The distributive property justified by an area model, its four versions with the factor on either side and a plus or minus inside, distributing a negative factor and preserving signs, using the property backwards for mental arithmetic, and the errors that come from failing to reach every term inside the bracket.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 2 — Properties of Real Numbers
The Distributive Property
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.6 The Distributive Property §2.6, pp. 100-106 — the lesson these objectives are drawn from
Warm-up
You have used this property in mental arithmetic for years without a name for it.
Discussion prompt
Work out 6 times 21 in your head. Describe exactly what you did — most people split one of the numbers.
Hint: Almost nobody multiplies twenty-one directly. What did you split it into?
Answer:
\[ 6(21) = 6(20 + 1) = 6(20) + 6(1) = 120 + 6 = 126 \]
You almost certainly split twenty-one into twenty and one, multiplied each by six, and added. That is the distributive property, used in the direction that makes arithmetic easy. This lesson names it and then uses it in the other direction, where a letter sits inside the bracket and the split is forced on you.
Concept
The distributive property says that multiplying a sum by a number gives the same result as multiplying each term of the sum separately and then adding. To distribute means to give something to each member of a group.
distributive property — The product of a and the quantity b plus c equals ab plus ac. The factor outside the bracket multiplies every term inside it.
\[ a(b + c) = ab + ac \]
It is the only property in this chapter that connects addition and multiplication rather than describing one of them alone.
Figure (svg): The factor 5 shown being shared out to both terms inside a bracket by two curved arrows
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.6 The Distributive Property §2.6, pp. 100-101
Section
Section 1
Concept
A rectangle of width three and length x plus two has an area that can be computed in two ways: as one rectangle, or as two rectangles side by side. Both expressions measure the same region, so they must be equal.
Figure (svg): One rectangle of width 3 and length x plus 2 shown beside the same rectangle split into two pieces of area 3x and 6
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.6 The Distributive Property §2.6, pp. 100-100 — Example 1, Use an Area Model
Picture it
The same region, with a line drawn in it or not.
Figure (svg): One rectangle of width 3 and length x plus 2 shown beside the same rectangle split into two pieces of area 3x and 6
Drawing the dividing line changes nothing about how much area there is. That is the entire argument, and it is why the property is a fact rather than a convention.
Worked example
This is Example 1 from the textbook. Width three, length x plus two.
\[ \text{Find the area of a rectangle of width } 3 \text{ and length } x + 2, \text{ in two ways.} \]
Compute the area as one rectangle
Why: Area is length times width, so three times the whole length.
\[ 3(x + 2) \]
Split the rectangle at the join
Why: One piece has length x and the other has length two, and both keep the width of three.
Compute the two areas and add them
Why: Three times x and three times two.
\[ 3(x) + 3(2) \]
Set the two expressions equal
Why: Both measure the same region, so they must be equal.
\[ 3(x + 2) = 3 x + 6 \]
Figure (svg): One rectangle of width 3 and length x plus 2 shown beside the same rectangle split into two pieces of area 3x and 6
\[ 3(x + 2) = 3(x) + 3(2) = 3x + 6 \]
Verify: substitute a value for x
Why: At x equal to 4 the length is six, so the area is three times six, which is eighteen. And three times four plus six is twelve plus six, also eighteen. The two expressions agree, as they must if they describe the same rectangle.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.6 The Distributive Property §2.6, pp. 100-100
Picture it
The area model is not a memory aid — it is the proof.
Figure (svg): One rectangle of width 3 and length x plus 2 shown beside the same rectangle split into two pieces of area 3x and 6
The equation on the pink strip is what the two pictures together say. Any doubt about whether a distribution was done correctly can be settled by drawing the rectangle and counting the pieces.
Worked example
Guided Practice 1 and 2. Width three, length x plus seven.
\[ \text{Write two expressions for the area of a rectangle of width } 3 \text{ and length } x + 7. \]
Write the area as one rectangle
Why: Three times the whole length.
\[ 3(x + 7) \]
Split the length into its two parts
Why: One piece of length x and one of length seven.
