Basic Algebra/Working with Numbers/Distributive Property

Vocabulary

Sum: The resulting quantity obtained by the addition of two or more terms.
Real Number: An element of the set of all rational and irrational numbers. All of these numbers can be expressed as decimals.

Term: A term is a number or a variable or the product of a number and a variable(s).

Monomial: An algebraic expression consisting of one term.

Binomial: An algebraic expression consisting of two terms.

Trinomial: An algebraic expression consisting of three terms.

Polynomial: An algebraic expression consisting of two or more terms.

Like Terms: Like terms are expressions that have the same variable(s) and the same exponent on the variable(s). Remember that constant terms are all like terms. This follows from the definition because all constant terms can be seen to have a variable with an exponent of zero.

Lesson

The distributive property is the short name for "the distributive property of multiplication over addition", although you will be using it to distribute multiplication over subtraction as well. When you are simplifying or evaluating you follow the order of operations. Sometimes you are unable to simplify any further because you cannot combine like terms. This is when the distributive property comes in handy.

Natural Language

When you first learned about multiplication it was described as grouping. You used multiplication as a way to condense the multiple addition of the same quantity. If you wanted to add $3+3+3+3$ you could think about it as four groups of three items.

|ooo| + |ooo| + |ooo| + |ooo|

You have 12 items. This is where $4\cdot 3$ comes in. So as you moved on you took this idea to incorporate variables as well. $3x$ is three groups of x.

$\underbrace {x+x+x} _{\text{three groups of x}}=3x$

And $3(2x)$ is three groups of $2x$ and $2x$ is $x+x$

$\underbrace {(x+x)+(x+x)+(x+x)} _{\text{three groups of 2x}}=3(2x)$

This gives you six x's or 6x. Now we need to take this idea and extend it even further. If you have $3(x+1)$ you might try to simplify using the order of operations first. This would have you do the addition inside the parentheses first. However, x and 1 are not like terms so the addition is impossible. We need to look at this expression differently if we are going to simplify it. What you have is $3(x+1)$ or in other words you have three groups of $(x+1)$

$\underbrace {(x+1)+(x+1)+(x+1)} _{\text{three groups of (x+1)}}=3(x+1)$

Here you can collect like terms. You have three x's and three 1's.

$\underbrace {x+x+x} _{\text{3x}}+\underbrace {1+1+1} _{\text{3}}=3x+3$

So you started with $3(x+1)$ and ended with $3x+3$

$3(x+1)=3x+3$

The last equation might make it easier to see what the distributive property says to do.

$3(x+1)=3(x)+3(1)=3x+3$

You are taking the multiplication by 3 and distributing that operation across the terms being added in the parentheses. You multiply the x by 3 and you multiply the 1 by 3. Then you just have to simplify using the order of operations.

What Is Coming Next

After you learn about the distributive property you will know how to multiply a monomial by a polynomial. Next, you can use this information to understand how to multiply a polynomial by a polynomial. You will probably move on to multiplying a binomial times a binomial. This will show up in something like (x+2)(3x+5). You can think of a problem like this as x(3x+5) + 2(3x+5). Breaking up the first binomial like this allows you to use your knowledge of the distributive property. Once you understand this use of the distributive property you can extend this understanding even further to justify the multiplication of any polynomial with any polynomial.

Sometimes while you are attempting to isolate a variable in an equation or inequality you will need to use the distributive property. You already know that you use inverse operations to isolate your desired variable, but before you do that you need to combine like terms that are on the same side of the equation (or inequality). Now there might be a step even before that. You will need to see if the distributive property needs to be used before you can combine like terms then proceed to use inverse operations to isolate a variable.

Word to the Wise

Remember that you still have the order of operations. If you can evaluate operations in a straightforward manner it is usually in your best interest to do so. The distributive property is like a back door to the order of operations for when you get stuck because you do not have like terms. Of course when you are dealing with only constant terms everything you encounter is like terms. The trouble happens when you introduce variables. This means that some terms cannot be combined. Remember that variables take the place of real numbers (at least in Algebra 1) so the same rules that govern real numbers will also govern the variables that hold their place and vice versa. You can use the distributive property even when you do not need to.

Example Problems

Example Problem #1:

Simplify $2(x+4)$

Solution to Example Problem #1:

Normally, to follow the order of operations you would add the two terms in the parenthesis first, then do the multiplication by. This does not work for this expression because x and 4 are unlike terms so you cannot combine them. We use the distributive property to help us find a way around the order of operations while still being sure that we keep the value of the express.

We distribute the multiplication by 2 across the addition. We will have 2 multiplied by x and 2 multiplied by 4.

$2(x)+2(4)$

Now we just need to finish the multiplication. $2(4)$ is equal to 8.

$2x+8$

We are done because we just have two terms being added and we cannot add them because they are not like terms.

Example Problem #2:

Simplify $3x(2x-4)$

Solution to Example Problem #2:

Since the terms inside the parentheses are not like terms we cannot combine them. We can use the distributive property to multiply by $3x$ .

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

This is the first example with subtraction in it. You keep this operation between the two terms just like we kept the addition between the two terms in the previous example. The next step is to multiply

$3x(2x)-3x(4)=6x^{2}-12x$

In order to complete the previous step you will already need to know how to multiply monomials.

To summarize all the steps...

$3x(2x-4)=3x(2x)-3x(4)=6x^{2}-12x$

Example Problem #3:

Solve for $x$ in $2(x+10)=60$

Solution to Example Problem #3:

To solve for a variable you must isolate it on one side of the equation. We need to get the $x$ out of the parentheses. Since we cannot go through the order of operations and just add x plus 10 then multiply by 2, we will have to use the distributive property. First, distribute the multiplication by 2 across the addition inside the parentheses.

$2(x+10)=60$

$2(x)+2(10)=60$

Now you can multiply

$2(x)+2(10)=60$

$2x+20=60$

Now we can work on getting the $x$ on one side by itself. You need to do the order of operations backwards so we can "undo" what is "being done to" $x$ . To get rid of adding 20 you need to subtract 20. And remember that an equation sets up a relationship that we need to preserve. If you subtract 20 from one side you need to subtract 20 from the other side as well to keep the balance.