Distributive Property: Expand and Factor Algebraic Expressions - Grade 6

Grade 6 examples and questions on how to use the distributive property in algebra to expand and factor algebraic expressions. Detailed solutions and explanations are presented below.

Use the Distributive Property to Expand Algebraic Expressions

For real numbers \(a\), \(b\), and \(c\), the distributive property is given by:

a(b + c) = ab + ac

Let’s evaluate the numerical expression \(2(3+4)\) in two different ways.

Method 1: Using the Order of Operations

Evaluate the parentheses first:

$$2(3+4) = 2(7) = 14$$

Method 2: Using the Distributive Property

Distribute the multiplication over addition:

$$2(3+4) = (2)(3) + (2)(4) = 6 + 8 = 14$$

Both methods give the same result. However, when variables are involved, you cannot simply calculate the inside of the parentheses. The distributive property becomes essential.

For example:

$$2(x+6) = (2)(x) + (2)(6) = 2x + 12$$

Since \(x\) is a variable, you cannot simplify \(x+6\) directly; the distributive property must be used to expand the expression.

Use the Distributive Property to Factor Algebraic Expressions

The distributive property can also be applied in reverse to factor expressions. You look for a common factor in all terms and pull it outside the parentheses.

ab + ac = a(b + c)

Example: Factor \(3x+6\)

Notice that both \(3x\) and \(6\) share a common factor of \(3\):

$$3x+6 = 3(x) + 3(2)$$

Applying the distributive property in reverse yields:

$$3(x+2)$$

Practice Questions

Question 1: Expand the following expressions

  1. \(2(x+2)\)
  2. \(3(a+4)\)
  3. \(4(3+b)\)
  4. \(5(3+n)\)
  5. \(2(a+b)\)
  6. \(2(x+y+4)\)
  7. \((7+b)4\)
View Solutions for Question 1

Multiply the outer term by every term inside the parentheses:

  1. \(2(x+2) = 2(x) + 2(2) = 2x + 4\)
  2. \(3(a+4) = 3(a) + 3(4) = 3a + 12\)
  3. \(4(3+b) = 4(3) + 4(b) = 12 + 4b\)
  4. \(5(3+n) = 5(3) + 5(n) = 15 + 5n\)
  5. \(2(a+b) = 2(a) + 2(b) = 2a + 2b\)
  6. \(2(x+y+4) = 2(x) + 2(y) + 2(4) = 2x + 2y + 8\)
  7. \((7+b)4 = 4(7) + 4(b) = 28 + 4b\)

Question 2: Expand and simplify

  1. \(3(x+1)+3\)
  2. \(5(1+n)+6\)
  3. \(5(a+2)+2(a+3)\)
  4. \(2(1+b)+6(b+2)+4\)
  5. \(2(a+b)+3(a+b)\)
View Solutions for Question 2

First distribute, then group and combine like terms:

  1. \(3(x+1)+3 = 3x + 3 + 3 = 3x + 6\)
  2. \(5(1+n)+6 = 5 + 5n + 6 = 5n + 11\)
  3. \(5(a+2)+2(a+3) = (5a + 10) + (2a + 6) = 7a + 16\)
  4. \(2(1+b)+6(b+2)+4 = (2 + 2b) + (6b + 12) + 4 = 8b + 18\)
  5. \(2(a+b)+3(a+b) = 2a + 2b + 3a + 3b = 5a + 5b\) (Note: You could also treat \((a+b)\) as a single unit: \(2(a+b) + 3(a+b) = 5(a+b) = 5a + 5b\))

Question 3: Factor the following expressions

  1. \(2x+4\)
  2. \(3x+3\)
  3. \(4a+12\)
  4. \(21+7b\)
  5. \(15+5x\)
  6. \(\dfrac{x}{2}+\dfrac{1}{2}\)
View Solutions for Question 3

Identify the greatest common factor in each term, then pull it out of the parentheses:

  1. \(2x+4 = 2(x) + 2(2) = 2(x + 2)\)
  2. \(3x+3 = 3(x) + 3(1) = 3(x + 1)\)
  3. \(4a+12 = 4(a) + 4(3) = 4(a + 3)\)
  4. \(21+7b = 7(3) + 7(b) = 7(3 + b)\)
  5. \(15+5x = 5(3) + 5(x) = 5(3 + x)\)
  6. \(\dfrac{x}{2}+\dfrac{1}{2} = \dfrac{1}{2}(x) + \dfrac{1}{2}(1) = \dfrac{1}{2}(x + 1)\)

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