MYP Inquiry Framework
Key Concept Form & Equivalence
Related Concepts Representation, Simplification, Patterns
Global Context Identities and relationships (Structural balance & optimization)
Statement of Inquiry Rewriting an algebraic expression reveals equivalent forms and can simplify a problem.
1.1

Algebraic Terminology & Expanding Brackets

  • Anatomy of an Expression: A term consists of a numerical coefficient multiplied by one or more variables raised to non-negative integer powers (e.g., in $-7x^2 y$, the coefficient is $-7$, variables are $x$ and $y$, degree is $2+1=3$). Constant terms have no variable component.
  • The Distributive Law: Multiplication distributes over addition and subtraction: $$a(b + c) = ab + ac \quad \text{and} \quad a(b - c) = ab - ac$$
  • Expanding Binomial Products (FOIL): $$(a + b)(c + d) = \underbrace{ac}_{\text{First}} + \underbrace{ad}_{\text{Outer}} + \underbrace{bc}_{\text{Inner}} + \underbrace{bd}_{\text{Last}}$$
  • Grid (Area) Method: Visualising expansion as finding the total area of a rectangle partitioned into sub-rectangles with side lengths $(a + b)$ and $(c + d)$.
  • Expanding Higher Products: When multiplying three or more binomials $(x+a)(x+b)(x+c)$, first expand $(x+a)(x+b) = x^2 + (a+b)x + ab$, then distribute the third factor $(x+c)$ across every term of the resulting trinomial.
c d a b ac ad bc bd Total Area: (a + b)(c + d) = ac + ad + bc + bd
1.2

Special Products: Difference of Squares & Perfect Squares

  • Difference of Two Squares (DOTS Expansion): $$(a - b)(a + b) = a^2 - b^2$$ The two cross-terms cancel out: $+ab - ab = 0$.
  • Perfect Square Expansions: $$(a + b)^2 = a^2 + 2ab + b^2$$ $$(a - b)^2 = a^2 - 2ab + b^2$$ Warning: $(a \pm b)^2 \ne a^2 \pm b^2$. The middle term $\pm 2ab$ must never be omitted!
  • Square of a Trinomial (Extended): $$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$$
  • Cube of a Binomial (Extended): $$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$ $$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$$
1.3

Factorisation Hierarchy: HCF, Grouping & Trinomials

  • Factorisation rule: Always check for the Highest Common Factor (HCF) first before attempting other methods (e.g., in $6x^3 - 24x$, extract $6x(x^2 - 4) = 6x(x-2)(x+2)$).
  • Factorising by Grouping in Pairs (4 Terms): Pair terms to extract common binomial factors: $$ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y)$$ Watch sign flips when grouping negative terms: $2x^3 - 3x^2 - 8x + 12 = x^2(2x - 3) - 4(2x - 3) = (x^2 - 4)(2x - 3) = (x-2)(x+2)(2x-3)$.
  • Factoring Monic Trinomials ($x^2 + bx + c$): Find two integers $p$ and $q$ such that: $$p + q = b \quad \text{and} \quad p \cdot q = c$$ Then: $x^2 + bx + c = (x + p)(x + q)$.
  • Factoring Non-Monic Trinomials ($ax^2 + bx + c, \, a \ne 1$) via the "$ac$ Method":
    1. Calculate product $P = a \cdot c$ and sum $S = b$.
    2. Find two numbers $p, q$ such that $p \cdot q = ac$ and $p + q = b$.
    3. Split the linear term: $ax^2 + px + qx + c$.
    4. Factorise by grouping in pairs.
  • Sum and Difference of Cubes (Extended): $$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$ $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$ Mnemonic (SOAP): Same sign, Opposite sign, Always Positive.
1.4

Algebraic Fractions: Simplification & Operations

  • Simplifying Algebraic Fractions: Factorise both numerator and denominator completely, then cancel common factors: $$\frac{x^2 - 9}{x^2 + 5x + 6} = \frac{(x-3)(x+3)}{(x+2)(x+3)} = \frac{x-3}{x+2} \quad (x \ne -2, -3)$$ Rule: You can only cancel factors (multiplicative), never individual additive terms!
  • Multiplication & Division: $$\frac{A}{B} \times \frac{C}{D} = \frac{AC}{BD}, \qquad \frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \times \frac{D}{C} = \frac{AD}{BC}$$
  • Addition & Subtraction with Lowest Common Denominator (LCD): Find the LCD of algebraic denominators, convert each fraction to equivalent forms, and combine numerators carefully distributing negative signs: $$\frac{2}{x-1} - \frac{3}{x+2} = \frac{2(x+2) - 3(x-1)}{(x-1)(x+2)} = \frac{2x+4 - 3x + 3}{(x-1)(x+2)} = \frac{7 - x}{(x-1)(x+2)}$$
  • Restriction Domains: Values of variables that cause any denominator in the original or simplified expression to equal zero are strictly undefined (e.g., $x \ne 1, x \ne -2$).