Write the two areas and add
Why: Three times x and three times seven.
\[ 3 x + 21 \]
State the algebraic statement
Why: The two expressions are equal because they measure the same area.
\[ 3(x + 7) = 3 x + 21 \]
Figure (svg): The solution to Worked example the guided-practice rectangle shown as a ladder of expressions, one row per algebraic move
\[ 3(x + 7) = 3x + 21 \]
Verify: substitute a value for x
Why: At x equal to 5 the length is twelve and the area is thirty-six. The second expression gives fifteen plus twenty-one, which is also thirty-six. Two agreeing values from two structurally different expressions is exactly what the property claims.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.6 The Distributive Property §2.6, pp. 100-100
Trap
\[ 3(x + 2) \]
Split the rectangle both ways, giving four small rectangles, and add all four areas
Why: Splitting once helped, so splitting again feels like more of the same.
The width was never a sum, so there is nothing to split there. Cutting it anyway produces pieces whose areas do not correspond to the terms in the expression.
\[ 3(x + 2) = 3x + 6 \]
Split only the side that is written as a sum
Why: The bracket marks the side with two parts; the other factor is a single quantity and stays whole.
Chapter 10 will multiply two brackets together, and there both sides really are sums and both really do get split — giving four pieces. Here only one side is a sum.
Translation
Each rectangle has a width and a length written as a sum.
Match the pairs
Why: In every case the width multiplies each part of the length separately. The last one is worth studying: the width is the letter and the length is a sum of numbers, which shows that the property does not care which of the two factors is the variable. Both sides could also be checked by adding the numbers inside first — x times ten is ten x, and four x plus six x is also ten x.
Elimination
A rectangle has width 4 and length x plus 5.
Eliminate the wrong options
Which two expressions both give its area?
Survives elimination: A
Why: Splitting the length into x and five gives two rectangles of area four x and twenty, which total four x plus twenty. Testing at a specific value settles it: at x equal to 1 the true area is twenty-four, and only the first option's second expression gives that.
Socratic
The area model does more than illustrate the property.
Discussion prompt
Explain why counting one rectangle's area in two ways establishes that 3 times the quantity x plus 2 equals 3x plus 6 for every value of x, not just for the value in the drawing.
Hint: Ask what the drawing assumed about x.
Answer:
The drawing never fixed a value for x. It only assumed that the length is x plus two, whatever x is, and that the region has one area however you choose to count it. Neither assumption depends on x being any particular number.
That is what makes it a proof rather than an example. A picture drawn with x equal to four would only establish the case x equals four; a picture drawn with the length labelled x plus two establishes every case at once. The limitation is that x has to be positive for the picture to make sense — the algebra holds for negatives too, but the rectangle does not.
Section
Section 2
Concept
To remove brackets from a product, multiply the factor outside by every term inside and add the results. The factor may sit on either side of the bracket, which gives two of the four versions.
\[ a(b + c) = ab + ac \qquad (b + c)a = ba + ca \]
The second version is the first one with the factors swapped, which the commutative property permits.
Figure (svg): The factor shown on the left of the bracket and on the right, giving the same result
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.6 The Distributive Property §2.6, pp. 101-101 — The Distributive Property box and Example 2
Picture it
Two arrangements, one answer.
Figure (svg): The factor shown on the left of the bracket and on the right, giving the same result
Only the first version has to be learned. The one with the factor on the right follows from it because a product may be written in either order.
Worked example
This is Example 2 from the textbook. One has the factor on the left and one on the right.
\[ \text{Rewrite without parentheses: } \; 2(x + 5) \; \text{ and } \; (1 + 2n)8. \]
Distribute the 2 to each term of the first
Why: Two times x and two times five.
\[ 2(x) + 2(5) \]
Multiply
Why: Two x plus ten.
\[ 2 x + 10 \]
Distribute the 8 to each term of the second
Why: The factor is on the right, which changes nothing about which terms it reaches.
\[ (1) 8 + (2 n) 8 \]
Multiply
Why: Eight plus sixteen n.
\[ 8 + 16 n \]
Figure (svg): The solution to Worked example two distributions shown as a ladder of expressions, one row per algebraic move
\[ 2(x + 5) = 2x + 10 \qquad (1 + 2n)8 = 8 + 16n \]
Verify: substitute a value into each
Why: At x equal to 3 the first bracket is eight, so the product is sixteen, and two times three plus ten is also sixteen. At n equal to 2 the second bracket is five, so the product is forty, and eight plus thirty-two is also forty. Both distributions check out.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.6 The Distributive Property §2.6, pp. 101-101
Sorting
Check whether the outside factor reached every term.
