The distributive property with variables
A number multiplies a sum whose parts cannot be added together, as in 4(x + 6). How do you carry out the multiplication anyway?
Where you are 12 in Expressions
When you reach for it
The order of operations says to work inside brackets first. In 4(10 + 6) that is easy: 10 + 6 = 16, and 4 \cdot 16 = 64. In 4(x + 6) it is impossible. The terms x and 6 are not like terms, so x + 6 is already as simple as it gets, and there is no single number to multiply by 4.
The distributive property gives another way in: multiply the 4 into each term separately.
Reach for it whenever a factor multiplies a bracketed sum or difference that cannot be simplified, and you want the expression without brackets: to see what it contains term by term, to collect it with other terms, or to solve an equation such as 3(x + 4) = 27. The bracketed form is called the factored form, because it is written as a product. The result is the expanded form, because it is written as a sum of terms. The two give the same value for every value of the letter, so they are equivalent expressions. This lesson goes from factored to expanded and stops there.
The procedure
The property is the one met with plain numbers in the distributive property: for all real numbers a, b and c,
A letter stands for a number, so the property applies to expressions with letters in them exactly as it does to numbers. This lesson takes two parts. Part 1 has a positive number outside a sum, whole, fraction or decimal. Part 2 has a negative outside and subtraction inside. A short look ahead at the end puts a letter outside the bracket.
Part 1: a number outside a sum
Think of 4(x + 6) as the area of a rectangle 4 tall and x + 6 wide. A line splits it into a part x wide and a part 6 wide.
Its area can be found two ways. As a whole, height times width, it is 4(x + 6). Part by part, it is 4 \cdot x = 4x for the left piece and 4 \cdot 6 = 24 for the right one, so 4x + 24 in total. It is the same rectangle either way, so the two areas are equal. The picture shows why the property holds for positive lengths; the property itself holds for every real number.
The outside factor is handed out to every term inside, which is what the word distribute means. Three facts follow, and the examples use all of them.
The factor may stand on either side. By the commutative property of multiplication, (b + c)a = a(b + c), so
For example (y + 2) \cdot 5 = 5y + 10. When a term is a number times a letter, the number is written first.
A coefficient inside is multiplied, not added to. In 3(2x + 7), the first product is 3 \cdot 2x. By the associative property of multiplication that is (3 \cdot 2)x = 6x, so 3(2x + 7) = 6x + 21.
Every term inside is multiplied, however many there are. Treat b + c as a single number and use the property twice:
So 2(x + 3y + 5) = 2x + 6y + 10: three terms inside, three products.
The outside number may be a fraction or a decimal. Nothing in the property asks for a whole number. Each product is worked as fractions or decimals are always multiplied, and a fraction can often be simplified before multiplying out: \displaystyle \frac{3}{4} \cdot 8x = \frac{3 \cdot 8}{4}x = 6x.
Because the two forms are equal for every value of the letter, any one value checks the work. At x = 5, 4(5 + 6) = 44 and 4 \cdot 5 + 24 = 44. Agreement at one value does not prove the two forms equivalent, but disagreement at one value proves they are not, so a check catches a slip.
Part 2: negatives and subtraction
A negative number outside. The property holds for negative a too, and each product takes its sign from the sign rules for products:
The negative sign is part of the factor, so it multiplies every term, not only the first.
A minus sign alone in front of a bracket. The opposite of a number is -1 times it, -a = (-1)a, so -(x + 5) means -1 \cdot (x + 5):
Every term inside changes sign. The opposite of a sum is the sum of the opposites.
Subtraction inside. Subtraction is adding the opposite, so a difference is a sum and the property applies to it:
So 5(x - 2) = 5x - 10. With a negative outside as well, two negatives meet in the second product, and a negative times a negative is positive:
In every case there is one product for each term inside the bracket. Counting the terms before multiplying says how many terms the expanded form will have before any of them are collected.
Looking ahead: a letter outside
Everything above stays with linear expressions: each term is a number, or a number times a single letter. The same property also works when the outside factor is a letter, since every letter stands for a number, but then the products go past linear:
Here x \cdot x = x^2 by the laws of exponents, and x \cdot 3 = 3x by the commutative property. The area picture still works: a rectangle x tall and x + 3 wide splits into an x by x square and an x by 3 strip.
When the outside factor is a number times a letter, such as 2a, each product is a product of two monomials: multiply the numbers, then the letters.
Worked examples
Examples 1 to 3 go with Part 1, Examples 4 to 6 with Part 2, and Example 7 is the look ahead.