Worked Examples (Criterion A & C)

Example 1 Non-Monic Trinomial Factoring with HCF

Problem: Fully factorise the expression: $12x^3 - 22x^2 - 20x$.

Step 1: Extract the Highest Common Factor (HCF)

The coefficients $12, -22, -20$ share a common factor of $2$, and all terms share $x$:

$$12x^3 - 22x^2 - 20x = 2x(6x^2 - 11x - 10)$$
Step 2: Factorise the Quadratic $6x^2 - 11x - 10$ using the $ac$ method

Here $a = 6$, $b = -11$, $c = -10$. Compute product $ac = 6 \times (-10) = -60$.

Find two numbers with product $-60$ and sum $-11$: These are $-15$ and $+4$, since $(-15)(4) = -60$ and $-15 + 4 = -11$.

Step 3: Split the middle term and factor by grouping $$6x^2 - 15x + 4x - 10 = 3x(2x - 5) + 2(2x - 5) = (3x + 2)(2x - 5)$$
Step 4: Combine with the initial factor $$\mathbf{12x^3 - 22x^2 - 20x = 2x(3x + 2)(2x - 5)}$$ Criterion A: Full factorization achieved with method accuracy
Example 2 Algebraic Fraction Addition with Sign Inversion

Problem: Simplify into a single simplified fraction: $\dfrac{3}{x^2 - 4} - \dfrac{2}{2 - x}$.

Step 1: Recognise sign inversion in the denominator

Note that $(2 - x) = -(x - 2)$. Therefore:

$$-\frac{2}{2 - x} = -\frac{2}{-(x - 2)} = +\frac{2}{x - 2}$$
Step 2: Factorise denominators to determine LCD

$x^2 - 4 = (x - 2)(x + 2)$ by difference of squares. The LCD of $(x-2)(x+2)$ and $(x-2)$ is $(x-2)(x+2)$.

Step 3: Combine numerators over the common denominator $$\frac{3}{(x-2)(x+2)} + \frac{2(x+2)}{(x-2)(x+2)} = \frac{3 + 2x + 4}{(x-2)(x+2)} = \frac{2x + 7}{(x-2)(x+2)}$$

Restrictions: $x \ne 2$ and $x \ne -2$.

Criterion C: Clear mathematical notation with stated domain restrictions

MYP 4–5 Practice Exercises: Algebraic Expressions

Tier 1 (Core skills), Tier 2 (Intermediate Reasoning), and Tier 3 (Extended & Criterion B/D Investigations). Click each question to reveal step-by-step method-marked solutions.

Question 1 Tier 1: Standard 3 Marks

Expand and simplify the algebraic expression: $4(2x - 3) - 3(x^2 - 5x + 2)$.

View Full Method Markscheme
[M1] Distribute both brackets carefully: $8x - 12 - 3x^2 + 15x - 6$ (mind the double negative: $-3 \times (-5x) = +15x$).
[M1] Group like terms by descending powers: $-3x^2 + (8x + 15x) + (-12 - 6)$.
[A1] Final simplified expression: $\mathbf{-3x^2 + 23x - 18}$.
Question 2 Tier 1: Standard 3 Marks

Expand and simplify the product: $(3x - 5)(2x + 7)$.

View Full Method Markscheme
[M1] Apply FOIL distribution: $(3x)(2x) + (3x)(7) + (-5)(2x) + (-5)(7)$.
[A1] Intermediate terms: $6x^2 + 21x - 10x - 35$.
[A1] Collect middle linear terms: $\mathbf{6x^2 + 11x - 35}$.
Question 3 Tier 1: Standard 3 Marks

Factorise completely: $50x^3 - 72xy^2$.