Sort into buckets
Sort each statement by whether it is correct.
All three errors have the same shape: the first term was multiplied and the second was not. Checking specifically that the last term inside the bracket got multiplied catches the whole family at once.
Worked example
Guided Practice 3 and 4. Reach every term.
\[ \text{Rewrite without parentheses: } \; 5(n + 3) \; \text{ and } \; (2p + 6)3. \]
Distribute 5 to each term
Why: Five times n and five times three.
\[ 5 n + 15 \]
Distribute 3 to each term of the second
Why: Three times two p and three times six.
\[ (2 p) 3 + (6) 3 \]
Multiply the second
Why: Six p plus eighteen.
\[ 6 p + 18 \]
Check that every term inside was reached
Why: Two terms inside each bracket, two terms in each answer.
Figure (svg): The solution to Worked example two from guided practice shown as a ladder of expressions, one row per algebraic move
\[ 5(n + 3) = 5n + 15 \qquad (2p + 6)3 = 6p + 18 \]
Verify: count the terms before and after
Why: Each bracket held two terms and each answer has two terms. A missing term in the answer is the commonest error here, and comparing the counts catches it in a second.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.6 The Distributive Property §2.6, pp. 101-101
Error analysis
The student removed brackets from four products. Two are wrong.
Annotate
On: \( 2(x + 5) = 2x + 10 \qquad 3(x + 2) = 3x + 2 \qquad (2p + 6)3 = 6p + 18 \qquad 5(n + 3) = 5n + 3 \)
The reliable check here is substitution rather than counting. One value of the variable separates a correct distribution from an incomplete one immediately.
Faded example
The first term is distributed. Supply the second.
Fill in the blanks
5(n + 3) = 5(n) + 5(3) = 5n + 15
Why: The factor five multiplies both terms inside, so the second product is five times three, which is fifteen. Writing the intermediate line with both products shown is what makes it obvious whether every term was reached — the error is invisible once the answer is compressed to a single line.
Elimination
The expression is the quantity 2p plus 6, all times 3.
Eliminate the wrong options
Which is correct?
Survives elimination: A
Why: Three multiplies both terms: three times two p is six p, and three times six is eighteen. Substituting p equal to one settles it — the bracket is eight, so the product is twenty-four, and only the first option gives twenty-four.
Prediction
The factor can be written before or after the bracket.
Predict first
Do 8 times the quantity x plus 4, and the quantity x plus 4 times 8, give the same result?
Correct: Yes, both give 8x + 32.
\[ 8(x + 4) = 8x + 32 \qquad (x + 4)8 = 8x + 32 \]
Why: Multiplication is commutative, so writing the factor before or after the bracket does not change the product. That is exactly why the four listed versions of the distributive property reduce to one: the other three follow from the first together with commutativity and the subtraction rule.
Section
Section 3
Concept
The property works over a subtraction as well as an addition. The two remaining versions handle that case, and the minus sign inside the bracket is preserved in the answer.
\[ a(b - c) = ab - ac \qquad (b - c)a = ba - ca \]
This follows from the addition version together with the subtraction rule: b minus c is b plus the opposite of c.
Figure (svg): The four versions of the distributive property, two with addition and two with subtraction
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.6 The Distributive Property §2.6, pp. 101-101 — The Distributive Property box and Example 3
Picture it
Two choices — which side the factor sits on, and which sign is inside.
Figure (svg): The four versions of the distributive property, two with addition and two with subtraction
Four rows, but only one thing to learn. The factor reaches every term, and each term keeps the sign it had.
