Polynomials

A polynomial in $x$ is an expression of the form $a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$ , where the coefficients $a_i$ are real numbers and $n$ is a non-negative integer. The non-negative integer condition matters: $x^{1/2}$ and $x^{-1}$ are not polynomial terms, so $\sqrt{x} + 3$ and $1/x$ are not polynomials. This lesson develops the algebra of polynomials at A-Level depth — arithmetic, division (with and without remainder), end-behaviour, multiplicity of roots, sketching, and the rational root theorem — preparing the ground for the factor and remainder theorems in lesson 6 and the algebraic-fraction work in lesson 7.

Spec mapping

Edexcel 9MA0 specification, Pure Mathematics Paper 1 — Section 2.4 / 2.5 covers manipulation of polynomials algebraically, including expanding brackets and collecting like terms; simplification of rational expressions including by factorising and cancelling, and algebraic division (by linear expressions only) (refer to the official specification document for exact wording). Polynomials are foundational: they appear directly in 9MA0-01 Paper 1 (algebraic division, factorisation, sketching), in 9MA0-02 Paper 2 (turning points via differentiation of cubics and quartics), and indirectly in 9MA0-03 Paper 3 Statistics & Mechanics (polynomial regression in S2 contexts; quartic energy functions in M2 contexts). The Edexcel formula booklet does not list the binomial coefficients used in polynomial expansions or the rational root theorem — both must be carried in your head.

Terminology and core vocabulary

A polynomial $p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$ has the following parts:

Degree $n$ — the highest power of $x$ with non-zero coefficient.
Leading coefficient $a_n$ — the coefficient of the highest-degree term. The sign of $a_n$ controls the long-term behaviour as $x \to \pm \infty$ .
Constant term $a_0$ — the value of $p(0)$ , which is the $y$ -intercept of the curve $y = p(x)$ .
Monic polynomial — one whose leading coefficient is $1$ . Many factorisation results assume monic form because non-monic leading coefficients shift the rational-root analysis.
Root (or zero) — any value $\alpha$ such that $p(\alpha) = 0$ . By the factor theorem (lesson 6), $\alpha$ is a root iff $(x - \alpha)$ is a factor.

The first few degrees have standard names: degree 1 is linear, degree 2 is quadratic (lesson 2), degree 3 is cubic, degree 4 is quartic, degree 5 is quintic. The Abel–Ruffini theorem (1824) — first proved by Niels Henrik Abel after Paolo Ruffini's earlier incomplete attempt — establishes that the general quintic has no solution by radicals; we revisit this in the going-further section.

Polynomial arithmetic

Polynomials behave under addition, subtraction and multiplication just like ordinary numbers, except you must collect like terms by power of $x$ .

Addition / subtraction. Add coefficients of the same power. If $p(x) = 2x^3 - x + 4$ and $q(x) = -x^3 + 2x^2 + 5$ , then

$p(x) + q(x) = (2 - 1)x^3 + 2x^2 - x + (4 + 5) = x^3 + 2x^2 - x + 9.$

Multiplication. Apply the distributive law and collect:

$(x^2 + 2x - 1)(x - 3) = x^3 - 3x^2 + 2x^2 - 6x - x + 3 = x^3 - x^2 - 7x + 3.$

A useful identity check: the degree of a product equals the sum of degrees, $\deg(pq) = \deg(p) + \deg(q)$ . If you multiply a cubic by a quadratic and your answer is degree 4, you have lost a term — go back and check.

Substitution. Polynomial values can be evaluated at any real number; e.g. $p(2) = 2(8) - 2 + 4 = 18$ for $p(x) = 2x^3 - x + 4$ . Substitution is the engine behind both the factor theorem ( $p(\alpha) = 0 \iff (x - \alpha) \mid p(x)$ ) and the rational root theorem (which restricts which $\alpha$ to test).

Polynomial long division

For 9MA0 you must be confident dividing a polynomial by a linear divisor $(x - c)$ — the spec is explicit on this restriction. The result is a quotient $q(x)$ and a remainder $r$ (a constant when the divisor is linear) such that

$p(x) = (x - c)\, q(x) + r.$

When $r = 0$ the divisor is a factor.

Worked example 1 — long division with zero remainder

Divide $p(x) = x^3 - 4x^2 + x + 6$ by $(x - 2)$ .

Step 1. Confirm the division is exact by checking $p(2)$ :

$p(2) = 8 - 16 + 2 + 6 = 0.$

So $(x - 2)$ is a factor and the remainder will be $0$ .

Step 2. Divide leading terms: $x^3 \div x = x^2$ . Multiply back: $x^2 (x - 2) = x^3 - 2x^2$ . Subtract:

$(x^3 - 4x^2) - (x^3 - 2x^2) = -2x^2.$

Bring down the next term to give $-2x^2 + x$ .