View Full Method Markscheme
[M1] Extract highest common factor $2x$: $2x(25x^2 - 36y^2)$.
[M1] Identify difference of two squares: $25x^2 = (5x)^2$ and $36y^2 = (6y)^2$.
[A1] Fully factored form: $\mathbf{2x(5x - 6y)(5x + 6y)}$.
Question 4 Tier 2: Intermediate 3 Marks

Factorise the quadratic trinomial: $x^2 - 13x + 42$.

View Full Method Markscheme
[M1] Identify product $c = 42$ and sum $b = -13$. Both factors must be negative.
[M1] Test pairs of factors of $42$: $(-1, -42)$, $(-2, -21)$, $(-3, -14)$, $(-6, -7)$. Notice $(-6) + (-7) = -13$.
[A1] Factored expression: $\mathbf{(x - 6)(x - 7)}$.
Question 5 Tier 2: Intermediate 4 Marks

Factorise completely by grouping in pairs: $3x^3 - 5x^2 - 12x + 20$.

View Full Method Markscheme
[M1] Group first two and last two terms: $(3x^3 - 5x^2) - (12x - 20)$.
[M1] Extract common monomials: $x^2(3x - 5) - 4(3x - 5)$.
[A1] Factor out binomial $(3x - 5)$: $(x^2 - 4)(3x - 5)$.
[A1] Decompose $(x^2 - 4)$ as difference of squares: $\mathbf{(x - 2)(x + 2)(3x - 5)}$.
Question 6 Tier 2: Intermediate 4 Marks

Factorise the non-monic quadratic trinomial: $8x^2 + 10x - 3$.

View Full Method Markscheme
[M1] Calculate product $ac = 8 \times (-3) = -24$ and sum $b = 10$.
[M1] Identify factor pair: $+12$ and $-2$ since $12 \times (-2) = -24$ and $12 - 2 = 10$.
[M1] Split linear term and factor by grouping: $8x^2 + 12x - 2x - 3 = 4x(2x + 3) - 1(2x + 3)$.
[A1] Factored form: $\mathbf{(4x - 1)(2x + 3)}$.
Question 7 Tier 2: Intermediate 4 Marks

Simplify the algebraic fraction completely: $\dfrac{2x^2 - 8}{x^2 + 5x + 6} \times \dfrac{x^2 - 9}{4x - 8}$.

View Full Method Markscheme
[M1] Factorise all 4 components:
  • $2x^2 - 8 = 2(x^2 - 4) = 2(x - 2)(x + 2)$
  • $x^2 + 5x + 6 = (x + 2)(x + 3)$
  • $x^2 - 9 = (x - 3)(x + 3)$
  • $4x - 8 = 4(x - 2)$
[M1] Express as single product before cancellation: $$\frac{2(x-2)(x+2) \cdot (x-3)(x+3)}{(x+2)(x+3) \cdot 4(x-2)}$$
[M1] Cancel common factors: $(x-2)$, $(x+2)$, and $(x+3)$, and simplify $\frac{2}{4} = \frac{1}{2}$.
[A1] Result: $\mathbf{\dfrac{x - 3}{2}}$ with restrictions $x \ne -3, -2, 2$.
Question 8 Tier 3: Extended 5 Marks

Express as a single simplified fraction: $\dfrac{5}{x^2 - x - 6} - \dfrac{2}{x^2 - 4} + \dfrac{1}{x+2}$.

View Full Method Markscheme
[M1] Factorise each denominator: $$x^2 - x - 6 = (x - 3)(x + 2), \quad x^2 - 4 = (x - 2)(x + 2)$$
[M1] Determine Lowest Common Denominator (LCD): $\text{LCD} = (x - 3)(x - 2)(x + 2)$.
[M1] Convert all terms to LCD: $$\frac{5(x - 2) - 2(x - 3) + 1(x - 3)(x - 2)}{(x - 3)(x - 2)(x + 2)}$$
[M1] Expand numerator: $$(5x - 10) - (2x - 6) + (x^2 - 5x + 6) = 5x - 10 - 2x + 6 + x^2 - 5x + 6 = x^2 - 2x + 2$$
[A1] Since $x^2 - 2x + 2$ has discriminant $\Delta = 4 - 8 = -4 < 0$, it has no real factors. Final answer: $$\mathbf{\frac{x^2 - 2x + 2}{(x - 3)(x - 2)(x + 2)}} \quad (x \ne -2, 2, 3)$$
Question 9 Tier 3: Extended 5 Marks

Criterion B Investigation: Consider the expression $N = (k+1)^4 - (k-1)^4$ for any integer $k \ge 1$.
(a) By letting $u = (k+1)^2$ and $v = (k-1)^2$, use the difference of two squares to prove that $N = 8k(k^2 + 1)$.
(b) Deduce that $N$ is always divisible by $16$ for every integer $k \ge 1$.

