MYP Inquiry Framework
Key Concept Logic & Form
Related Concepts Equivalence, Generalization, Decomposition
Global Context Scientific and technical innovation (Algorithmic computing & polynomial cryptography)
Statement of Inquiry Decomposing complex polynomial expressions into linear and quadratic factors through deductive logic provides exact solutions to higher-degree systems.
7.1

Polynomial Anatomy & The Division Algorithm

  • Polynomial Definition: An expression of the form: $$P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 \quad (a_n \ne 0)$$ Where $n \in \mathbb{N}$ is the degree, $a_n$ is the leading coefficient, and $a_0$ is the constant term.
  • The Division Algorithm for Polynomials: When a dividend polynomial $P(x)$ is divided by a divisor polynomial $D(x)$ ($D(x) \ne 0$), there exist unique polynomials $Q(x)$ (the quotient) and $R(x)$ (the remainder) such that: $$\mathbf{P(x) = D(x) \cdot Q(x) + R(x)}$$ Where either $R(x) = 0$ or $\deg(R(x)) < \deg(D(x))$. In fractional form: $$\frac{P(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}$$
  • Degree Rule: If $\deg(P) = n$ and $\deg(D) = m$ ($n \ge m$), then $\deg(Q) = n - m$. If dividing by a linear divisor $D(x) = x - c$, the remainder $R$ is always a constant of degree $0$.
The Polynomial Division Theorem P(x) = D(x) · Q(x) + R Dividend Divisor Quotient Remainder If divisor is linear (x - c), remainder R is a constant: R = P(c)
7.2

The Remainder Theorem

  • Statement of the Theorem: When a polynomial $P(x)$ is divided by a linear divisor of the form $(x - c)$, the constant remainder is: $$\mathbf{R = P(c)}$$
  • Proof: By the division algorithm, $P(x) = (x - c)Q(x) + R$. Substituting $x = c$: $$P(c) = (c - c)Q(c) + R = 0 \cdot Q(c) + R = R \quad \blacksquare$$
  • Generalisation for $(ax - b)$: When $P(x)$ is divided by $(ax - b)$, the remainder is: $$R = P\left(\frac{b}{a}\right)$$
7.3

The Factor Theorem

  • Statement of the Theorem: A linear binomial $(x - c)$ is a factor of the polynomial $P(x)$ if and only if the remainder is zero: $$\mathbf{(x - c) \text{ is a factor of } P(x) \iff P(c) = 0}$$
  • Root and Factor Duality: If $x = c$ is a root (zero) of the equation $P(x) = 0$, then $(x - c)$ is a factor of $P(x)$, and the graph of $y = P(x)$ passes through $(c, 0)$.
  • Rational Root Theorem (Heuristic for Guessing Roots): For a polynomial $a_n x^n + \dots + a_0$ with integer coefficients, any rational root $x = \frac{p}{q}$ (in simplest form) must satisfy: $$p \text{ is a factor of the constant term } a_0, \quad \text{and } q \text{ is a factor of the leading coefficient } a_n$$ For monic polynomials ($a_n = 1$), test integer factors of the constant term $a_0$: $\pm 1, \pm 2, \pm 3, \dots$
7.4

Completely Factoring Cubics & Curve Sketching

  • Full 3-Step Cubic Factoring Algorithm:
    1. Find First Linear Factor: Use Factor Theorem to test small integer divisors of the constant term until $P(c) = 0$. This identifies factor $(x - c)$.
    2. Perform Polynomial Division: Divide $P(x)$ by $(x - c)$ using long division or synthetic division to obtain quadratic quotient $Q(x) = ax^2 + bx + d$.
    3. Decompose the Quadratic Factor: Factorise $Q(x)$ via $ac$ method or find remaining roots via quadratic formula: $$P(x) = (x - c)(x - r_2)(x - r_3)$$
  • Sketching Cubics: Identify real $x$-intercepts, $y$-intercept $(0, P(0))$, and end behaviour:
    • If $a > 0$: As $x \to \infty, y \to \infty$; as $x \to -\infty, y \to -\infty$ (starts bottom-left, finishes top-right).
    • If $a < 0$: Starts top-left, finishes bottom-right.
    • Repeated factor $(x - c)^2$: graph touches $x$-axis and turns around (tangent).

Worked Examples (Criterion A & C)

Example 1 Unknown Coefficients via Remainder Theorem

Problem: When $P(x) = 2x^3 + kx^2 - 5x + 6$ is divided by $(x - 2)$, the remainder is $16$. Find the value of constant $k$.

