The Distributive Property

What does the distributive property say?

Multiplying a sum or difference by a number is equivalent to multiplying each term by that number and then adding or subtracting the products. Every term inside the group receives the complete outside factor, including its sign.

For real numbers \(a,b,c\):

\[a(b+c)=ab+ac,\qquad a(b-c)=ab-ac.\]

For example, \(4(10+3)=4\cdot10+4\cdot3=40+12=52\). The grouped form and expanded form describe the same value. These rules apply to positive and negative numbers, fractions, decimals, and variable expressions with real-number values.

All examples, self-checks, and the diagram below were independently created for this lesson.

See the rule in a rectangle

A rectangle with height 3 units and width \(4+2\) units has area \(3(4+2)=18\) square units. Splitting the width into 4 and 2 gives two rectangles with areas 12 and 6. Their combined area remains 18.

The height multiplies both widths. Writing \(3(4+2)=3\cdot4+2\) would leave the second piece without its height factor and give the wrong area.

Expanding: a reliable routine

1. Identify the complete factor multiplying the group.
2. Read every term inside the group with its sign. Rewrite subtraction as addition of a negative term if helpful.
3. Multiply the outside factor by each term once.
4. Simplify the products, then combine like terms if possible.
5. Check that no term or sign was lost. A numerical substitution can help detect an error.

The rule also works with a factor on the right: \((b+c)a=ba+ca\). For three terms, \(a(b+c+d)=ab+ac+ad\). Expansion is not merely deleting parentheses.

Worked example: a positive factor

Problem: Expand \(5(2x+3)\).

\[5(2x+3)=5(2x)+5(3)=10x+15.\]

The coefficient of \(x\) becomes \(5\cdot2=10\), and the constant becomes \(5\cdot3=15\). Do not stop at \(10x+3\).

For \(x=2\), the original gives \(5(4+3)=35\) and the expanded form gives \(20+15=35\). This check supports the calculation; the distributive identity is what justifies equality for every real \(x\).

Worked example: a negative factor and subtraction

Problem: Expand \(-3(2x-5)\).

Treat the terms inside as \(2x\) and \(-5\). Then

\[-3(2x-5)=(-3)(2x)+(-3)(-5)=-6x+15.\]

The second product is positive because a negative times a negative is positive. At \(x=1\), both \(-3(2-5)\) and \(-6+15\) equal 9.

Do not write \(-6x-15\). The minus sign in \(-5\) must participate in the multiplication.

Worked example: an implied factor of negative one

Problem: Simplify \(-(x-4)\).

The minus sign before the group means multiplication by \(-1\):

\[-(x-4)=(-1)x+(-1)(-4)=-x+4.\]

Similarly, \(-(x+4)=-x-4\). Both terms change sign because both are multiplied by \(-1\).

For \(10-(2x-3)\), the 10 is outside the group being negated:

\[10-(2x-3)=10-2x+3=13-2x.\]

Do not negate the leading 10; it is not inside those parentheses.

Worked example: a fractional factor

Problem: Expand \(\frac34(8x-12)\).

\[\frac34(8x-12)=\frac34\cdot8x-\frac34\cdot12=6x-9.\]

Multiplying by a fraction still applies to every term. Here the factors of 4 in 8 and 12 make the products convenient. At \(x=2\), both forms give 3.

For a decimal example, \(0.2(15x+10)=3x+2\). Decimals and fractions follow the same property; the coefficient does not need to be a whole number.

Worked example: expand, then combine like terms

Problem: Simplify \(4(x+3)-2(x-5)\).

The complete second factor is \(-2\), not merely 2:

\[4(x+3)-2(x-5)=4x+12-2x+10.\]

Combine the variable terms and the constants:

\[(4x-2x)+(12+10)=2x+22.\]

At \(x=3\), the original is \(4(6)-2(-2)=24+4=28\); the simplified form is \(6+22=28\). Expansion uses distribution, while regrouping and combining compatible terms use other number properties.

Worked example: more than two terms

Problem: Expand \(-2(3x-y+4)\).

\[(-2)(3x)+(-2)(-y)+(-2)(4)=-6x+2y-8.\]

The implicit coefficient of \(-y\) is \(-1\). There are three terms, so there must be three distributed products before any combining. At \(x=1\), \(y=2\), both forms equal \(-10\).

Use distribution for mental arithmetic

Example: \(7\cdot98=7(100-2)=700-14=686\).

Example: \(6\cdot2.5=6(2+0.5)=12+3=15\).

Choose a convenient sum or difference without changing the number. This is exact arithmetic, not estimation.

Reverse distribution: factor out a common factor

Distribution works in both directions. Factoring rewrites a sum of products as one product containing a group:

\[ab+ac=a(b+c).\]

Problem: Factor \(12x+18\) using the greatest common positive integer factor of its coefficients.

