The Distributive Property is the third and final property of addition and multiplication that we will be learning about. We will take what we learned from the Associative and Commutative Properties and apply them in part of what will become the Distributive Property. Build a solid understanding of these three properties and you will have a solid base to expand your algebra knowledge on!

If a, b, and c are all real numbers then : a(b + c) = a(b) + a(c)

This may seem contradicting to the Order of Operations, stating brackets are to be equated first, however, through these following examples, we will understand the Distributive Property.

We will first start by applying the Distributive Property to problems just involving numbers. We will see how we can use the rule to make calculations much easier to do without a calculator.

Write the folllowing terms as an equal addition of two “easier” numbers:

i) 15

We can choose any numbers we want, (like 7 + 8 or 9 + 6) but generally there are easier numbers to work with,

Lets chose 10 and 5, so:

15 = 10 + 5

ii) 22

20 + 2

iii) 17, use subtraction,

Think what easy number is 17 close to? 20. So:

17 = 20 - 3

iv) 14, write this number as an addition of two numbers and a subtraction of two numbers

Think of the two “easier” numbers to work with around 14, they’d be 10 and 20, now figure out what youd have to add and subtract to those numbers to get 14.

14 = 10 + 4

14 = 20 - 6

Now, use the distributive property to multiply these more complex numbers, notice we are using some of the same numbers from the examples above. Do not use a calculator and show your work.

i) 3(15)

The distributive property states a(b + c) = a(b) + a(c),

So, we want to rearrange 3(15) to look like this,

first let's start by making the number 15 “easier”,

15 = 10 + 5, so sub that in to the expression,

3(10 + 5),

[ Below is the mathematical way of showing how the distributive property works

3(10 + 5) = (10 + 5) + (10 + 5) + (10 + 5) = (using commutative and associative properties) =

10 + 10 + 10 + 5 + 5 + 5 =

3(10) + 3(5)

notice, this is the same as 3(10 + 5) ]

So, starting from the top,

3(15) =

3(10 + 5) =

3(10) + 3(5) =

30 + 15 =

45

ii) 4(22)

Make 22 “easier”,

22 = 20 + 2, now sub that in,

4(20 + 2) =

use the distributive property, a(b + c) = a(b) + a(c)

4(20) + 4 (2) =

80 + 8 = 88

iii) 5(17), (using subtraction)

5(17) =

5(20 - 3) =

There is an easy spot to make a common mistake here, make sure that the negative is included, when expanding the brackets,

5(20) - 5(3) =

[If it helps, you can turn (20 - 3) into an addition of negatives,

So (20 - 3) = (20 + (-3)),

So, 5(20 + (-3)) =

5(20) + 5(-3) =

5(20) - 5(3) ]

5(20) - 5(3) =

100 - 15 =

85

iv) 4(14), use the distributive property using both the addtion and subtraction expressions from a)iv) and show that they are both equal.

First, lets do the addition,

4(14) =

4(10 + 4) =

4(10) + 4(4) =

40 + 16 =

56

Now, for the subtraction,

So, 14 = 20 - 6

4(14) =

4(20 - 6) =

Remember not to make the common mistake!

4(20) - 4(6) =

80 - 24 =

56

Notice how the addition and subtraction methods are both equal! Cool, right? This is a simple example of how we can use numbers, manipulate them and rearrange them in ways it allows us to calculate what we don't know and, largely, better understand the word around us.

Express the following expressions as binomials (a binomial is an algebraic expression of the sum or the difference of two terms)

3(x + 4)

With the introduction of variables, do not be put off by what might look more complicated. Stick to the definition a(b +c) = a(b) + a(c)

A method that might help, write the Distributive Property on top and write the expression you are working with below. i.e.

a(b + c) = a(b) + a(c) and below it write your expression

3(x + 4),

If we equate the a to the 3, the b to the x, and the c to the 4, we can follow the definition that we wrote above, subbing in

3 where it says a,

x where it says b,

and 4 where it says c.

a(b + c) = a(b) + a(c)

3(x + 4) = 3(x) + 3(4) ]

3x + 3(4) =

3x + 12

-5(2x - 7)

Pay attention to the negatives and keep the negatives with you when you expand the expression.