Distributing a whole number
Expand (a) 7(2y + 5) \qquad (b) (k + 9) \cdot 4 \qquad (c) 3(2a + b + 6)
Solution(a) Multiply each term inside by 7. The first product is 7 \cdot 2y = (7 \cdot 2)y = 14y, by the associative property of multiplication, and the second is 7 \cdot 5 = 35:
(b) The factor stands on the right, which changes nothing: (b + c)a = ba + ca. So k \cdot 4 + 9 \cdot 4, and with the number written first,
(c) Three terms inside means three products: 3 \cdot 2a = 6a, 3 \cdot b = 3b and 3 \cdot 6 = 18.
The terms 6a, 3b and 18 are unlike, so nothing more can be collected.
Check (a) at y = 1: 7(2 + 5) = 49, and 14 + 35 = 49.
A fraction or a decimal outside
Expand (a) \displaystyle \frac{2}{3}(9n + 6) \qquad (b) 0.4(5m + 1.5) \qquad (c) \displaystyle \frac{1}{4}(8 + 3w)
Solution(a) Multiply each term by \displaystyle \frac{2}{3}, simplifying before multiplying out: \displaystyle \frac{2}{3} \cdot 9n = \frac{2 \cdot 9}{3}n = 6n and \displaystyle \frac{2}{3} \cdot 6 = 4.
(b) 0.4 \cdot 5m = 2m and 0.4 \cdot 1.5 = 0.6:
(c) \displaystyle \frac{1}{4} \cdot 8 = 2, and \displaystyle \frac{1}{4} \cdot 3w = \frac{3}{4}w:
Not every product comes out whole, and that is fine: \displaystyle \frac{3}{4}w is a term with a fractional coefficient. By the commutative property of addition the answer may also be written \displaystyle \frac{3}{4}w + 2.
Counting identical packs
Briseis fills 6 identical gift packs. Each pack holds g granola bars and 2 apples.
(a) Write the total number of items in two ways, and show the two are equal.
(b) How many items are there when g = 3?
Solution(a) Count pack by pack: each pack holds g + 2 items, and there are 6 packs, so 6(g + 2). Count kind by kind: there are 6g granola bars and 6 \cdot 2 = 12 apples, so 6g + 12. The distributive property says these are the same number:
The two forms say different things about the same total. The factored form shows 6 equal packs; the expanded form shows how many of each kind there are.
(b) At g = 3, the factored form gives 6(3 + 2) = 30 and the expanded form gives 18 + 12 = 30. There are 30 items.
A negative outside
Expand (a) -4(3x + 5) \qquad (b) -(y + 8) \qquad (c) \displaystyle -\frac{1}{2}(6m + 10)
Solution(a) The factor is -4, and it multiplies both terms. A negative times a positive is negative:
(b) A minus sign alone in front of a bracket is a factor of -1: -(y + 8) = (-1)y + (-1)(8), so
(c) \displaystyle \left(-\frac{1}{2}\right)(6m) = -3m and \displaystyle \left(-\frac{1}{2}\right)(10) = -5:
Check (a) at x = 1: -4(3 + 5) = -32, and -12 - 20 = -32.
Subtraction inside
Expand (a) 6(2n - 7) \qquad (b) -3(5 - 2p) \qquad (c) -(9 - w)
Solution(a) Read the difference as adding the opposite, 2n + (-7). Then 6 \cdot 2n = 12n and 6 \cdot (-7) = -42:
(b) Here both the factor and the second term are negative:
The second product is positive because a negative times a negative is positive. By the commutative property of addition, with each term keeping its own sign, this is also 6p - 15.
(c) The factor is -1: (-1)(9) = -9 and (-1)(-w) = w, so
Check (b) at p = 4: -3(5 - 8) = -3(-3) = 9, and 6 \cdot 4 - 15 = 9.
Finding the missing piece
Find what goes in each box so that the statement is true for every value of the letter.
(a) \square(2x - 5) = -8x + 20 \qquad (b) 6(\square - 2) = 18y - 12 \qquad (c) -(\square) = 7 - a
Solution(a) The missing factor times 2x must give -8x, so the factor is -4. It must also work for the second term, and it does: (-4)(-5) = 20. So -4(2x - 5) = -8x + 20. Both products have to agree; if only one had matched, no number would fit the box.
(b) 6 times the missing term must give 18y. Since 6 \cdot 3y = 18y, the term is 3y. The other product checks: 6 \cdot (-2) = -12. So 6(3y - 2) = 18y - 12.
(c) The minus sign changes the sign of every term inside, so the inside must hold the opposite of each term of 7 - a, which is -7 + a. Written with the letter first, the box holds a - 7. Check: -(a - 7) = -a + 7 = 7 - a.
