Combining Like Terms and Factoring | Grade 6 Math | LevelKite

You already know how to multiply with the distributive property: $3(x + 2) = 3x + 6$ . Here $x$ is a variable, a letter that stands for a number.

The two sides are equivalent expressions. They have the same value for every number that $x$ can stand for, not just for one number you happen to try. In this lesson, a variable can stand for any number unless the problem says otherwise.

Today you will write equivalent expressions in two new ways. First, you will shorten an expression by adding parts of the same kind, such as turning $3x + 4x$ into $7x$ . Second, you will run the distributive property backward, such as turning $3x + 6$ back into $3(x + 2)$ . Both moves change how an expression looks. Neither changes its value.

A term is a part of an expression that is added or subtracted. In $4x + 9 + 3x + 2$ , the terms are $4x$ , $9$ , $3x$ and $2$ .

In the term $4x$ , the number $4$ is the coefficient: the number that multiplies the variable. So $4x$ means $4 \cdot x$ . A term with no variable, such as $9$ , is a constant.

Like terms have exactly the same variable part. The variable part is everything in the term except the coefficient, such as the $x$ in $4x$ . Constants have no variable part, so all constants are like terms with each other.

Why can you add like terms? Think of $3x + 4x$ as 3 groups of $x$ plus 4 groups of $x$ . That makes 7 groups of $x$ , whatever number $x$ stands for.

The distributive property says this in symbols. In both terms a number is multiplied by $x$ , so $x$ is a shared factor: one of the things being multiplied. Multiplication works in either order, so it does not matter that $x$ is written second. You add the coefficients and keep the variable part:

3x + 4x = (3 + 4)x = 7x

Are these like terms?

Terms	Like terms?	Why
$5x$ and $6x$	Yes	Same variable part, $x$ . So $5x + 6x = (5 + 6)x = 11x$ .
$x$ and $8x$	Yes	$x$ means $1x$ . So $x + 8x = (1 + 8)x = 9x$ .
$9$ and $2$	Yes	Both are constants. So $9 + 2 = 11$ .
$6x$ and $6$	No	$6x$ has the variable part $x$ . The constant $6$ has none.
$2x$ and $2y$	No	$x$ and $y$ are different variables. They can stand for different numbers.
$5x$ and $5x^2$	No	$x^2$ means $x \cdot x$ . That is a different variable part from $x$ .

Key concept

One property, two directions

The distributive property is true for all numbers $a$ , $b$ and $c$ . Read one way, it multiplies: $a(b + c) = ab + ac$ . Read the other way, it pulls a shared factor out of every term.

Combining like terms pulls out the shared variable part and adds the coefficients.
Factoring means writing an expression as a product. It pulls out a shared number and leaves the rest in parentheses.

Either way you get an equivalent expression. It has the same value as the original for every number the variable can stand for. A property proves this for every number at once. Testing one number cannot.

ab + ac = a(b + c)

Worked example

Combine like terms

Let $x$ stand for any number. Write $4x + 9 + 3x + 2$ as an equivalent expression with as few terms as possible.

Find the like terms. 4x and 3x have the same variable part, x. The constants 9 and 2 are like terms too.
Move like terms next to each other. You can add numbers in any order and any grouping. These are the commutative and associative properties of addition, so the value does not change.
$4x + 9 + 3x + 2 = 4x + 3x + 9 + 2$
Combine the x-terms. Both have the factor x, so the distributive property lets you add the coefficients: 4 plus 3 is 7.
$4x + 3x = (4 + 3)x = 7x$
Add the constants: 9 plus 2 is 11. Then stop. 7x and 11 are not like terms, so you cannot combine them.
$4x + 3x + 9 + 2 = 7x + 11$
Check with one value, x = 2. A check like this can catch a mistake. It cannot prove the expressions are equivalent, because one match is not every value. Steps 2 to 4 are the proof, because the properties hold for every number.
$4(2) + 9 + 3(2) + 2 = 8 + 9 + 6 + 2 = 25 \quad \text{and} \quad 7(2) + 11 = 14 + 11 = 25$

Answer

For every number $x$ , the expression $4x + 9 + 3x + 2$ is equivalent to $7x + 11$ .

Now look at the other direction, factoring. You turn a sum into a product: a number times an expression in parentheses.