View Full Method Markscheme
[M1] Write $N = u^2 - v^2 = (u - v)(u + v)$ where $u = (k+1)^2$ and $v = (k-1)^2$.
[M1] Evaluate $(u - v)$: $$(k+1)^2 - (k-1)^2 = (k^2 + 2k + 1) - (k^2 - 2k + 1) = 4k$$
[M1] Evaluate $(u + v)$: $$(k+1)^2 + (k-1)^2 = (k^2 + 2k + 1) + (k^2 - 2k + 1) = 2k^2 + 2 = 2(k^2 + 1)$$
[A1] Multiply the two parts: $N = (4k) \times 2(k^2 + 1) = \mathbf{8k(k^2 + 1)}$. [Proved]
[R1] For part (b): If $k$ is even, $k = 2m$, so $8k = 16m$, making $N$ divisible by $16$. If $k$ is odd, $k^2$ is odd, so $(k^2 + 1)$ is even, meaning $(k^2 + 1) = 2p$. Then $N = 8k(2p) = 16kp$, which is also divisible by $16$. Thus $N$ is divisible by $16$ for all $k \ge 1$.
Question 10 Tier 3: Extended 5 Marks

Factorise the cubic expression using the sum/difference of cubes identity: $27x^3 - 64y^6$.

View Full Method Markscheme
[M1] Identify cubic terms: $27x^3 = (3x)^3$ and $64y^6 = (4y^2)^3$.
[M1] Recall difference of cubes identity: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.
[M1] Substitute $a = 3x$ and $b = 4y^2$:
  • $a - b = 3x - 4y^2$
  • $a^2 = (3x)^2 = 9x^2$
  • $ab = (3x)(4y^2) = 12xy^2$
  • $b^2 = (4y^2)^2 = 16y^4$
[A1] Complete factorised form: $\mathbf{(3x - 4y^2)(9x^2 + 12xy^2 + 16y^4)}$.

Essential Algebraic Identities

  • FOIL Expansion: $(a+b)(c+d) = ac + ad + bc + bd$
  • Difference of Squares: $(a-b)(a+b) = a^2 - b^2$
  • Perfect Square Expansions: $(a \pm b)^2 = a^2 \pm 2ab + b^2$
  • Square of Trinomial: $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
  • Sum of Cubes: $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
  • Difference of Cubes: $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$
  • Monic Factoring: $x^2 + (p+q)x + pq = (x+p)(x+q)$
  • $ac$ Method: For $ax^2 + bx + c$, find $p, q$ where $p \cdot q = ac$ and $p + q = b$, then group.

Top Common MYP Pitfalls

1. The "Freshman's Dream" Error

Writing $(a + b)^2 = a^2 + b^2$. Always remember the middle cross term: $(a + b)^2 = a^2 + 2ab + b^2$.

2. Illegal Fraction Cancellation

Cancelling terms across plus/minus signs, such as simplifying $\frac{x + 5}{5}$ to $x + 1$. You can ONLY cancel factors when the entire numerator and denominator are multiplicative products!

3. Missing the Negative Sign Distribution

In subtractions like $-(3x - 7)$, writing $-3x - 7$ instead of $-3x + 7$. In fractions: $\frac{A}{B} - \frac{C + D}{E} = \frac{AE - (C+D)B}{BE}$. The brackets around $(C+D)$ are vital.

4. Forgetting to Extract HCF First

Attempting quadratic formula or heavy factoring before taking out a common constant or variable. Always pull out the HCF first!

Interactive Mini Test: Algebraic Expressions

Answer the questions to see your score and feedback.

Question 1 of 5 Expansion of Perfect Square

Which expression is equivalent to $(3x - 4y)^2$?

Question 2 of 5 Factoring by Difference of Squares

What is the complete factorisation of $18x^4 - 32$?

Question 3 of 5 Non-Monic Trinomial Factoring

Which of the following is the factorised form of $6x^2 + 7x - 5$?

Question 4 of 5 Algebraic Fraction Simplification

Simplify the algebraic fraction: $\dfrac{x^2 - 25}{2x^2 + 10x}$.

Question 5 of 5 Sum of Cubes Factorisation

Which expression represents the complete factorisation of $x^3 + 8$?