Working from the expanded form back to the brackets is the same property read from right to left. That direction is factoring out a common factor.
Looking ahead: a letter outside
Expand (a) y(y - 7) \qquad (b) 2a(5a - 4b)
Solution(a) y \cdot y = y^2, and y \cdot (-7) = -7y:
The two terms have different powers of y, so they cannot be collected.
(b) Each product is a product of monomials. 2a \cdot 5a = (2 \cdot 5)(a \cdot a) = 10a^2, and 2a \cdot (-4b) = -8ab, with the letters in alphabetical order:
Check (a) at y = 10: 10(10 - 7) = 30, and 100 - 70 = 30.
Boundaries
The property spreads a factor over a sum or difference, not over a product. In 3(4x) there is a single term, and 3(4x) = (3 \cdot 4)x = 12x by the associative property. Multiplying the 3 into both factors would give 3 \cdot 4 \cdot 3 \cdot x = 36x, which multiplies by 9 instead of 3. In the same way 2(ab) = 2ab, not 2a \cdot 2b.
Multiplication distributes over addition, not the other way round. Addition does not spread over a product: 5 + 2 \cdot 3 = 11, while (5 + 2)(5 + 3) = 56.
A power of a bracket is not a number outside it. (x + 3)^2 means (x + 3)(x + 3): both factors are sums, and every term of one multiplies every term of the other. It is not x^2 + 9; at x = 1 the bracket gives 16 and x^2 + 9 gives 10. That product is multiplying two binomials.
A sum divided by a number splits; a number divided by a sum does not. Dividing by 3 is multiplying by \displaystyle \frac{1}{3}, so \displaystyle \frac{6x + 9}{3} = 2x + 3. But \displaystyle \frac{12}{x + 2} is not \displaystyle \frac{12}{x} + \frac{12}{2}: at x = 2 the first is 3 and the second is 12.
A minus sign in front of a bracket is a factor of -1, and like any factor it reaches every term inside.
The area picture needs positive lengths; the property does not. A rectangle cannot be -3 tall, but the sign rules for products carry the property to every real number, as Part 2 shows.
Expanded is not always better. The two forms are equal, and the job decides which is more useful. 6(g + 2) shows that the total in Example 3 is a multiple of 6; 6g + 12 shows how many of each kind there are.
Where this goes next
Opening brackets is usually the first step of something larger. In 5(y - 1) + 2y, once the bracket is expanded there are like terms to collect, and distributing, then combining takes that next step. Read backwards, 14y + 21 = 7(2y + 3) is factoring out a common factor, and the two directions check each other.
Equations with brackets are solved with the same move. An equation such as 3(x + 4) = 27 can be solved by dividing first or distributing first, and longer equations with brackets on one or both sides are cleared by distributing before collecting like terms on each side.
The move also builds bigger products. In (x + 2)(x + 5) the whole bracket x + 5 is distributed over x + 2, giving x(x + 5) + 2(x + 5), and each half is a bracket times a single factor, as in this lesson. That is multiplying two binomials. Finally, choosing between the factored and the expanded form to say something about a situation is rewriting an expression to show what it means.
Exercises
Part 1: a whole number outside
Expand.
- 9(m + 4)
- 5(3x + 8)
- (h + 6) \cdot 7
- 4(2a + 3b + 1)
Part 1: fractions and decimals outside
Expand.
- \displaystyle \frac{1}{2}(4t + 10)
- \displaystyle \frac{3}{5}(10p + 15)
- 0.3(20k + 7)
- \displaystyle 12\left(\frac{1}{4}x + \frac{5}{6}\right)
Part 1: areas and packs
- Zephyr’s poster is 50 cm tall. It is split into a picture part z cm wide and a text part 30 cm wide, side by side, both the full height of the poster. Write the area of the poster in two ways, one with brackets and one without, and explain why they are equal.
- Yolanda fills 8 identical bags. Each bag holds r red beads and b blue beads. Write the total number of beads with brackets and in expanded form, and find the total when r = 12 and b = 9.
Part 2: a negative outside
Expand.
- -3(x + 7)
- -(a + 9)
- \displaystyle -\frac{1}{2}(8y + 6)
Part 2: subtraction inside
Expand.
- 8(c - 5)
- -4(3n - 2)
- -(6 - 5b)
- -0.5(4 - 10h)
- Find the number that goes in the box: \square(3x - 4) = -15x + 20.
Looking ahead: a letter outside
- Expand q(q + 8).