Take $3x + 6$ . It is 3 groups of $x$ plus 3 groups of $2$ . Together that makes 3 groups of $x + 2$ , so $3x + 6 = 3 \cdot x + 3 \cdot 2 = 3(x + 2)$ .

The $3$ is a common factor: a number that is a factor of every term. In this lesson, common factors are whole numbers greater than 1. The largest one is the greatest common factor. If the terms share no common factor greater than 1, as in $4x + 9$ , there is no such number to pull out.

Worked example

Factor with the greatest common factor

Let $x$ stand for any number. Factor $12x + 18$ using the greatest common factor.

Find the common factors of 12 and 18. Factors of 12: 1, 2, 3, 4, 6, 12. Factors of 18: 1, 2, 3, 6, 9, 18. Besides 1, both lists contain 2, 3 and 6. The greatest common factor is 6. Using it means nothing more will be left to pull out.
Divide each term by 6 to find its other factor. You may divide because 6 is not zero.
$12x \div 6 = 2x \qquad 18 \div 6 = 3$
Rewrite each term as 6 times its other factor. The value does not change, because 6 times 2x is 12x and 6 times 3 is 18.
$12x + 18 = 6 \cdot 2x + 6 \cdot 3$
Both terms now show the factor 6. Use the distributive property backward to write the 6 once, in front of the parentheses.
$6 \cdot 2x + 6 \cdot 3 = 6(2x + 3)$
Check by multiplying back. The distributive property holds for every number x, so getting the original expression back proves the two are equivalent.
$6(2x + 3) = 6 \cdot 2x + 6 \cdot 3 = 12x + 18$
Look inside the parentheses. 2 and 3 share no common factor greater than 1, so you are done. A smaller common factor also works, but it leaves more to do. For example, 2(6x + 9) is equivalent too, yet 6x and 9 still share the factor 3.
$2(6x + 9) = 2 \cdot 3(2x + 3) = 6(2x + 3)$

Answer

For every number $x$ , the expression $12x + 18$ is equivalent to $6(2x + 3)$ .

Worked example

Combine like terms, then factor

An art club places two orders for paintbrushes. Let $b$ be the price of one brush in dollars, so $b$ is a positive number. The first order is 6 brushes plus 9 dollars for shipping. The second order is 4 brushes plus 6 dollars for shipping. Write the total cost in dollars with as few terms as possible. Then factor it.

Write the total cost. Each brush costs b dollars, so 6 brushes cost 6b dollars and 4 brushes cost 4b dollars. Every term is an amount in dollars.
$6b + 9 + 4b + 6$
Move like terms next to each other. You can add in any order, so the total does not change.
$6b + 9 + 4b + 6 = 6b + 4b + 9 + 6$
Combine like terms with the distributive property. Add the coefficients: 6 plus 4 is 10. Add the constants: 9 plus 6 is 15. That is 10 brushes and 15 dollars of shipping in all.
$6b + 4b + 9 + 6 = (6 + 4)b + (9 + 6) = 10b + 15$
Find the greatest common factor of 10 and 15. Factors of 10: 1, 2, 5, 10. Factors of 15: 1, 3, 5, 15. The greatest common factor is 5.
Divide each term by 5, which is not zero: 10b divided by 5 is 2b, and 15 divided by 5 is 3. Then use the distributive property backward to pull out the 5.
$10b + 15 = 5 \cdot 2b + 5 \cdot 3 = 5(2b + 3)$
Check by multiplying back. The distributive property holds for every value of b, so this proves the factored form is equivalent to 10b + 15.
$5(2b + 3) = 5 \cdot 2b + 5 \cdot 3 = 10b + 15$

Answer

The total cost is $10b + 15$ dollars, which is equivalent to $5(2b + 3)$ dollars for every price $b$ . The factored form shows the total as 5 equal parts, each worth 2 brushes plus 3 dollars of shipping.

Takeaway: combining like terms and factoring are one idea. When terms share a factor, the distributive property lets you pull it out, as in $4x + 3x = (4 + 3)x = 7x$ and $12x + 18 = 6(2x + 3)$ . Combine only like terms. Factor out only a nonzero number that divides every term.

Each rewrite is an equivalent expression, with the same value as the original for every number the variable can stand for. One test value can expose a mistake, but only the properties prove equivalence.

Next, you will use properties to write equivalent expressions in more ways. You will also use substitution to test whether expressions are equivalent, and you will find mistakes in rewritten expressions.