What is remainder theorem?

When f(x) is divided by (x - a), the remainder is f(a).

What is wrong region for a quadratic inequality?

(k - 3)(k + 1) > 0 means k 3, not -1 < k < 3.

Pearson Edexcel Edexcel 2019-style practice: Express 14 - sqrt(12) in the form a + bsqrt(3), where a and b are rational numbers.

First simplify the surd: sqrt(12) = 2sqrt(3), so the expression is 14 - 2sqrt(3) (M1 for sqrt(12) = 2sqrt(3)). Rationalise by multiplying top and bottom by the conjugate 4 + 2sqrt(3) (M1): 14 - 2sqrt(3) × 4 + 2sqrt(3)4 + 2sqrt(3) = 4 + 2sqrt(3)16 - 12 = 4 + 2sqrt(3)4 (A1). Simplify: 4 + 2sqrt(3)4 = 1 + 12sqrt(3), so a = 1 and b = 12 (A1). Markers reward the conjugate, correct expansion of the difference of squares in the denominator, and writing the final answer in the requested form.

EnglandMathsSyllabus dot point

How do you manipulate expressions, solve equations and inequalities, and transform graphs?

Algebra and functions including indices and surds, quadratics, simultaneous equations, inequalities, polynomials, graphs, functions and transformations, the binomial expansion and partial fractions.

A focused answer to the Edexcel A-Level Mathematics algebra and functions content, covering indices and surds, quadratics, the discriminant, simultaneous equations, inequalities, polynomials, graphs, function notation, transformations, the binomial expansion and partial fractions.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

Edexcel wants fluent algebra: working with indices and surds, solving quadratics every way, using the discriminant, solving simultaneous equations and inequalities, dividing and factorising polynomials, sketching and transforming graphs, using function notation including composite and inverse functions, expanding binomials, and splitting expressions into partial fractions. This content underpins almost every other Pure topic, so accuracy here protects marks everywhere else.

The answer

Indices and surds

The laws of indices let you combine powers, and you rationalise surds by multiplying by the conjugate.

A surd is simplified by extracting square factors, for example $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$ . Fractional indices and surds are the same idea written two ways, so $8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$ .

Quadratics and the discriminant

You should solve quadratics by factorising, completing the square, and the quadratic formula, and use the discriminant to determine the number of real roots.

Completing the square rewrites $ax^2 + bx + c$ as $a\left(x + \dfrac{b}{2a}\right)^2 + c - \dfrac{b^2}{4a}$ , which reveals the vertex (turning point) directly.

Solve

2x^2 - 7x + 3 = 0

by factorising and confirm with the formula

step 1 Factorise

Look for two numbers multiplying to $ac = 2 \times 3 = 6$ and adding to $-7$ ; these are $-6$ and $-1$ . Split the middle term: $2x^2 - 6x - x + 3 = 0$ .

step 2 Group

$2x(x - 3) - 1(x - 3) = 0$ , so $(2x - 1)(x - 3) = 0$ .

step 3 Read off the roots

$2x - 1 = 0 \Rightarrow x = \dfrac{1}{2}$ , or $x - 3 = 0 \Rightarrow x = 3$ .

step 4 Confirm with the formula

$x = \dfrac{7 \pm \sqrt{49 - 24}}{4} = \dfrac{7 \pm 5}{4}$ , giving $x = 3$ or $x = \dfrac{1}{2}$ , which agrees.

Simultaneous equations and inequalities

A linear and a quadratic equation are solved by substitution. Inequalities are solved by finding critical values and testing the sign in each region; for quadratic inequalities a quick sketch of the parabola is the safest method.

Polynomials, the factor and remainder theorems

To factorise a cubic, find one root by inspection then divide.

Graphs, functions and transformations

A function has one output per input. Composite functions apply one then another, written $fg(x) = f(g(x))$ ; an inverse undoes a function and reflects its graph in the line $y = x$ .