Step 1: Apply the Remainder Theorem

The remainder upon division by $(x - 2)$ is $P(2)$. Set $P(2) = 16$.

Step 2: Evaluate $P(2)$ $$P(2) = 2(2)^3 + k(2)^2 - 5(2) + 6 = 2(8) + 4k - 10 + 6 = 16 + 4k - 4 = 12 + 4k$$
Step 3: Solve for $k$ $$12 + 4k = 16 \implies 4k = 4 \implies \mathbf{k = 1}$$ Criterion A: Direct application of remainder theorem without costly long division
Example 2 Full Cubic Factorisation & Solving

Problem: Factorise completely and solve: $P(x) = x^3 - 4x^2 - 7x + 10 = 0$.

Step 1: Test factors of $10$ ($\pm 1, \pm 2, \pm 5, \pm 10$)

Test $x = 1$: $P(1) = (1)^3 - 4(1)^2 - 7(1) + 10 = 1 - 4 - 7 + 10 = 0$.

By the Factor Theorem, $\mathbf{(x - 1)}$ is a factor.

Step 2: Divide $P(x)$ by $(x - 1)$

Using polynomial long division:

$$(x^3 - 4x^2 - 7x + 10) \div (x - 1) = x^2 - 3x - 10$$
Step 3: Factorise the quadratic quotient $x^2 - 3x - 10$ $$x^2 - 3x - 10 = (x - 5)(x + 2)$$
Step 4: State complete factorisation and solutions $$P(x) = \mathbf{(x - 1)(x - 5)(x + 2)}$$

Roots to $P(x) = 0$: $\mathbf{x = 1, \quad x = 5, \quad x = -2}$.

Criterion C: Logical systematic workflow from root testing to complete factorisation

MYP Extended Practice Exercises: Polynomials

Tier 1 (Standard Division & Theorems), Tier 2 (Multi-Step Factoring), and Tier 3 (Extended Proofs & Simultaneous Theorems).

Question 1 Tier 1: Standard 3 Marks

Find the remainder when $P(x) = 3x^3 - 5x^2 + 2x - 7$ is divided by $(x - 2)$ without performing long division.

View Full Method Markscheme
[M1] Apply Remainder Theorem: $R = P(2)$.
[M1] Substitute $x = 2$: $P(2) = 3(2)^3 - 5(2)^2 + 2(2) - 7 = 3(8) - 5(4) + 4 - 7$.
[A1] Calculate: $24 - 20 + 4 - 7 = \mathbf{1}$.
Question 2 Tier 1: Standard 3 Marks

Use the Factor Theorem to determine whether $(x + 3)$ is a factor of $P(x) = x^3 + 2x^2 - 5x - 6$.

View Full Method Markscheme
[M1] Test $x = -3$: Evaluate $P(-3)$.
[M1] $P(-3) = (-3)^3 + 2(-3)^2 - 5(-3) - 6 = -27 + 2(9) + 15 - 6 = -27 + 18 + 15 - 6 = 0$.
[A1] Since $P(-3) = 0$, by the Factor Theorem, $\mathbf{(x + 3) \text{ is indeed a factor}}$.
Question 3 Tier 1: Standard 3 Marks

Perform polynomial division to find the quotient and remainder: $(2x^3 - 7x^2 + 9x - 5) \div (x - 2)$.

View Full Method Markscheme
[M1] Divide leading terms: $\frac{2x^3}{x} = 2x^2$. Subtract $2x^2(x - 2) = 2x^3 - 4x^2$, leaving $-3x^2 + 9x$.
[M1] Next term: $\frac{-3x^2}{x} = -3x$. Subtract $-3x(x - 2) = -3x^2 + 6x$, leaving $3x - 5$. Next term: $+3$. Subtract $3(x - 2) = 3x - 6$, leaving $+1$.
[A1] Quotient: $\mathbf{Q(x) = 2x^2 - 3x + 3}$, Remainder: $\mathbf{R = 1}$.
Question 4 Tier 2: Intermediate 4 Marks

Find the value of $a$ such that $(2x - 1)$ is a factor of $P(x) = 4x^3 + ax^2 - 5x + 2$.

