Lesson 1.5 · Unit 1 · A-SSE.1, A-SSE.2

The distributive property, and the minus sign that must reach every term

The distributive property is the single most used rule in Algebra 1, and the single most common place to lose a sign. When a negative sits in front of a bracket, it multiplies everything inside, not just the first thing. Almost every student knows that and a large fraction still forget it under time pressure.

The method
  1. The property: \( a(b + c) = ab + ac \). The factor outside multiplies every term inside, without exception.
  2. Subtraction inside is addition of a negative. \( a(b - c) = ab - ac \), which follows from the same rule once \( b - c \) is read as \( b + (-c) \).
  3. A negative outside multiplies every term. \( -3(x - 4) = -3x + 12 \). Both signs change, because both terms are multiplied by a negative.
  4. A bare minus in front of a bracket means \( -1 \) times it. \( -(x - 7) = -x + 7 \). This is the case students most often get wrong, because the \( -1 \) is invisible.
  5. Like terms have the same variables raised to the same powers. \( 5x \) and \( -2x \) are like. \( 5x \) and \( 5x^2 \) are not. \( 5x \) and \( 5y \) are not. Constants are like each other.
  6. Combining like terms adds the coefficients only. The variable part is unchanged: \( 7x - 3x = 4x \), not \( 4x^2 \) and not 4.
  7. Distribute first, then combine. Doing it in the other order is the usual source of confusion in expressions with several brackets.
  8. Check by substituting a value. Pick any convenient number, evaluate the original and the simplified form, and confirm they agree. Avoid 0 and 1, which can hide errors.

Where students lose marks: distributing a negative to the first term only. \( -4(3x + 1) \) is \( -12x - 4 \), not \( -12x + 4 \). Write the sign onto both products as you go rather than fixing them afterward.

Worked example

The problem. Simplify \( 5(2x - 3) - 4(3x + 1) \), then verify the result.

Step one: identify what multiplies each bracket, including its sign. The first bracket is multiplied by \( +5 \). The second is multiplied by \( -4 \), not by \( 4 \). Taking the minus sign as part of the multiplier is the whole trick.

Step two: distribute the 5 across the first bracket. \( 5 \times 2x = 10x \) and \( 5 \times (-3) = -15 \). So the first bracket becomes \( 10x - 15 \).

Step three: distribute the \( -4 \) across the second bracket, term by term. \( -4 \times 3x = -12x \). Then \( -4 \times 1 = -4 \). So the second bracket becomes \( -12x - 4 \). Both terms are negative, because both were multiplied by a negative.

Step four: write the whole expression with no brackets left. \[ 10x - 15 - 12x - 4 \]

Step five: group the like terms. The \( x \) terms are \( 10x \) and \( -12x \). The constants are \( -15 \) and \( -4 \). Keeping each term's sign attached as you move it is what the commutative property permits.

Step six: combine. \( 10x - 12x = -2x \), since \( 10 - 12 = -2 \). And \( -15 - 4 = -19 \). The simplified expression is \[ -2x - 19 \]

Step seven: verify by substitution. Take \( x = 3 \), avoiding 0 and 1. Original: \( 5(2(3) - 3) - 4(3(3) + 1) = 5(3) - 4(10) = 15 - 40 = -25 \). Simplified: \( -2(3) - 19 = -6 - 19 = -25 \). They agree.

Step eight: see what the common error would have produced. A student who distributed the \( -4 \) to the \( 3x \) only would get \( 10x - 15 - 12x + 4 \), simplifying to \( -2x - 11 \). At \( x = 3 \) that gives \( -17 \), not \( -25 \). The substitution check catches the error in about ten seconds, which is why it is worth doing every time.

Practice · 10 problems

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

  1. Expand \( 4(x + 6) \).
    Show the full solution

    \( 4x + 24 \)

  2. Expand \( -3(x + 5) \).
    Show the full solution

    Both terms are multiplied by \( -3 \). \( -3x - 15 \)

  3. Expand \( -2(4x - 7) \).
    Show the full solution

    \( -2 \times 4x = -8x \) and \( -2 \times (-7) = +14 \). \( -8x + 14 \)

  4. Simplify \( 9y - 4y + 2y \).
    Show the full solution

    All like terms; add the coefficients: \( 9 - 4 + 2 = 7 \). \( 7y \)

  5. Simplify \( -(x - 9) \).
    Show the full solution

    The bare minus is \( -1 \) times the bracket. \( -x + 9 \)

  6. Simplify \( 3(2x + 4) - 2(x - 5) \).
    Show the full solution

    First bracket: \( 6x + 12 \). Second: \( -2 \times x = -2x \) and \( -2 \times (-5) = +10 \), giving \( -2x + 10 \). Together: \( 6x + 12 - 2x + 10 \). Combining: \( 6x - 2x = 4x \) and \( 12 + 10 = 22 \). \( 4x + 22 \)

  7. Simplify \( 5 - 2(3x - 4) + x \).
    Show the full solution

    Distribute the \( -2 \): \( -2 \times 3x = -6x \) and \( -2 \times (-4) = +8 \). So the expression is \( 5 - 6x + 8 + x \). Combine \( x \) terms: \( -6x + x = -5x \). Combine constants: \( 5 + 8 = 13 \). \( -5x + 13 \)

  8. Simplify \( 4(2a - 3b) - 3(a - 2b) \).
    Show the full solution

    First bracket: \( 8a - 12b \). Second: \( -3 \times a = -3a \) and \( -3 \times (-2b) = +6b \), giving \( -3a + 6b \). Together: \( 8a - 12b - 3a + 6b \). The \( a \) terms give \( 8a - 3a = 5a \); the \( b \) terms give \( -12b + 6b = -6b \). These cannot be combined with each other, since \( a \) and \( b \) are different variables. \( 5a - 6b \)

  9. A student simplifies \( 7 - 3(x - 2) \) to \( 4x - 8 \). Find both errors and give the correct answer.
    Show the full solution

    Two things went wrong. First, they subtracted 3 from 7 before distributing, treating the expression as \( 4(x - 2) \); multiplication must be done before subtraction. Second, they distributed a positive 3 rather than the \( -3 \) that actually sits in front of the bracket. Correctly: \( -3 \times x = -3x \) and \( -3 \times (-2) = +6 \), so the expression is \( 7 - 3x + 6 = -3x + 13 \). \( -3x + 13 \); they subtracted before distributing and dropped the negative sign

  10. Explain why \( 3x + 4y \) cannot be simplified to \( 7xy \), using a substitution to make the point.
    Show the full solution

    Terms can only be combined when they have identical variable parts, because combining is an application of the distributive property in reverse: \( 3x + 4x = (3 + 4)x \) works precisely because the same \( x \) can be factored out. With \( 3x + 4y \) there is no common variable factor, so nothing can be pulled out and the terms stay separate. A substitution settles it: at \( x = 2 \) and \( y = 5 \), the original is \( 3(2) + 4(5) = 6 + 20 = 26 \), while \( 7xy \) gives \( 7(2)(5) = 70 \). Different values for the same inputs means the two expressions are not equal. They are unlike terms with no common factor; at \( x = 2, y = 5 \) the original is 26 while \( 7xy \) is 70

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