Polynomials are the foundation of nearly every algebra question in IB Mathematics Analysis & Approaches HL. The HL extension over SL is significant — beyond curve sketching and quadratic formulas, you also need the Conjugate Root Theorem, Vieta's formulas, double-root conditions, polynomial division, and the symmetric-function identities for transformed roots. Together these account for two or three multi-mark questions on every Paper 1 and a regular feature in Section B.
What makes polynomials tricky is the trap density: a single sign error in Vieta's, a forgotten conjugate, or a constant remainder where a linear remainder was needed will collapse the rest of a six-mark question. This cheatsheet condenses every formula, method, and trap from SL 2.5–2.7 and AHL 2.12–2.16 into one page you can revise from.
§1 — Polynomial Structure AHL 2.12
Definition & notation
$P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0, \quad a_n \neq 0$
- Degree $= n$ (highest power); leading coefficient $= a_n$
- Constant term $= a_0$; monic iff $a_n = 1$
- Degree 1: linear; 2: quadratic; 3: cubic; 4: quartic
End behaviour (arms)
| Degree | $a_n > 0$ | $a_n < 0$ |
|---|---|---|
| Odd | Left down, right up | Left up, right down |
| Even | Both up | Both down |
Root multiplicity — graph behaviour at $x = a$
| Multiplicity | Graph at root | Factor |
|---|---|---|
| 1 (simple) | Crosses the axis | $(x-a)$ |
| 2 (double) | Touches and bounces — axis is tangent | $(x-a)^2$ |
| 3 (triple) | Crosses with an inflection (S-shape) | $(x-a)^3$ |
| $2k$ (even) | Always bounces — never crosses | $(x-a)^{2k}$ |
| $2k+1$ (odd) | Always crosses (flatter for higher mult.) | $(x-a)^{2k+1}$ |
§2 — Quartic Graph Cases AHL 2.12
$a_4 > 0$ — arms both up
The five canonical shapes are: 4 distinct real roots; one double root + 2 simple roots; two double roots; 2 real + 2 complex roots; no real roots (graph stays above the $x$-axis).
$a_4 < 0$ — arms both down
The four canonical shapes (mirror of the $a_4 > 0$ cases): 4 distinct real roots; one double root + 2 simple; 2 real + 2 complex; no real roots (graph stays below the $x$-axis).
§3 — Complex Roots in Polynomials AHL 1.14 / 2.12
Conjugate Root Theorem
If $P(x)$ has real coefficients and $\alpha = a + bi$ ($b \neq 0$) is a root, then $\bar\alpha = a - bi$ is also a root.
Real quadratic factor: $(x - \alpha)(x - \bar\alpha) = (x - a)^2 + b^2$.
Consequence: a real polynomial of odd degree always has at least one real root.
Quadratic discriminant
- $\Delta > 0$: two distinct real roots
- $\Delta = 0$: one repeated real root
- $\Delta < 0$: two complex conjugate roots: $x = \dfrac{-b \pm i\sqrt{4ac - b^2}}{2a}$
Using the Conjugate Root Theorem — 4-step method
Given a complex root $\alpha = a + bi$ of a real polynomial $P(x)$ of degree $n$:
- Write $\bar\alpha = a - bi$ immediately (second root).
- Form the real quadratic factor: $(x - a)^2 + b^2$.
- Divide $P(x)$ by this quadratic (long division or coefficient comparison).
- Factorise the remaining polynomial to find all other roots.
§4 — Remainder & Factor Theorems AHL 2.12
Remainder Theorem
Dividing by $(ax - b)$: substitute $x = b/a$. Key use: find an unknown coefficient given a remainder.
Factor Theorem
$(x - a)$ is a factor of $P(x) \iff P(a) = 0$. More generally, $(ax - b)$ is a factor $\iff P\!\left(\dfrac{b}{a}\right) = 0$.
Rational root candidates: $\dfrac{p}{q}$ where $p \mid a_0$ and $q \mid a_n$.
Double-root condition — when $(x - a)^2$ divides $P(x)$
$(x - a)^2$ divides $P(x)$ exactly $\iff$ both of these hold:
$$P(a) = 0 \qquad \text{AND} \qquad P'(a) = 0$$
This gives two simultaneous equations for two unknowns. Differentiate $P(x)$ directly — do not attempt long division first.
§5 — Polynomial Division AHL 2.12
Division algorithm
| Divisor | Remainder form | Method |
|---|---|---|
| Linear $(x - a)$ | Constant $R$ | Synthetic or long division |
| Quadratic | Linear $mx + c$ | Long division only |
| Cubic | Quadratic $ax^2 + bx + c$ | Long division only |
Critical: always insert $0x^k$ for missing powers before dividing.