Step 3. Divide: $-2x^2 \div x = -2x$ . Multiply back: $-2x(x - 2) = -2x^2 + 4x$ . Subtract:

$(-2x^2 + x) - (-2x^2 + 4x) = -3x.$

Bring down the $+6$ to give $-3x + 6$ .

Step 4. Divide: $-3x \div x = -3$ . Multiply back: $-3(x - 2) = -3x + 6$ . Subtract:

$(-3x + 6) - (-3x + 6) = 0.$

Quotient $q(x) = x^2 - 2x - 3$ , remainder $0$ .

Step 5. Factorise the quadratic quotient: $x^2 - 2x - 3 = (x - 3)(x + 1)$ .

Conclusion:

$x^3 - 4x^2 + x + 6 = (x - 2)(x - 3)(x + 1).$

Roots are $x = 2,\ 3,\ -1$ . Verify by substitution: $p(3) = 27 - 36 + 3 + 6 = 0$ ✓, $p(-1) = -1 - 4 - 1 + 6 = 0$ ✓.

Worked example 2 — long division with non-zero remainder

Divide $p(x) = 2x^3 + 3x^2 - 5x + 4$ by $(x + 2)$ .

Step 1. Test $p(-2)$ :

$p(-2) = 2(-8) + 3(4) - 5(-2) + 4 = -16 + 12 + 10 + 4 = 10.$

By the remainder theorem (lesson 6) the remainder will be $10$ .

Step 2. Divide $2x^3 \div x = 2x^2$ ; multiply: $2x^2(x + 2) = 2x^3 + 4x^2$ ; subtract: $(2x^3 + 3x^2) - (2x^3 + 4x^2) = -x^2$ . Bring down: $-x^2 - 5x$ .

Step 3. Divide $-x^2 \div x = -x$ ; multiply: $-x(x + 2) = -x^2 - 2x$ ; subtract: $(-x^2 - 5x) - (-x^2 - 2x) = -3x$ . Bring down: $-3x + 4$ .

Step 4. Divide $-3x \div x = -3$ ; multiply: $-3(x + 2) = -3x - 6$ ; subtract: $(-3x + 4) - (-3x - 6) = 10$ .

Quotient $q(x) = 2x^2 - x - 3$ , remainder $10$ . We can write the result either as

$2x^3 + 3x^2 - 5x + 4 = (x + 2)(2x^2 - x - 3) + 10$

or, dividing through, as

$\dfrac{2x^3 + 3x^2 - 5x + 4}{x + 2} = 2x^2 - x - 3 + \dfrac{10}{x + 2}.$

The fact that the remainder agrees with $p(-2)$ is the remainder theorem, which lesson 6 develops in full.

Exam tip. When 9MA0 asks "divide $p(x)$ by $(x - c)$ and state the quotient and remainder", the most efficient check is to evaluate $p(c)$ : if you get $0$ you know the division is exact, and if not you have the remainder before doing any algebra. Use this every time as a self-check.

End behaviour: sign × parity

The behaviour of $y = p(x)$ as $x \to \pm \infty$ depends on only two things: the sign of the leading coefficient $a_n$ and the parity of the degree $n$ (odd or even). For very large $|x|$ the leading term dominates all others.

Leading coefficient	Degree parity	As $x \to -\infty$	As $x \to +\infty$	Typical shape
$a_n > 0$	even	$+\infty$	$+\infty$	"U-shape" — both arms up
$a_n > 0$	odd	$-\infty$	$+\infty$	rises left to right
$a_n < 0$	even	$-\infty$	$-\infty$	"∩-shape" — both arms down
$a_n < 0$	odd	$+\infty$	$-\infty$	falls left to right

This four-row table is the first thing to write down when sketching any polynomial. It pins the orientation of the curve before you worry about turning points or roots.

Root multiplicity and graphical implications

If $(x - \alpha)^k$ divides $p(x)$ but $(x - \alpha)^{k+1}$ does not, then $\alpha$ is a root of multiplicity $k$ . Multiplicity controls the shape of the curve at the $x$ -axis crossing.

Multiplicity $k$	Name	Behaviour at $x = \alpha$
1	simple root	curve crosses the $x$ -axis transversally
2	double root	curve touches the $x$ -axis (turning point on the axis)
3	triple root	curve crosses with a point of inflection on the axis

A useful diagnostic: $\alpha$ is a root of multiplicity $\geq 2$ iff $p(\alpha) = 0$ and $p'(\alpha) = 0$ . This connects directly to lesson 6 (factor theorem) and to the differentiation work later in 9MA0-01 (calculus stationary points).

Worked example 3 — verifying a double root

Show that $p(x) = x^3 - 3x + 2$ has a double root at $x = 1$ and identify the third root.