View Full Method Markscheme
[M1] For $(2x - 1)$ to be a factor, $P\left(\frac{1}{2}\right) = 0$.
[M1] Substitute $x = \frac{1}{2}$: $$P\left(\frac{1}{2}\right) = 4\left(\frac{1}{8}\right) + a\left(\frac{1}{4}\right) - 5\left(\frac{1}{2}\right) + 2 = 0$$
[M1] Simplify: $\frac{1}{2} + \frac{a}{4} - \frac{5}{2} + 2 = 0 \implies \frac{a}{4} + (0) = 0$.
[A1] Thus $\frac{a}{4} = 0 \implies \mathbf{a = 0}$.
Question 5 Tier 2: Intermediate 4 Marks

Factorise completely: $P(x) = x^3 - 2x^2 - 11x + 12$.

View Full Method Markscheme
[M1] Test $x = 1$: $P(1) = 1 - 2 - 11 + 12 = 0 \implies (x - 1)$ is a factor.
[M1] Divide by $(x - 1)$: $x^3 - 2x^2 - 11x + 12 = (x - 1)(x^2 - x - 12)$.
[M1] Factorise quadratic $x^2 - x - 12$: factors of $-12$ summing to $-1$ are $-4$ and $+3$.
[A1] Full factorisation: $\mathbf{(x - 1)(x - 4)(x + 3)}$.
Question 6 Tier 2: Intermediate 4 Marks

When polynomial $P(x) = x^4 - 3x^2 + ax + b$ is divided by $(x - 1)$, the remainder is $4$. When divided by $(x + 2)$, the remainder is $-8$. Find the values of $a$ and $b$.

View Full Method Markscheme
[M1] From $(x - 1)$: $P(1) = 1 - 3 + a + b = 4 \implies a + b = 6$. (Eq 1)
[M1] From $(x + 2)$: $P(-2) = (-2)^4 - 3(-2)^2 + a(-2) + b = 16 - 12 - 2a + b = 4 - 2a + b = -8 \implies -2a + b = -12$. (Eq 2)
[M1] Subtract Eq 2 from Eq 1: $(a - (-2a)) + (b - b) = 6 - (-12) \implies 3a = 18 \implies a = 6$.
[A1] Find $b$: $6 + b = 6 \implies b = 0$. Solutions: $\mathbf{a = 6, \; b = 0}$.
Question 7 Tier 2: Intermediate 4 Marks

Find all real solutions to the cubic equation: $2x^3 + 3x^2 - 8x + 3 = 0$.

View Full Method Markscheme
[M1] Test $x = 1$: $2(1) + 3(1) - 8(1) + 3 = 0 \implies (x - 1)$ is a factor.
[M1] Divide by $(x - 1)$: $2x^3 + 3x^2 - 8x + 3 = (x - 1)(2x^2 + 5x - 3)$.
[M1] Factorise quadratic: $2x^2 + 5x - 3 = (2x - 1)(x + 3)$.
[A1] Roots are $\mathbf{x = 1, \quad x = \frac{1}{2}, \quad x = -3}$.
Question 8 Tier 3: Extended 5 Marks

A polynomial $P(x)$ leaves a remainder of $3$ when divided by $(x - 1)$, and a remainder of $7$ when divided by $(x - 3)$.
(a) State the maximum possible degree of the remainder when $P(x)$ is divided by $(x - 1)(x - 3)$.
(b) Determine this remainder in the form $R(x) = Ax + B$.

View Full Method Markscheme
[M1] Part (a): Since the divisor is quadratic ($\deg = 2$), the remainder must have degree at most $1$: $\mathbf{\deg(R) \le 1}$.
[M1] Write $P(x) = (x - 1)(x - 3)Q(x) + (Ax + B)$.
[M1] Apply $P(1) = 3$: $(1 - 1)(1 - 3)Q(1) + (A(1) + B) = 3 \implies A + B = 3$.
[M1] Apply $P(3) = 7$: $3A + B = 7$.
[A1] Subtract equations: $2A = 4 \implies A = 2$. Then $2 + B = 3 \implies B = 1$. The remainder is $\mathbf{R(x) = 2x + 1}$.
Question 9 Tier 3: Extended 5 Marks

Criterion B Investigation: Consider the polynomial $P_n(x) = x^n - 1$ where $n \in \mathbb{Z}^+$.
(a) Prove using the Factor Theorem that $(x - 1)$ is a factor of $x^n - 1$ for all integers $n \ge 1$.
(b) Prove that $(x + 1)$ is a factor of $x^n - 1$ if and only if $n$ is an even integer.