Synthetic division (linear divisors only)
To divide by $(x - c)$: write the coefficients in a row, bring down $a_n$, multiply by $c$, add to the next coefficient, repeat. The last entry equals the remainder $R = P(c)$; the remaining entries are the quotient coefficients.
Quadratic divisor — remainder shortcut
Divide by $(x - a)(x - b)$; let $R = mx + c$. Then $P(a) = ma + c$ and $P(b) = mb + c$ — solve simultaneously.
Divide by $(x - a)^2$; let $R = mx + c$. Then $P(a) = ma + c$ and $P'(a) = m$.
§6 — Factoring Higher-Degree Polynomials AHL 2.12
Method selector
| Situation | Best method |
|---|---|
| No information given | Factor Theorem trial: $\pm p/q$ where $p \mid a_0$, $q \mid a_n$ |
| A linear factor is given | Divide immediately: synthetic division |
| A quadratic factor is given | Long division by the quadratic |
| Form of factorisation known | Coefficient comparison — faster than dividing |
| Complex root given | Write conjugate; form $(x - a)^2 + b^2$; divide |
| "Divisible by $(x - a)^2$" | $P(a) = 0$ and $P'(a) = 0$ simultaneously |
| Cubic — after one root found | Reduce to quadratic; then use formula or factorisation |
Over $\mathbb{R}$ vs over $\mathbb{C}$
Over $\mathbb{R}$: write as a product of linear and irreducible quadratic factors. $(x^2 + bx + c)$ is irreducible when $\Delta < 0$.
Over $\mathbb{C}$: every factor is linear. A degree $n$ polynomial $=$ exactly $n$ linear factors. Irreducible quadratics split over $\mathbb{C}$ into conjugate linear factors.
Grouping shortcut
Look for a common-factor structure: $ax^3 + bx^2 + ax + b = x^2(ax + b) + (ax + b) = (ax + b)(x^2 + 1)$. This works when the middle terms share a ratio with the outer terms — saves time on Paper 1 cubics.
§7 — Vieta's Formulas (Sum & Product of Roots) AHL 2.12
General Vieta's (formula booklet)
For $a_n x^n + \cdots + a_0 = 0$ with roots $r_1, \ldots, r_n$:
Quadratic and cubic forms
Quadratic $ax^2 + bx + c = 0$ with roots $\alpha, \beta$: $\alpha + \beta = -b/a$, $\alpha\beta = c/a$.
Cubic $ax^3 + bx^2 + cx + d = 0$ with roots $\alpha, \beta, \gamma$:
- $\alpha + \beta + \gamma = -b/a$
- $\alpha\beta + \beta\gamma + \gamma\alpha = c/a$
- $\alpha\beta\gamma = -d/a$
Symmetric function identities — derive without solving
Let $S = \alpha + \beta$ and $P = \alpha\beta$. Then:
- $\alpha^2 + \beta^2 = S^2 - 2P$
- $(\alpha - \beta)^2 = S^2 - 4P$
- $\alpha^3 + \beta^3 = S^3 - 3SP$
- $\alpha^2\beta + \alpha\beta^2 = SP$
- $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{S}{P}, \quad \dfrac{1}{\alpha}\cdot\dfrac{1}{\beta} = \dfrac{1}{P}$
§8 — Transformations & Root Analysis SL 2.11 / AHL 2.12
Effect of transformations on roots
| Transformation | New roots | Vieta impact |
|---|---|---|
| $g(x) = f(x - k)$ | $\alpha_i + k$ | New sum $=$ old sum $+ nk$ |
| $g(x) = f(x + k)$ | $\alpha_i - k$ | New sum $=$ old sum $- nk$ |
| $g(x) = f(kx)$, $k > 0$ | $\alpha_i / k$ | New sum $=$ old sum$/k$; new product $=$ old$/k^n$ |
| $g(x) = c\,f(x)$ | $\alpha_i$ unchanged | Only leading coeff changes |
| $g(x) = f(ax + b)$ | $(\alpha_i - b)/a$ | New sum $= (\text{old sum} - nb)/a$; new product $=$ old$/a^n$ |
| $g(x) = f(x) + c$ | Must re-solve $f(x) = -c$ | No shortcut |
Master substitution trick
To find the roots of $g(x) = f(ax + b)$: set $ax + b = \alpha_i$ for each original root $\alpha_i$, so $x = \dfrac{\alpha_i - b}{a}$. Each original root $\to$ one new root. No expansion needed.
Leading coefficient warning
If $f(x) = a_n x^n + \cdots$ and $g(x) = f(kx)$, then the leading coefficient of $g$ is $a_n \cdot k^n$, not $a_n$. This affects all Vieta calculations for $g(x)$. Always recompute $a_n$ for the transformed polynomial.