- Expand -k(k - 9).
- Wendeline’s garden plot is a square bed w meters on each side, with a path 2 meters wide running along one side of it, so the whole plot is a rectangle w meters by w + 2 meters. Write its area with brackets, expand it, and say what each term of the expanded form measures.
Reasoning
- Use an area picture to explain why 3(x + 6) equals 3x + 18 and not 3x + 6.
- Explain why a(b - c - d) = ab - ac - ad for all numbers a, b, c and d.
- Explain why 2(3x) is 6x and not 6 \cdot 2x, while 2(3 + x) is 6 + 2x.
- Show that -2(x - 3) and 2(3 - x) are equivalent expressions.
Answers
- 9m + 36. The products are 9 \cdot m and 9 \cdot 4.
- 15x + 40. 5 \cdot 3x = 15x and 5 \cdot 8 = 40.
- 7h + 42. The factor on the right multiplies each term: h \cdot 7 = 7h and 6 \cdot 7 = 42.
- 8a + 12b + 4. Three terms inside, three products: 8a, 12b and 4.
- 2t + 5. \displaystyle \frac{1}{2} \cdot 4t = 2t and \displaystyle \frac{1}{2} \cdot 10 = 5.
- 6p + 9. \displaystyle \frac{3}{5} \cdot 10p = \frac{30}{5}p = 6p and \displaystyle \frac{3}{5} \cdot 15 = 9.
- 6k + 2.1. 0.3 \cdot 20k = 6k and 0.3 \cdot 7 = 2.1.
- 3x + 10. \displaystyle 12 \cdot \frac{1}{4}x = 3x and \displaystyle 12 \cdot \frac{5}{6} = 10.
- 50(z + 30) = 50z + 1500 square centimeters. As a whole, the poster is 50 tall and z + 30 wide. Part by part, the picture is 50z and the text part is 50 \cdot 30 = 1500. Both describe the area of the same poster, and the distributive property says they are equal.
- 8(r + b) = 8r + 8b, and the total is 168 beads. Counted bag by bag, 8 bags of r + b; counted color by color, 8r red and 8b blue. At r = 12, b = 9: 8 \cdot 21 = 168, and 96 + 72 = 168.
- -3x - 21. (-3)x = -3x and (-3)(7) = -21.
- -a - 9. -(a + 9) = (-1)(a + 9), and -1 multiplies both terms.
- -4y - 3. \displaystyle \left(-\frac{1}{2}\right)(8y) = -4y and \displaystyle \left(-\frac{1}{2}\right)(6) = -3.
- 8c - 40. 8 \cdot c = 8c and 8 \cdot (-5) = -40.
- -12n + 8. (-4)(3n) = -12n and (-4)(-2) = 8.
- 5b - 6. (-1)(6) = -6 and (-1)(-5b) = 5b, so -6 + 5b, which is 5b - 6.
- 5h - 2. (-0.5)(4) = -2 and (-0.5)(-10h) = 5h, so -2 + 5h, which is 5h - 2.
- -5. The number times 3x must be -15x, so it is -5, and the second product agrees: (-5)(-4) = 20.
- q^2 + 8q. q \cdot q = q^2 and q \cdot 8 = 8q.
- -k^2 + 9k. (-k)(k) = -k^2 and (-k)(-9) = 9k.
- w(w + 2) = w^2 + 2w square meters. The term w^2 is the area of the square bed, and 2w is the area of the path, which is 2 meters wide and w meters long.
- Draw a rectangle 3 tall and x + 6 wide, and split it into a part x wide and a part 6 wide. The parts have areas 3x and 3 \cdot 6 = 18, so the whole area 3(x + 6) is 3x + 18. The expression 3x + 6 would count the right part as 6, as if it were only 1 tall: it multiplies the x by 3 but not the 6.
- Subtraction is adding the opposite, so b - c - d = b + (-c) + (-d), a sum of three terms. The factor a multiplies each of them: ab + a(-c) + a(-d). By the sign rules a(-c) = -ac and a(-d) = -ad, so the result is ab - ac - ad.
- 3x is a single term, a product, so there is nothing to distribute over: by the associative property, 2(3x) = (2 \cdot 3)x = 6x. Writing 6 \cdot 2x = 12x would multiply by 2 twice. 3 + x is a sum of two terms, so the 2 multiplies each of them: 2 \cdot 3 + 2 \cdot x = 6 + 2x.
- Expand both. -2(x - 3) = -2x + 6, and 2(3 - x) = 6 - 2x, which is -2x + 6 by the commutative property of addition. They have the same expanded form, so they are equal for every x.