Worked example
This is Example 3 from the textbook. The first has a negative factor outside.
\[ \text{Rewrite without parentheses: } \; -3(1 - y) \; \text{ and } \; (2x - 4)\tfrac{1}{2}. \]
Distribute negative 3 to each term of the first
Why: Negative three times one, and negative three times negative y.
\[ -3(1) - (-3) (y) \]
Multiply, watching the signs
Why: Negative three, and negative three times negative y is positive three y.
\[ -3 + 3 y \]
Distribute one half to each term of the second
Why: One half times two x, and one half times negative four.
\[ (\frac{1}{2}) (2 x) - (\frac{1}{2}) (4) \]
Multiply
Why: One half of two x is x; one half of four is two.
\[ x - 2 \]
Figure (svg): Distributing over a subtraction, shown with the sign preserved on the second term
\[ -3(1 - y) = -3 + 3y \qquad (2x - 4)\tfrac{1}{2} = x - 2 \]
Verify: substitute a value into the first
Why: At y equal to 4 the bracket is negative three, so the product is nine. And negative three plus three times four is negative three plus twelve, also nine. The sign on the second term came out positive, which is what distributing a negative over a subtraction must produce.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.6 The Distributive Property §2.6, pp. 101-101
Discrimination
Predict the signs before computing anything.
Sort into buckets
Sort each product by the signs of the two terms in its expansion.
Worked example
Guided Practice 5 and 6. Carry each sign with its factor.
\[ \text{Rewrite without parentheses: } \; -2(x + 5) \; \text{ and } \; (3y - 9)\tfrac{2}{3}. \]
Distribute negative 2 to each term of the first
Why: Negative two times x and negative two times five.
\[ -2 x - 10 \]
Note that both terms became negative
Why: The factor was negative and both terms inside were positive, so both products are negative.
Distribute two thirds to each term of the second
Why: Two thirds of three y is two y; two thirds of nine is six.
\[ 2 y - 6 \]
Keep the minus sign in the second answer
Why: The bracket held a subtraction, and the positive factor preserves it.
\[ 2 y - 6 \]
Figure (svg): The solution to Worked example two from guided practice shown as a ladder of expressions, one row per algebraic move
\[ -2(x + 5) = -2x - 10 \qquad (3y - 9)\tfrac{2}{3} = 2y - 6 \]
Verify: substitute into the first
Why: At x equal to 1 the bracket is six and the product is negative twelve. And negative two minus ten is also negative twelve. Both terms had to become negative for that to work, which confirms the factor reached both.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.6 The Distributive Property §2.6, pp. 101-101
Trap
\[ -2(x + 5) \]
Distribute the 2 to both terms and put the minus sign back at the front
Why: The minus sign feels attached to the whole expression rather than to the factor.
\[ -2x + 10 \quad \text{(wrong)} \]
Only the first term became negative. At x equal to one the true value is negative twelve, and this expression gives eight.
\[ -2(x + 5) = -2x - 10 \]
Treat the factor as negative two, sign included, and multiply both terms by it
Why: The sign is part of the factor, not a decoration in front of it, so it distributes along with the digits.
A quick check: if the factor is negative and both terms inside are positive, both terms of the answer must be negative. Any mixed-sign answer signals that the sign was left behind.
Faded example
The first product is done. Supply the second.
Fill in the blanks
-3(1 - y) = -3(1) + (-3)(-y) = -3 + 3y
Why: Negative three times negative y has two negative factors, so the product is positive three y. The sign of the second term flipped because both the factor and the term were negative, which is exactly the counting rule from Lesson 2.5 applied inside a distribution.
Elimination
The expression is negative 4 times the quantity y minus 2.
Eliminate the wrong options
Which is right?
Survives elimination: A
Why: Negative four times y is negative four y, and negative four times negative two is positive eight. Substituting y equal to 3 confirms it: the bracket is one, so the product is negative four, and negative twelve plus eight is also negative four.
Socratic
The textbook lists four versions and says the last three follow from the first.
Discussion prompt
Show how the version with a subtraction inside follows from the version with an addition, using the subtraction rule from Lesson 2.4. Then say which property gives the versions with the factor on the right.
Hint: Rewrite b minus c as a sum first.
Answer:
\[ a(b - c) = a(b + (-c)) = ab + a(-c) = ab - ac \]
The subtraction rule turns b minus c into b plus the opposite of c, which is a sum, so the addition version applies directly. The product a times the opposite of c is the opposite of ac, giving the minus sign in the answer.
The versions with the factor on the right come from the commutative property of multiplication, since the quantity b plus c times a is the same product as a times the quantity b plus c. So one statement plus two earlier facts generates all four, which is why only the first is worth memorising.