View Full Method Markscheme
[M1] Part (a): Substitute $x = 1$ into $P_n(x)$: $P_n(1) = 1^n - 1 = 1 - 1 = 0$ for any integer $n \ge 1$.
[A1] By the Factor Theorem, because $P_n(1) = 0$, $(x - 1)$ is always a factor for all $n \ge 1$. [Proved]
[M1] Part (b): By the Factor Theorem, $(x + 1)$ is a factor if and only if $P_n(-1) = 0$.
[M1] Evaluate: $P_n(-1) = (-1)^n - 1$.
[R1] If $n$ is even, $(-1)^n = +1$, so $1 - 1 = 0$, confirming $(x + 1)$ is a factor. If $n$ is odd, $(-1)^n = -1$, giving $-1 - 1 = -2 \ne 0$. Thus $(x + 1)$ is a factor if and only if $n$ is even. [Proved]
Question 10 Tier 3: Extended 5 Marks

A cubic polynomial $P(x)$ has a turning point at $(2, 0)$, a root at $x = -1$, and passes through the point $(0, 12)$.
(a) Write down the general factorised form of $P(x)$ using its roots and repeated root.
(b) Determine the equation of $P(x)$ in expanded polynomial form.

View Full Method Markscheme
[M1] Since the turning point lies on the $x$-axis at $(2, 0)$, the line $y = 0$ is tangent to the curve, meaning $(x - 2)$ is a repeated factor of multiplicity $2$.
[M1] The root at $x = -1$ gives factor $(x + 1)$. General form: $P(x) = a(x - 2)^2(x + 1)$.
[M1] Substitute point $(0, 12)$: $12 = a(0 - 2)^2(0 + 1) = a(4)(1) \implies 4a = 12 \implies a = 3$.
[M1] Expand: $P(x) = 3(x^2 - 4x + 4)(x + 1) = 3(x^3 + x^2 - 4x^2 - 4x + 4x + 4) = 3(x^3 - 3x^2 + 4)$.
[A1] Expanded polynomial: $\mathbf{P(x) = 3x^3 - 9x^2 + 12}$.

Extended Polynomial Formulas

  • Division Algorithm: $P(x) = D(x) \cdot Q(x) + R(x)$
  • Remainder Theorem: Remainder upon division by $(x - c)$ is $R = P(c)$
  • Generalised Remainder: Division by $(ax - b)$ yields $R = P\left(\dfrac{b}{a}\right)$
  • Factor Theorem: $(x - c) \text{ is a factor of } P(x) \iff P(c) = 0$
  • Rational Root Theorem: If $\frac{p}{q}$ is a root, $p \mid a_0$ and $q \mid a_n$
  • Sum of Cubes: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
  • Difference of Cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Top Common MYP Pitfalls

1. Sign Error in Divisor Substitution

When dividing by $(x + 4)$, evaluating $P(+4)$ instead of $P(-4)$. Setting $x + 4 = 0$ gives $x = -4$!

2. Missing Terms in Long Division

Dividing $x^3 - 7x + 6$ without inserting a $0x^2$ placeholder column. Always align terms by descending powers!

3. Subtraction Sign Error in Long Division

Forgetting to subtract the entire line during long division, turning $(-7x^2 - (-4x^2))$ into $-11x^2$ instead of $-3x^2$.

4. Not Testing Fractional Candidates

When factoring non-monic cubics like $2x^3 + \dots$, forgetting that roots can be fractions such as $\pm \frac{1}{2}$.

Interactive Mini Test: Polynomials (Extended)

Answer the questions to see your score and feedback.

Question 1 of 5 Remainder Theorem Calculation

What is the remainder when $P(x) = 2x^3 - 3x + 5$ is divided by $(x + 1)$?

Question 2 of 5 Factor Theorem Application

If $(x - 2)$ is a factor of $P(x) = x^3 - kx^2 + 4$, what is the value of $k$?

Question 3 of 5 Degree of Quotient

When a degree-5 polynomial is divided by a degree-2 polynomial, what is the degree of the quotient?

Question 4 of 5 Linear Divisor with Coefficient

To find the remainder when $P(x)$ is divided by $(3x - 2)$, which value must be substituted into $P(x)$?

Question 5 of 5 Cubic Curve Feature

If a cubic polynomial has a factor $(x - 4)^2$, how does its graph interact with the $x$-axis at $x = 4$?