NCEA Level 1Manipulating and simplifying expressionsAS 91947

Expressions: 7 exam questions solved step by step

What this lesson covers

  • An expression can be rewritten in many equivalent forms, and each rewrite must keep its value for every value of the variable.
  • Index laws follow from what a power means: a product of repeated factors.
Key idea

7 questions, four tools

  • 1. Expand: every term by every term.
  • 2. Index laws for powers.
  • 3. Fractions: factorise, then cancel.
  • 4. Check with a number.
Your turn

Question 1 · Achieved

  • Expand and simplify (2x + 3)²
Worked example

Q1: (2x + 3)²

  • Squared: times itself.
  • Keep the middle term.
  1. (2x + 3)(2x + 3)times itself
  2. 4x² + 6x + 6x + 9four products
  3. 4x² + 12x + 9collect like terms
  4. 4x² + 9lost the middle term
  5. x = 1: 5² = 25
  6. x = 1: 4 + 12 + 9 = 25
  7. 4x² + 12x + 9
Your turn

Question 2 · Achieved

  • Simplify (3x²y)² × 2xy³
Worked example

Q2: (3x²y)² × 2xy³

  • Bracket first.
  • Multiply: add indices.
  1. (3x²y)² = 3² × (x²)² × y²
  2. = 9x⁴y²
  3. 9x⁴y² × 2xy³
  4. 9 × 2 = 18numbers multiply
  5. x⁴ × x¹ = x⁵add the indices
  6. y² × y³ = y⁵add the indices
  7. 18x⁵y⁵
Your turn

Question 3 · Merit

  • Expand and simplify
  • (x − 5)² − (x + 2)(x − 6)
  • Careful: the minus is for the whole product.
Worked example

Q3: (x − 5)² − (x + 2)(x − 6)

  • Expand each part alone.
  • Bracket it, flip signs.
  1. (x − 5)² = x² − 10x + 25
  2. (x + 2)(x − 6) = x² − 6x + 2x − 12
  3. = x² − 4x − 12
  4. x² − 10x + 25 − (x² − 4x − 12)
  5. = x² − 10x + 25 − x² + 4x + 12
  6. = −6x + 37collect like terms
  7. x = 1: 16 − (3)(−5) = 31
  8. x = 1: −6 + 37 = 31
Your turn

Question 4 · Merit

  • Simplify 12a⁵b² ÷ (2a²b)³
  • Give your answer with positive indices.
Worked example

Q4: 12a⁵b² ÷ (2a²b)³

  • Cube the 2 as well.
  • Divide: subtract indices.
  1. (2a²b)³ = 8a⁶b³cube everything
  2. 12a⁵b² ÷ 8a⁶b³
  3. 12 ÷ 8 = 3/2numbers divide
  4. a⁵ ÷ a⁶ = a⁻¹subtract the indices
  5. b² ÷ b³ = b⁻¹subtract the indices
  6. = (3/2)a⁻¹b⁻¹
  7. = 3/(2ab)a⁻¹ = 1/a
  8. a = 2, b = 1: 384 ÷ 512 = 0.75
  9. 3/(2 × 2 × 1) = 0.75
Your turn

Question 5 · Merit, in context

  • A square section has sides of (x + 6) m.
  • A shed on it is (x + 2) m by (x − 1) m.
  • The rest is lawn.
  • Find the lawn area, fully simplified.
Worked example

Q5: area of the lawn

  • Section minus shed.
  • Answer with units.
  1. Lawn = (x + 6)² − (x + 2)(x − 1)
  2. (x + 6)² = x² + 12x + 36
  3. (x + 2)(x − 1) = x² + x − 2
  4. x² + 12x + 36 − x² − x + 2every sign flips
  5. = 11x + 38collect like terms
  6. x = 4: 10² − 6 × 3 = 82
  7. x = 4: 44 + 38 = 82
  8. Lawn area = (11x + 38) m²
Your turn

Question 6 · Excellence

  • Simplify (2x² − 18) / (x² + x − 6)
  • Factorise the top and the bottom first.
Worked example

Q6: an algebraic fraction

  • x² is not a factor.
  • Cancel whole brackets.
  1. Top: 2x² − 18 = 2(x² − 9)HCF 2
  2. = 2(x + 3)(x − 3)difference of squares
  3. Bottom: x² + x − 6
  4. = (x + 3)(x − 2)3 × −2, 3 + −2
  5. (x + 3) is on top and bottom
  6. 2(x − 3)/(x − 2), x ≠ −3, 2
  7. x = 4: 14/14 = 1
  8. x = 4: 2(1)/2 = 1
Your turn

Question 7 · Excellence

  • Show that the difference between
  • the squares of two consecutive
  • odd numbers is always a multiple of 8.
Worked example

Q7: a proof with algebra

  • Use n for every case.
  • Finish with a sentence.
  1. Let them be 2n − 1, 2n + 1n a whole number
  2. (2n + 1)² − (2n − 1)²
  3. (2n + 1)² = 4n² + 4n + 1
  4. (2n − 1)² = 4n² − 4n + 1
  5. 4n² + 4n + 1 − 4n² + 4n − 1flip every sign
  6. = 8ncollect like terms
  7. 8n = 8 × n, a multiple of 8
  8. n = 3: 7² − 5² = 24 = 8 × 3
Summary

The tools that solved all seven

  • Expand, use the index laws, factorise, and check.
StepHow
Square itthe bracket times itself
Subtractbracket it, flip every sign
Indicesadd, subtract or multiply
Negativea⁻¹ means 1/a
Fractionsfactorise, then cancel
Check itsubstitute a number

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