Binomial expansion and partial fractions

The binomial theorem expands $(a + b)^n$ ; for a positive integer $n$ the expansion terminates.

Examples in context

Express

\dfrac{11x + 1}{(x - 1)(2x + 1)}

in partial fractions

step 1 Set up the form

Write $\dfrac{11x + 1}{(x - 1)(2x + 1)} = \dfrac{A}{x - 1} + \dfrac{B}{2x + 1}$ , so $11x + 1 = A(2x + 1) + B(x - 1)$ .

step 2 Substitute strategic values

Let $x = 1$ : $12 = 3A$ , so $A = 4$ .

Let $x = -\dfrac{1}{2}$ : $11\left(-\dfrac{1}{2}\right) + 1 = B\left(-\dfrac{3}{2}\right)$ , so $-\dfrac{9}{2} = -\dfrac{3}{2}B$ , giving $B = 3$ .

step 3 Write the answer

$\dfrac{11x + 1}{(x - 1)(2x + 1)} = \dfrac{4}{x - 1} + \dfrac{3}{2x + 1}$ .

Try this

Q1. Solve $2x^2 - 5x - 3 = 0$ by factorising. [3 marks]

Cue. $(2x + 1)(x - 3) = 0$ , so $x = -\dfrac{1}{2}$ or $x = 3$ .

Q2. Express $x^2 + 8x + 3$ in the form $(x + a)^2 + b$ . [2 marks]

Cue. $(x + 4)^2 - 16 + 3 = (x + 4)^2 - 13$ .

Q3. Find the set of values of $x$ for which $x^2 - x - 12 < 0$ . [3 marks]

Cue. $(x - 4)(x + 3) < 0$ , so $-3 < x < 4$ (between the roots).

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20194 marksExpress

\dfrac{1}{4 - \sqrt{12}}

in the form

a + b\sqrt{3}

, where

a

and

b

are rational numbers.

Show worked answer →

First simplify the surd: $\sqrt{12} = 2\sqrt{3}$ , so the expression is $\dfrac{1}{4 - 2\sqrt{3}}$ (M1 for $\sqrt{12} = 2\sqrt{3}$ ).

Rationalise by multiplying top and bottom by the conjugate $4 + 2\sqrt{3}$ (M1):
$\dfrac{1}{4 - 2\sqrt{3}} \times \dfrac{4 + 2\sqrt{3}}{4 + 2\sqrt{3}} = \dfrac{4 + 2\sqrt{3}}{16 - 12} = \dfrac{4 + 2\sqrt{3}}{4}$ (A1).

Simplify: $\dfrac{4 + 2\sqrt{3}}{4} = 1 + \dfrac{1}{2}\sqrt{3}$ , so $a = 1$ and $b = \dfrac{1}{2}$ (A1).

Markers reward the conjugate, correct expansion of the difference of squares in the denominator, and writing the final answer in the requested form.

Edexcel 20215 marksThe equation

x^2 + (k + 3)x + (2k + 3) = 0

has two distinct real roots. Find the set of possible values of

k

Show worked answer →

Two distinct real roots require a positive discriminant: $b^2 - 4ac > 0$ (M1).

Here $a = 1$ , $b = k + 3$ , $c = 2k + 3$ , so $(k + 3)^2 - 4(2k + 3) > 0$ (M1).

Expand: $k^2 + 6k + 9 - 8k - 12 > 0$ , giving $k^2 - 2k - 3 > 0$ (A1).

Factorise: $(k - 3)(k + 1) > 0$ (M1).

The critical values are $k = 3$ and $k = -1$ ; the quadratic is positive outside the roots, so $k < -1$ or $k > 3$ (A1).

Markers reward forming the discriminant inequality, correct expansion, factorising, and the correct outside region from a positive coefficient of $k^2$ .

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Mathematics (9MA0) specification — Pearson Edexcel (2017)