Section
Section 4
Concept
The property can be read from right to left as well as left to right. Splitting one factor into a sum or difference of two convenient numbers turns a hard product into two easy ones.
\[ 6(12.95) = 6(13 - 0.05) = 78 - 0.30 = 77.70 \]
This is what you were already doing in the warm-up, now with a name and a written form.
Figure (svg): Six CDs at 12.95 each computed mentally by splitting the price into 13 minus 0.05
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.6 The Distributive Property §2.6, pp. 102-102 — Example 5, on the cost of six CDs
Picture it
Neither of the two easy products needs a calculator.
Figure (svg): Six CDs at 12.95 each computed mentally by splitting the price into 13 minus 0.05
Thirteen was chosen because six times thirteen is a fact you already know, and five hundredths because six of them is thirty hundredths. The split is chosen to make both pieces easy, not to be tidy.
Worked example
Example 5 in spirit. The split is the decision that matters.
\[ \text{Compute } 6(12.95) \text{ mentally.} \]
Find a nearby number that is easy to multiply by six
Why: Thirteen is five hundredths above 12.95, and six thirteens is a familiar fact.
\[ 12.95 = 13 - 0.05 \]
Rewrite the product using the split
Why: Six times the quantity thirteen minus five hundredths.
\[ 6(13 - 0.05) \]
Distribute
Why: Six times thirteen is seventy-eight; six times five hundredths is thirty hundredths.
\[ 78 - 0.30 \]
Subtract
Why: Seventy-eight minus thirty cents is 77.70.
\[ 77.70 \]
Figure (svg): Six CDs at 12.95 each computed mentally by splitting the price into 13 minus 0.05
\[ 6(12.95) = 6(13) - 6(0.05) = 78 - 0.30 = 77.70 \]
Verify: estimate independently
Why: Six CDs at about thirteen dollars each is about seventy-eight dollars, and the true price is a little under that because each CD is a few cents cheaper than thirteen. The answer of 77.70 sits exactly where the estimate says it should.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.6 The Distributive Property §2.6, pp. 102-102
Translation
Match each product with the split that makes it easiest.
Match the pairs
Why: Every split picks the nearest round number and records the small difference as a correction. Two of these split with a plus and two with a minus, depending on whether the awkward number sits just above or just below the round one. The correction term is always the one people forget to multiply.
Worked example
The same product can be split several ways, and some choices are much better than others.
\[ \text{Compute } 7(98) \text{ mentally, choosing the split yourself.} \]
Look for the nearest round number
Why: One hundred is two above ninety-eight, and seven hundreds is trivial.
\[ 98 = 100 - 2 \]
Rewrite and distribute
Why: Seven times one hundred minus seven times two.
\[ 700 - 14 \]
Subtract
Why: Seven hundred minus fourteen is six hundred and eighty-six.
\[ 686 \]
Compare with a worse split
Why: Splitting as ninety plus eight would give 630 plus 56, which is correct but needs two harder multiplications and a harder addition.
Figure (svg): The solution to Worked example choosing a good split shown as a ladder of expressions, one row per algebraic move
\[ 7(98) = 7(100) - 7(2) = 700 - 14 = 686 \]
Verify: check with the alternative split
Why: Seven times ninety is 630 and seven times eight is 56, totalling 686 — the same answer. Both splits are legal, and the point of choosing well is speed rather than correctness.
Trap
\[ 6(12.95) = 6(13 - 0.05) = 78 - 0.05 \]
Multiply the round part by six and subtract the correction as it stands
Why: The correction is small, so it feels like a final adjustment rather than another term to distribute.
\[ = 77.95 \quad \text{(wrong)} \]
The correction has to be multiplied too. Each of the six CDs is five cents cheaper, so the total saving is thirty cents rather than five.
\[ 6(12.95) = 6(13) - 6(0.05) = 78 - 0.30 = 77.70 \]
Distribute to both pieces of the split, exactly as with any bracket
Why: The split created a bracket, and every term inside a bracket gets multiplied.
Sense-checking helps: six items each five cents cheaper must save six times five cents. Saying the saving in words catches the error before the arithmetic does.
Estimation
The estimate comes from the round part of the split alone.