Step 1 — check $p(1) = 0$ :

$p(1) = 1 - 3 + 2 = 0. \quad \checkmark$

Step 2 — check $p'(1) = 0$ :

$p'(x) = 3x^2 - 3, \qquad p'(1) = 3 - 3 = 0. \quad \checkmark$

Both $p(1) = 0$ and $p'(1) = 0$ , so $x = 1$ is a root of multiplicity at least 2.

Step 3 — extract the factor $(x - 1)^2$ . Divide $p(x)$ by $(x - 1)$ to get a quadratic, then divide again:

Long division of $x^3 + 0x^2 - 3x + 2$ by $(x - 1)$ gives quotient $x^2 + x - 2$ and remainder $0$ .

Now divide $x^2 + x - 2$ by $(x - 1)$ : quotient $x + 2$ , remainder $0$ .

So $p(x) = (x - 1)^2 (x + 2)$ .

Step 4 — verify the multiplicity rigorously. Write $p(x) = (x - 1)^2 q(x)$ with $q(x) = x + 2$ . Compute $q(1) = 1 + 2 = 3$ . Because $q(1) \neq 0$ , the multiplicity of $x = 1$ is exactly $2$ , not $3$ . The remaining root is $x = -2$ (from $q(x) = 0$ ), of multiplicity $1$ .

Sanity-check by expansion:

$(x - 1)^2 (x + 2) = (x^2 - 2x + 1)(x + 2) = x^3 + 2x^2 - 2x^2 - 4x + x + 2 = x^3 - 3x + 2. \quad \checkmark$

The graph of $y = p(x)$ therefore touches the $x$ -axis at $x = 1$ (turning point on the axis) and crosses the $x$ -axis at $x = -2$ . As $x \to \pm \infty$ the curve behaves like $x^3$ : down on the left, up on the right.

Sketching cubics and quartics

A polynomial sketch at A-Level requires the following labels: every root (with multiplicity behaviour shown), the $y$ -intercept, and the long-term direction at both ends. Turning-point coordinates between roots are not required for a "sketch" command word, but the shape between roots must be drawn correctly.

Algorithm for a cubic $p(x) = ax^3 + bx^2 + cx + d$ .

Note the sign of $a$ and use the end-behaviour table (odd parity) to fix the left and right directions.
Find the $y$ -intercept: $y = d$ .
Find roots by factorisation, the rational root theorem, or inspection.
For each root, determine multiplicity (1, 2, or 3) and use the multiplicity table to draw correct local behaviour.
Connect smoothly. A cubic has at most two turning points; a "wiggle" between three distinct roots is mandatory.

Algorithm for a quartic $p(x) = ax^4 + bx^3 + cx^2 + dx + e$ .

Identical, except even parity changes the end behaviour, and a quartic has at most three turning points. A quartic with four distinct real roots resembles a "W" (if $a > 0$ ) or "M" (if $a < 0$ ).

Guided sketch for $p(x) = x^3 - 3x + 2 = (x - 1)^2 (x + 2)$ .

Leading coefficient $+1$ , degree $3$ (odd): curve falls from $-\infty$ on the left, rises to $+\infty$ on the right.
$y$ -intercept $(0, 2)$ .
Root at $x = -2$ (multiplicity 1): curve crosses the axis.
Root at $x = 1$ (multiplicity 2): curve touches the axis with a turning point on it.
Connect: the curve rises from $-\infty$ , crosses the axis at $x = -2$ heading upward, reaches a local maximum somewhere in $(-2, 1)$ , falls to touch the axis at $x = 1$ , then rises away to $+\infty$ . The local minimum is exactly the touching point at $(1, 0)$ .

Expressing a polynomial as a product from a known root

A workflow you will use again in lesson 6 (factor theorem) and lesson 7 (algebraic fractions): given one root, factor it out and reduce to a lower-degree problem.

Procedure. If you know $p(\alpha) = 0$ , then $(x - \alpha)$ is a factor. Divide $p(x)$ by $(x - \alpha)$ to obtain a quotient one degree lower; factorise that quotient by quadratic or cubic methods.

Example. Given that $x = 3$ is a root of $p(x) = x^3 - 6x^2 + 11x - 6$ , fully factorise $p(x)$ .

Divide by $(x - 3)$ . Leading-term work gives quotient $x^2 - 3x + 2$ with remainder $0$ . Factorise the quadratic: $x^2 - 3x + 2 = (x - 1)(x - 2)$ . So

$p(x) = (x - 3)(x - 1)(x - 2).$

Polynomials

Polynomials

Spec mapping

Terminology and core vocabulary

Polynomial arithmetic

Polynomial long division

Worked example 1 — long division with zero remainder

Worked example 2 — long division with non-zero remainder

End behaviour: sign × parity

Root multiplicity and graphical implications

Worked example 3 — verifying a double root

Sketching cubics and quartics

Expressing a polynomial as a product from a known root

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