Chapter 4 · II · Algebra

Exploring Algebraic Identities Worksheet

7 questions Geometric identities Factorisation and tiles Cubic identities Rational-expression restrictions

Try each question before opening its solution.

Name: ________________________   Class: __________   Date: __________

Foundation

Simplifying rational expressions with exclusions · 1 practice mark

Question 1

The expression (x² − 9)/(x − 3) simplifies to:

  1. x − 3 for all x
  2. x + 3 for x ≠ 3
  3. x + 3 for all x
  4. x² + 3
Show step-wise answer

(b) x + 3 for x ≠ 3

  1. Factor x² − 9 = (x − 3)(x + 3).
  2. Cancel only when the original denominator is non-zero; retain x ≠ 3.

Difference of squares · 1 practice mark

Question 2

(x + 4)(x − 4) equals:

  1. x² + 16
  2. x² − 16
  3. x² − 8x + 16
  4. x² + 8x + 16
Show step-wise answer

(b) x² − 16

  1. Expand: x² − 4x + 4x − 16.
  2. The middle terms cancel, leaving x² − 16.

Practice

Identities and geometric models · Simplifying rational expressions with exclusions · 3 practice marks

Question 3

Simplify (a³ − b³)/(a − b), stating its restriction, and verify the resulting identity by multiplication.

Show step-wise answer

a² + ab + b² for a ≠ b.

  1. Multiply (a − b)(a² + ab + b²).
  2. The mixed terms cancel, leaving a³ − b³.
  3. Divide by a − b only when a ≠ b; this restriction remains after simplification.

Sum of cubes · 3 practice marks

Question 4

Factor x³ + 27 and verify your factors by multiplying.

Show step-wise answer

(x + 3)(x² − 3x + 9).

  1. Use a³ + b³ = (a + b)(a² − ab + b²) with a = x, b = 3.
  2. Multiply: x³ − 3x² + 9x + 3x² − 9x + 27.
  3. The x² and x terms cancel, giving x³ + 27.

Challenge

Identities and geometric models · Factorisation and algebra tiles · 4 practice marks

Question 5

Expand (x + 3)² and factor x² + 6x + 9. Explain both using a square divided into an x-by-x square, two rectangles and a small square.

Show step-wise answer

(x + 3)² = x² + 6x + 9; the factors are (x + 3)(x + 3).

  1. A square of side x + 3 has area (x + 3)².
  2. The pieces have areas x², 3x, 3x and 9.
  3. Their sum is x² + 6x + 9, establishing the identity for the geometric model x ≥ 0.
  4. Algebraic expansion extends the identity to all real x; reversing the expansion gives the factors.

Identities and geometric models · Factorisation and algebra tiles · 4 practice marks

Question 6

Factor x² + 5x + 6 by splitting its middle term. Describe how algebra tiles form a rectangle and derive a² − b² from (a + b)(a − b).

Show step-wise answer

(x + 2)(x + 3); a² − b² = (a + b)(a − b).

  1. Numbers 2 and 3 have sum 5 and product 6, so write x² + 2x + 3x + 6.
  2. Group x(x + 2) + 3(x + 2) = (x + 2)(x + 3).
  3. For x ≥ 0, arrange one x² tile, five x tiles and six unit tiles as a rectangle of sides x + 2 and x + 3.
  4. Expand (a + b)(a − b): a² − ab + ab − b² = a² − b².

Tiles and excluded values · 4 practice marks

Question 7

Factor x² + 7x + 12 using a rectangular tile model, then simplify (x² + 7x + 12)/(x + 3). State the excluded value.

Show step-wise answer

Factors (x + 3)(x + 4); quotient x + 4 with x ≠ −3.

  1. Choose 3 and 4: their sum is 7 and their product is 12.
  2. For x ≥ 0, an x² tile, seven x tiles and twelve unit tiles form a rectangle with sides x + 3 and x + 4.
  3. Algebraically group x² + 3x + 4x + 12 = x(x + 3) + 4(x + 3).
  4. Cancel x + 3 only when it is non-zero. The original expression remains undefined at x = −3.