Predict first
Roughly what is 8 times 19.95?
Correct: A bit under 160.
\[ 8(19.95) = 8(20) - 8(0.05) = 160 - 0.40 = 159.60 \]
Why: Twenty is the nearby round number, and eight twenties is a hundred and sixty. Each of the eight items is five cents cheaper than twenty, so the true total is forty cents below, at 159.60. The round part gives the estimate and the correction gives the adjustment, which is the whole structure of the method.
Faded example
The split is chosen. Complete it.
Fill in the blanks
5(198) = 5(200) - 5(2) = 1000 - 10 = 990
Why: One hundred and ninety-eight is two below two hundred, so the correction is two, and five of those is ten. A thousand minus ten is nine hundred and ninety. The correction has to be multiplied by five just as the round part was, which is the step this kind of calculation most often loses.
Socratic
Several splits are always legal. Only some are useful.
Discussion prompt
Give two properties a good split should have, and then explain why splitting 98 as 50 plus 48 would be legal but useless when multiplying by 7.
Hint: Think about which multiplications you can do without writing anything down.
Answer:
A good split has one piece you can multiply instantly — a round number, usually a power of ten or a small multiple of one — and one piece that is small enough that multiplying it is also instant. The whole point is to replace one hard multiplication with two easy ones.
Splitting 98 as 50 plus 48 satisfies neither condition: seven fifties is 350 and seven forty-eights is 336, and the second of those is no easier than the original problem. The split is perfectly legal and produces the right answer, but it does not reduce the difficulty, which was the only reason to split at all.
Section
Section 5
Concept
Almost every error with this property is one of two things: failing to multiply one of the terms inside the bracket, or leaving a negative sign behind with the factor rather than distributing it.
The substitution check catches both errors and takes about ten seconds.
Figure (svg): Two columns contrasting distributing to only the first term with distributing to both
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.6 The Distributive Property §2.6, pp. 101-102
Picture it
The two expressions differ by a single multiplication.
Figure (svg): Two columns contrasting distributing to only the first term with distributing to both
Nine against five at x equal to one. A distribution error is never small, and one substitution exposes it without any need to re-examine the working.
Worked example
The fastest way to check a distribution is to test it at one value.
\[ \text{Is } 3(x + 2) = 3x + 2 \text{ correct? Test it at } x = 1 \text{ and at } x = 4. \]
Substitute 1 into the original
Why: The bracket is three, so the product is nine.
\[ 3(3) = 9 \]
Substitute 1 into the claimed answer
Why: Three plus two is five.
\[ 3(1) + 2 = 5 \]
Compare
Why: Nine and five are different, so the claim is false.
Confirm at a second value
Why: At x equal to four the original gives eighteen and the claim gives fourteen. The gap is four at both values, which is the unmultiplied term.
Figure (svg): Two columns contrasting distributing to only the first term with distributing to both
\[ 3(x + 2) = 3x + 6 \neq 3x + 2 \]
Verify: check the corrected version at the same values
Why: At x equal to one, three plus six is nine, matching the original. At x equal to four, twelve plus six is eighteen, also matching. The corrected version agrees at both values, which is what a correct distribution must do.
Sorting
Only a bracket containing a sum or difference calls for distribution.
Sort into buckets
Sort each expression by whether the distributive property applies.
Looking for a plus or minus sign inside the bracket is the whole test. Brackets are used for several purposes, and only one of them calls for distribution.
Worked example
The property extends to brackets with three or more terms — the factor still reaches all of them.
\[ \text{Rewrite } \; -2(3x - 4 + y) \; \text{ without parentheses.} \]
Count the terms inside
Why: Three of them: three x, negative four, and y.
Multiply each by negative 2
Why: Negative two times three x is negative six x; negative two times negative four is eight; negative two times y is negative two y.
\[ -6 x, +8, -2 y \]
Write the answer
Why: Three terms, each carrying the sign its multiplication produced.
\[ -6 x + 8 - 2 y \]
Check the count
Why: Three terms in, three terms out.
Figure (svg): The solution to Worked example a distribution with several terms shown as a ladder of expressions, one row per algebraic move
\[ -2(3x - 4 + y) = -6x + 8 - 2y \]
Verify: substitute values for both letters
Why: With x equal to 1 and y equal to 2, the bracket is three minus four plus two, which is one, so the product is negative two. And negative six plus eight minus four is also negative two. Two letters and one agreeing value is a genuine check on a three-term distribution.