It may help to write brackets around the -5.

(-5)(2x - 7) =

(-5)(2x) - (-5)(7) - remember both negatives that we are working with,

Use the commutative and associative properties,

(-5)(2)(x) - (-5)(7) =

-10(x) - (-35) =

-10x + 35

(or 35 - 10x if you want it to look nicer!)

x(2x + 1)

Now we have a variable on the outside of the bracket, but as always, that does not change our method or the definition; follow the steps.

x(2x + 1) =

x(2x) + x(1) =

2(x)(x) + x = (remember, (x)(x) = x^{2})

2x^{2} + x

3x(4 - 2x)

With 3x on the outside of the brackets, it may seem like we have two terms, the 3 and the x. However, it is important to treat it as just one number. Writing brackets around 3x may help.

(3x)(4 - 2x) =

(3x)(4) - (3x)(2x) =

Associative and Commutative Properties,

(3)(4)(x) - (3)(2)(x)(x) =

(12)(x) - (6)(x)(x) =

12x - 6x^{2}

Now we move on to the other part of the Distributive Property, Division over Addition,

Distributive Property - Division over Addition

If a, b, and c are all real numbers then, {b + c}/a = b/a + c/a

So what this is saying is, if multiple terms are being added on top and then divided, the terms on the top (numerator) can be split up individually and divided individually by the same denominator.

It's like saying if you want to split a burger into four, you can take the whole burger and divide it with a knife into 4, keeping all of the burger as one,

OR

You can split the burger into the top bun, the patty and the bottom bun, then divide each of those by 4, and then hand out each of the pieces to each person.

For these following quotients, express them as binomials using the distributive property. Note: some of your answers will contain coefficients as fractions. (Coefficient is the number multiplying a variable, e.g. 3x -> 3 is the coefficient)

{6x + 3} / 3

First, write the definition again,

{b + c}/a = b/a + c/a

Write the expression below it,

{b + c} / a = b / a + c / a

{6x + 3} / 3 =

In the same way with the multiplication, we can line up or values and expand, based off the definition. Expand our a term to the two terms in the brackets

{b + c} / a = b / a + c / a

{6x + 3} / 3 = (6x)/3 + (3)/3

Again, using commutative and associative properties,

((6)/(3))(x) + (3)/3 =

(2)x + 1=

2x + 1

[You can also think of {b + c} / a being the same as,

(1/a){b + c} =

(1/a)(b) + (1/a)(c) = b / a + c / a]

{30x - 25} / 5

Remembering the negative.

(30x)/5 - (25)/5 =

(30/5)(x) - (25/5) =

6x - 5

{4x - 24} / 8

{4x - 24} / 8 =

(4x)/8 - (24)/8 =

(4/8)x - (24/8) =

(½)x - 3

{-12x + 32} / 16

Remember the negatives.

(-12x)/16 + (32)/16 =

(-12/16)x + (32/16) =

(¾)x + 3

Keep in mind, that as we introduce more complex expressions involving negatives, fractions and variables, the definitions, methods and math does not change. Be able to identify what is being asked and apply it to our methods.

Use the distributive property to expand the following question and then simplify it using the associatative and commutative properties.

3(x + 4 + 2y) + 2(2x - 5 + 3y)

So we first multiply out of each set of brackets with the distributive property, still follow the same methods even with three terms in the brackets.

3(x + 4 + 2y) + 2(2x - 5 + 3y)

3(x) + 3(4) + 2(y) + 2(2x) + 2(-5) + 3(y)

3x + 12 + 2y + 4x -10 + 3y

Use the commutative and then the associative properties to Group together like terms.

(3x + 4x) + (12 - 10) + (2y + 3y)

7x + 2 + 5y