§9 — Special Root Structures AHL 2.12
Roots in arithmetic progression
Let the roots of a cubic be $a - d, a, a + d$. Then sum $= 3a = -b/a_3$, so $a = -\dfrac{b}{3a_3}$ — the middle root is always $-b/(3a_3)$, a root you get for free. Product $= a(a^2 - d^2) = -d_{\text{coeff}}/a_3$; use this to find $d$, then check the discriminant of the remaining quadratic for real/complex roots.
Roots in geometric progression
Let the roots of a cubic be $a/r, a, ar$. Then product $= a^3 = -d/a_3$, so the middle root is $\sqrt[3]{-d/a_3}$. Sum $= a(1/r + 1 + r) = -b/a_3$.
Other special structures
| Structure | What to write |
|---|---|
| One root is $k$ times another: $\beta = k\alpha$ | Vieta sum: $\alpha(1 + k) = -b/a$; product: $k\alpha^2 = c/a$. Divide. |
| Roots sum to zero | Coefficient of $x^{n-1}$ is zero: $a_{n-1} = 0$ |
| Roots are reciprocals | $\alpha \cdot (1/\alpha) = 1$, so product $= 1$, so $a_0 = (-1)^n a_n$ |
| One root is negative of another: $\beta = -\alpha$ | Product of that pair $= -\alpha^2 < 0$; use Vieta sign |
§10 — Exam Attack Plan All sections
When you see this in the question — reach for that:
| Question trigger | Reach for |
|---|---|
| "Remainder when $P(x) \div (x - a)$" | Compute $P(a)$ directly — no long division |
| "$(x - a)$ is a factor" | Write $P(a) = 0$ and solve |
| "Find all roots of cubic" | Factor Theorem trial $\to$ divide out $\to$ quadratic formula |
| "Exactly divisible by $(x - a)^2$" | $P(a) = 0$ AND $P'(a) = 0$ — two equations |
| "$P(x) \div (x - a)(x - b)$, find remainder" | Let $R = mx + c$; use $P(a)$ and $P(b)$ |
| "Complex root $a + bi$ given" | Write conjugate; form $(x - a)^2 + b^2$; divide |
| "Sum / product of roots" | Vieta's formulas — never solve for individual roots |
| "New equation with roots $f(\alpha), f(\beta)$" | Find new sum and product using symmetric identities |
| "Roots in AP" | Let $\{a - d, a, a + d\}$; sum gives $a$ immediately |
| "Roots in GP" | Let $\{a/r, a, ar\}$; product gives $a$ immediately |
| "$g(x) = f(ax + b)$, find roots" | Each new root $= (\alpha_i - b)/a$ |
| "Both roots positive" | Require $\alpha\beta > 0$ AND $\alpha + \beta > 0$ |
| "Show $P(x)$ has no real roots" | Prove discriminant $< 0$, or prove $P(x) > 0$ always |
| "Factorise over $\mathbb{R}$" | Keep irreducible quadratics; don't split into complex |
| "Factorise over $\mathbb{C}$" | Split everything into linear factors |
Worked Example — IB-Style Polynomial with a Complex Root
Question (HL Paper 1 style — 8 marks)
The polynomial $P(x) = 2x^4 - 3x^3 + ax^2 + bx + 10$ has $1 - 2i$ as a root, where $a, b \in \mathbb{R}$. Find the values of $a$ and $b$, and hence find all four roots of $P(x)$.
Solution
- Since $P$ has real coefficients and $1 - 2i$ is a root, by the Conjugate Root Theorem $1 + 2i$ is also a root. (R1)
- Form the real quadratic factor: $(x - 1)^2 + 2^2 = x^2 - 2x + 5$. (M1)(A1)
- Divide: $P(x) = (x^2 - 2x + 5)(2x^2 + cx + d)$ for some $c, d \in \mathbb{R}$. Expand and compare coefficients.
$x^3$ coeff: $-4 + c = -3 \Rightarrow c = 1$.
Constant: $5d = 10 \Rightarrow d = 2$. (M1)(A1) - Use the remaining coefficient comparisons to read off $a$ and $b$.
$x^2$ coeff: $10 - 2c + d = a$, so $a = 10 - 2 + 2 = 10$.
$x$ coeff: $5c - 2d = b$, so $b = 5 - 4 = 1$. (A1) - Solve $2x^2 + x + 2 = 0$: $\Delta = 1 - 16 = -15 < 0$, so $x = \dfrac{-1 \pm i\sqrt{15}}{4}$. (M1)
- The four roots are $1 + 2i,\; 1 - 2i,\; \dfrac{-1 + i\sqrt{15}}{4},\; \dfrac{-1 - i\sqrt{15}}{4}$. (A1)
Examiner's note: The two most common errors here are (a) writing the conjugate factor as $(x - 1)^2 - 4$ (wrong sign on $b^2$) and (b) forgetting that $P$ is degree 4, so there must be exactly four roots. If you only find two, you have stopped halfway and lost the final 3 marks.