Trap
\[ 3(2x) \]
Distribute the 3 to both symbols inside, giving 3 times 2 and 3 times x
Why: The bracket looks like the ones that get distributed, so the same move gets applied.
\[ 3(2x) = 6 \cdot 3x = 18x \quad \text{(wrong)} \]
There is nothing to distribute over. The bracket holds a single term, not a sum, so the three simply multiplies it once: six x.
\[ 3(2x) = 6x \]
Check that the bracket actually contains a sum or difference before distributing
Why: Distribution shares a factor over terms being added; with only one term there is nothing to share.
A quick test at x equal to one: the true value is six, and the wrong version gives eighteen. Substitution catches this as reliably as it catches a missed term.
Elimination
A student writes that 5 times the quantity n plus 3 equals 5n plus 3.
Eliminate the wrong options
Which check exposes the mistake most reliably?
Survives elimination: A
Why: Substituting one is decisive: the bracket becomes four so the original gives twenty, while the claimed answer gives eight. The other three checks all pass on the incorrect answer, which makes substitution the only one worth relying on — and it works for every distribution error, not just this one.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Error | What it looks like | How to catch it |
|---|---|---|
| missed a term | 3(x + 2) = 3x + 2 | substitute one value |
| left the sign behind | -2(x + 5) = -2x + 10 | check both terms are negative |
Both errors survive every check except substitution, which is why the substitution habit is worth more here than any amount of care while multiplying.
Socratic
The chapter lists a dozen properties. This one gets a whole lesson.
Discussion prompt
Explain what makes the distributive property different in kind from the commutative, associative and identity properties. Then predict one thing it will be needed for in a later chapter.
Hint: Look at how many operations each property mentions.
Answer:
Every other property in this chapter describes one operation on its own: how addition may be reordered, how multiplication may be regrouped, what leaves a sum or a product unchanged. The distributive property is the only one that says how addition and multiplication interact, which is why it cannot be derived from any of the others.
It will be needed everywhere. Lesson 2.7 uses it backwards to combine like terms, Chapter 3 uses it to clear brackets from equations before solving, and Chapter 10 uses it twice at once to multiply two brackets together. Almost every manipulation in the rest of the book is a distribution in one direction or the other.
Comparison
Fill the blanks from memory before you scroll back. All four say the same thing.
Comparison matrix
| Version | Statement | Example |
|---|---|---|
| factor left, addition | a(b + c) = ab + ac | 5(x + 2) = 5x + 10 |
| factor right, addition | (b + c)a = ba + ca | (x + 4)8 = 8x + 32 |
| factor left, subtraction | a(b - c) = ab - ac | 4(x - 7) = 4x - 28 |
| factor right, subtraction | (b - c)a = ba - ca | (x - 5)9 = 9x - 45 |
The last three follow from the first: the commutative property moves the factor to the other side, and the subtraction rule turns a difference into a sum.
Pattern
Whether the bracket holds two terms or five, and whether the factor is positive or negative, the same five moves cover it.
Step three says sign included, and step five is what catches you when it was not. Those two together account for almost every mark available in this lesson.
OpenStax Elementary Algebra 2e, §1.9 Properties of Real Numbers §1.9
Check
Basic distribution. Reach both terms.
Check your understanding
Rewrite 7 times the quantity x plus 3 without parentheses.
Answer: A
Why: Seven multiplies both terms: seven times x is seven x, and seven times three is twenty-one. Substituting x equal to one confirms it — the bracket is four, so the product is twenty-eight, and seven plus twenty-one is also twenty-eight.
Check
A negative factor. Carry the sign to both terms.
Check your understanding
Rewrite negative 3 times the quantity 2x minus 5 without parentheses.
Answer: A
Why: Negative three times two x is negative six x, and negative three times negative five has two negative factors and is therefore positive fifteen. Substituting x equal to one gives a bracket of negative three, so the product is nine, and negative six plus fifteen is also nine.
Check
Mental arithmetic. Choose a split.
Check your understanding
Use the distributive property to compute 4 times 9.99 mentally.