Both terms contain a factor of 6: \(12x=6(2x)\) and \(18=6(3)\). Therefore

\[12x+18=6(2x+3).\]

Check by expanding back: \(6(2x)+6(3)=12x+18\). Extracting a factor means dividing each term by it to obtain what remains inside.

If you factor \(5x+5\), write \(5(x+1)\), not \(5x\) or \(5(x+5)\). The constant 5 becomes \(5\cdot1\), so the 1 must remain.

A quantity model with units

Suppose four identical kits each contain \(n\) pencils and three erasers. The total item count, counting pencils and erasers together, is \(4(n+3)=4n+12\). The variable \(n\) is a nonnegative whole-number pencil count per kit.

The expanded form counts all pencils and all erasers separately. For \(n=5\), there are \(4\cdot5+12=32\) items. If asked for pencils only, the answer would be \(4n\), so define which quantity you are counting before adding unlike object types.

Distinguish distribution from other rules

Do not distribute into a product as if it were a sum: \(3(xy)=3xy\), not \((3x)(3y)=9xy\). The factor 3 is used once in the product.

Likewise, \((x+2)^2\) is not \(x^2+4\). Squaring means multiplying the whole sum by itself, not independently squaring its terms. At \(x=1\), the values would be 9 and 5, which disproves that proposed rule.

Division: where a related rewrite works and where it fails

For a nonzero real number \(d\), dividing an entire sum by \(d\) is multiplication by \(1/d\), so

\[\frac{b+c}{d}=\frac bd+\frac cd.\]

For example, \((12+8)/4=12/4+8/4=5\). But you cannot split a sum in the denominator in the same way: \(12/(2+4)=2\), while \(12/2+12/4=9\). Always identify which quantity the outside operation acts on.

Common mistakes

• Multiplying only the first term: Every term receives the outside factor.
• Dropping a signed factor: In \(-3(2x-5)\), distribute \(-3\), not 3.
• Missing the implicit negative one: \(-(x-4)=-x+4\).
• Negating terms outside the parentheses: Preserve the outer structure.
• Combining unlike terms: \(6x-9\) cannot be merged into one term without knowing x.
• Using a sum rule on a product or power: \(3(xy)\) and \((x+2)^2\) need different treatment.
• Factoring out a number without dividing every term: Expand back to check your factorization.
• Splitting a sum in a denominator: Distribution does not authorize that rewrite.

Check your understanding

These independently written self-checks do not change the separate Quizzes or Practice tabs.

1. Expand \(6(x+4)\).
2. Expand \(5(3x-2)\).
3. Expand \(-4(x-6)\).
4. Simplify \(-(2x+7)\).
5. Expand \(\frac12(10x-8)\).
6. Expand \(0.25(12x+20)\).
7. Simplify \(3(x+2)+2(x-5)\).
8. Expand \(-3(2a-b+1)\).
9. Factor \(15x+20\) using the greatest common positive integer factor of its coefficients.
10. Calculate \(8\cdot99\) using distribution.
11. Explain why \(2(xy)\) cannot be rewritten as \((2x)(2y)\).
12. Three identical packs each contain \(n\) cards and two stickers. Write the total item-count expression in grouped and expanded forms, then evaluate it for \(n=7\).

Answers and explanations

1. \(6x+24\): multiply both terms by 6.
2. \(15x-10\): multiply the coefficient 3 and the signed constant \(-2\) by 5.
3. \(-4x+24\): the product of \(-4\) and \(-6\) is positive.
4. \(-2x-7\): multiply both terms by \(-1\).
5. \(5x-4\).
6. \(3x+5\).
7. \(3x+6+2x-10=5x-4\).
8. \(-6a+3b-3\): three terms produce three products.
9. \(5(3x+4)\). Expanding back gives \(15x+20\).
10. \(8(100-1)=800-8=792\).
11. \(2(xy)=2xy\), but \((2x)(2y)=4xy\). For example, at \(x=y=1\), the values are 2 and 4.
12. \(3(n+2)=3n+6\); for \(n=7\), the total is \(21+6=27\) items.

Related lessons

• Variables and Algebraic Expressions: identify terms and coefficients
• Translating Words into Expressions: recognize a multiple of a sum
• Commutative, Associative, and Identity Properties: distinguish valid rewrites
• Combining Like Terms and Classifying Numbers: simplify after expanding
• Multiplying and Dividing Signed Numbers: track negative factors
• Greatest Common Factor and Least Common Multiple: find a common factor

Sources and authorship

The publicly accessible educational resource below was consulted for topic coverage and mathematical checks. This lesson's explanations, example settings, table, self-checks, and area diagram were independently created; textbook prose, exercises, and artwork are not reproduced. Public access does not imply unrestricted reuse: check the resource's current license before reproducing material.

• OpenStax Prealgebra 2e, 7.3: Distributive Property
• OpenStax Prealgebra 2e: preface and reuse conditions