c+d)$ by $a$ and then by $b$, giving us $(a+b)(c+d) = a(c+d) + b(c+d) = ac+ad + bc + bd$. Notice that the product has four terms, corresponding to the products of the first, outside, inside, and last pairs of terms (often remembered by the acronym "FOIL").

What is the product of $(x+3)$ and $(x-2)$?

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Using the distributive property, we get $(x+3)(x-2) = x^2 + 3x - 2x -6 = x^2 + x -6$. $_\square$

Expand $( x^2 + 2 ) ( x - 1 )$.

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We have

$$\begin{aligned} ( x^2 + 2 ) ( x -1 ) & = x^2 \times x + x^2 \times (-1) + 2 \times x + 2 \times (-1 ) \\ & = x^3 - x^2 + 2x - 2.\ _\square \end{aligned}$$

This generalizes to product of many more terms. Always be careful to ensure that you have obtained every possible pairing of the product. Be systematic and work through each term, before moving on to the next.

What is $( a + b + c) ( x + y + z )$?

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We take the first term in the first parenthesis and multiply it to every term in the second parenthesis, to obtain $ax + ay + az$.

We take the second term in the first parenthesis and multiply it to every term in the second parenthesis, to obtain $bx + by + bz$.

We take the third term in the first parenthesis and multiply it to every term in the second parenthesis, to obtain $cx + cy + cz$.

The final product is equal to the sum of all of these terms, which is

$$ax + ay + az + bx + by + bz + cx + cy + cz.\ _\square$$

Simplify $( x^2 + x + 1 ) (x^2 - x - 1 )$.

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Since all the coefficients are $1$ or $-1,$ we have to be very careful with our signs. Use a similar approach to the previous problem, where we took the first term and multiply throughout, before moving on to the second term. For this problem, we will obtain

$$\begin{aligned} & \, x^2 \times x^2 + x^2 \times ( - x ) + x^2 \times (-1) \\ & + x \times x^2 + x \times (-x) + x \times ( -1 ) \\ & + 1 \times x^2 + 1 \times (-x) + 1 \times (-1 ) \\ = & \, x^4 - x^3 - x^2 + x^3 - x^2 - x + x^2 - x - 1 \\ = & \, x^4 - x^2 - 2x - 1.\ _\square \end{aligned}$$

We start to see how something as simple as the distributive property could result in a lot of careless mistakes if one is not careful. Always be systematic in your approach of expanding.

The distributive property is especially useful when used to simplify an expression, rather than expand it. In these cases, it is not always obvious that it should be applied.

Find the value of

$$\frac{2015 \times 1337 + 2015 + 1337 + 1}{2016}.$$

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We want to avoid doing lots of calculation. Looking at the numerator, we make a clever observation that it might be the expansion of a product from the distributive property. In particular,

$$2015 \times 1337 + 2015 + 1337 + 1 = (2015 + 1) \times (1337 + 1) = 2016 \times 1338.$$

Thus, the answer is $\frac{2016 \times 1338}{2016} = 1338.\ _\square$

To make sure you've got that idea down, give this problem a shot:

Looking for an extra challenge? Try this example!