Answer: A
Why: Splitting as ten minus one hundredth gives four times ten, which is forty, minus four hundredths, which is 39.96. Each of the four items is a penny cheaper than ten, so the total saving is four pence.
Real world
A shop sells notebooks at 3.98 each. You buy seven of them, and there is a promotion giving 50 cents off each notebook.
Discussion prompt
Compute the total two ways: by finding the discounted price per notebook first, and by finding the full total and then the total discount. Show that the distributive property is what guarantees the two agree, and say which version of it each calculation uses.
Hint: One calculation multiplies a difference; the other subtracts two products.
Answer:
\[ \text{price first: } 7(3.98 - 0.50) = 7(3.48) = 24.36 \]
\[ \text{total first: } 7(3.98) - 7(0.50) = 27.86 - 3.50 = 24.36 \]
The two agree because they are the two sides of the distributive property over a subtraction. The first calculation is the bracketed form, seven times a difference; the second is the expanded form, a difference of two products.
In practice the second is easier mentally, because 7 times 3.98 can itself be split as 7 times 4 minus 7 times two hundredths. Real arithmetic often applies the property twice, which is exactly what makes it worth naming.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Does the distributive property apply to 3 times the quantity 2x, where the bracket holds only one term?
Correct: No, there is nothing to distribute over — the answer is 6x.
\[ 3(2x) = 6x \qquad \text{but} \qquad 3(2 + x) = 6 + 3x \]
Why: Distribution shares a factor over terms being added or subtracted, and a bracket containing a single term has no terms to share it over. The three simply multiplies the two x once, giving six x. Testing at x equal to one settles it immediately: the true value is six, and any distributed version gives eighteen. Brackets serve several purposes, and only the ones enclosing a sum or difference call for this property.
Explain it
They can multiply and have never seen a letter inside a bracket.
Discussion prompt
In no more than four sentences, explain what the distributive property lets you do, using the word share or give. Then tell them the one check that will catch any mistake they make with it, and why it works.
Hint: The check involves picking a number.
Answer:
A usable answer: when a number sits outside a bracket, it has to be shared out to everything inside. Three times the quantity x plus two means three lots of x and also three lots of two. Missing one of them is like paying for some of your shopping and not the rest.
The check is to pick any number for the letter and work out both versions. If they disagree, the distribution is wrong. It works because the two expressions are supposed to be equal for every value of the letter, so a single disagreement is enough to refute the claim — the same counterexample logic as Lesson 2.2.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Reaching every term is fixed by writing the intermediate line with all the products shown before compressing. Negative factors are fixed by predicting the sign pattern first — a negative over a sum gives two negative terms. The right-hand version is fixed by remembering it is the same statement with the factors swapped. Mental arithmetic is fixed by always multiplying the correction as well as the round part. And every one of the four is caught by substituting a value. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page draw a rectangle of width 4 and length x plus 3, write its area as one expression, then draw the dividing line and write it as two, and set the two equal. Underneath, write all four versions of the distributive property in symbols with one example each, and mark the one version the other three are derived from. In the middle, work one distribution with a negative factor over a subtraction, showing every product on its own line before you combine them. Near the bottom, take a product such as 6 times 12.95 and compute it mentally by splitting a factor, showing both pieces. Finally, in the margin, write a wrong distribution and the single substitution that exposes it.
Your marked version should be the first, with the factor on the left and a plus inside. If you marked one of the subtraction versions, look again at how the subtraction rule generates it from the addition one.
Recap
Five things, and the last one is the check that makes the other four safe.
| If the question says | Your first move is |
|---|---|
| Rewrite without parentheses | Count the terms inside the bracket |
| Distribute -2 over a sum | Expect both terms to come out negative |
| Write two expressions for the area | Draw the rectangle whole and split |
| Use mental math to find | Split the awkward factor into round plus correction |
| Is this simplification correct | Substitute one value into both forms |
Lesson 2.7 runs the distributive property backwards on expressions rather than on numbers: three x plus five x becomes the quantity three plus five, all times x, which is eight x. That move is called combining like terms, and it is the last piece of Chapter 2's toolkit.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.6 The Distributive Property §2.6, pp. 100-106 — everything on these slides traces back here
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