Identities and inequalities for Edexcel IGCSE Further Pure Mathematics: Methods and Worked Examples

Identities and inequalities bring together polynomial algebra, equation solving and optimisation, and they are examined heavily across both papers

This section of the edexcel igcse further pure mathematics specification tests your ability to manipulate polynomials, apply the factor and remainder theorems, solve systems of equations, and handle inequalities including linear programming. Each of these skills connects directly to other parts of the course: algebraic division feeds into curve sketching, the factor theorem is essential for solving cubic equations, and inequalities underpin feasible-region problems that combine algebra with geometry.

These edexcel igcse further pure mathematics identities and inequalities revision notes work through every sub-topic with full worked solutions. The approach is systematic: understand the method, see it applied, then identify the common errors that cost marks in the exam.

Simple algebraic division

Algebraic division is the process of dividing a polynomial by a linear factor such as (x + a), (x - a), (ax + b) or (ax - b). The result is a quotient and a remainder, just like long division with numbers.

Worked example: polynomial long division

Divide 2x³ + 5x² - 4x - 3 by (x + 3).

Set up the long division. Divide the leading term 2x³ by x to get 2x². Multiply (x + 3) by 2x² to get 2x³ + 6x². Subtract from the original: (2x³ + 5x²) - (2x³ + 6x²) = -x².

Bring down -4x to get -x² - 4x. Divide -x² by x to get -x. Multiply (x + 3) by -x to get -x² - 3x. Subtract: (-x² - 4x) - (-x² - 3x) = -x.

Bring down -3 to get -x - 3. Divide -x by x to get -1. Multiply (x + 3) by -1 to get -x - 3. Subtract: (-x - 3) - (-x - 3) = 0.

The quotient is 2x² - x - 1 with remainder 0. Since the remainder is 0, (x + 3) is a factor, and 2x³ + 5x² - 4x - 3 = (x + 3)(2x² - x - 1).

Common error. When dividing by (ax + b) where a is not equal to 1, students sometimes forget to divide by a at the leading-term stage. If dividing by (2x - 1), the first step is to divide the leading term by 2x, not by x. Missing that factor of 2 propagates errors through every subsequent step.

The factor and remainder theorems

The factor theorem

If f(a) = 0, then (x - a) is a factor of f(x). Conversely, if (x - a) is a factor of f(x), then f(a) = 0. This provides a quick test for factors: evaluate the polynomial at the candidate value. If you get zero, you have found a factor.

Worked example: factorising a cubic

Factorise f(x) = x³ - 2x² - 5x + 6 completely, given that (x - 1) is a factor.

Verify: f(1) = 1 - 2 - 5 + 6 = 0. Confirmed.

Divide x³ - 2x² - 5x + 6 by (x - 1) using long division or inspection to get x² - x - 6.

Factorise the quadratic: x² - x - 6 = (x - 3)(x + 2).

Therefore f(x) = (x - 1)(x - 3)(x + 2).

The remainder theorem

When f(x) is divided by (ax + b), the remainder is f(-b/a). This is faster than performing full long division when you only need the remainder.

Worked example: Find the remainder when f(x) = 2x³ + 3x² - x + 4 is divided by (2x - 1).

The remainder is f(1/2) = 2(1/8) + 3(1/4) - 1/2 + 4 = 1/4 + 3/4 - 1/2 + 4 = 1 - 1/2 + 4 = 4.5 = 9/2.

Solutions of equations

Simultaneous equations: one linear, one quadratic

The method is substitution. Rearrange the linear equation for one variable, substitute into the quadratic, and solve the resulting quadratic equation.

Worked example: Solve y = 2x + 1 and x² + y² = 10 simultaneously.

Substitute y = 2x + 1 into x² + y² = 10:

x² + (2x + 1)² = 10

x² + 4x² + 4x + 1 = 10

5x² + 4x - 9 = 0

(5x + 9)(x - 1) = 0

x = 1 or x = -9/5.

When x = 1: y = 2(1) + 1 = 3. When x = -9/5: y = 2(-9/5) + 1 = -18/5 + 5/5 = -13/5.

Solutions: (1, 3) and (-9/5, -13/5).

Cubic equations with at least one rational root

The standard approach: use the factor theorem to find one rational root by testing factors of the constant term divided by factors of the leading coefficient. Once you have one root, divide to reduce to a quadratic, then solve.

Worked example: Solve 2x³ - 3x² - 11x + 6 = 0.

Try x = 3: f(3) = 2(27) - 3(9) - 11(3) + 6 = 54 - 27 - 33 + 6 = 0. So (x - 3) is a factor.

Dividing gives 2x² + 3x - 2 = (2x - 1)(x + 2).

Therefore 2x³ - 3x² - 11x + 6 = (x - 3)(2x - 1)(x + 2) = 0.

Solutions: x = 3, x = 1/2, x = -2.

Inequalities

Linear inequalities

Solve as you would a linear equation, but remember: multiplying or dividing by a negative number reverses the inequality sign.

Worked example: Solve 3 - 2x > 7.

-2x > 4. Divide by -2 and reverse: x < -2.

Quadratic inequalities

Solve the corresponding equation first, then use the shape of the parabola to determine the intervals.

Worked example: Solve x² - 5x + 6 < 0.

Factorise: (x - 2)(x - 3) < 0. The roots are x = 2 and x = 3.

The parabola y = x² - 5x + 6 opens upward (positive coefficient of x²), so it is negative between the roots.

Solution: 2 < x < 3.

Linear inequalities in two variables and linear programming

A linear inequality in two variables (for example, 2x + 3y <= 12) defines a half-plane. The boundary line is 2x + 3y = 12, and the inequality tells you which side of the line to shade. In a linear programming problem, you combine several such inequalities to define a feasible region, then find the maximum or minimum of an objective function (such as P = 5x + 3y) by evaluating it at each vertex of the feasible region.

Worked example: Maximise P = 3x + 2y subject to x + y <= 6, 2x + y <= 10, x >= 0, y >= 0.

The vertices of the feasible region are found by solving pairs of boundary equations:

Vertex	How found	P = 3x + 2y
(0, 0)	Origin	0
(5, 0)	2x + y = 10 and y = 0	15
(4, 2)	x + y = 6 and 2x + y = 10	16
(0, 6)	x + y = 6 and x = 0	12

The maximum value of P is 16, at the point (4, 2).

Exam strategy for linear programming. Draw the feasible region accurately on graph paper, label every vertex with its coordinates, and tabulate the objective function at each vertex. This structured approach earns method marks even if you make an arithmetic error at one vertex. The identities and inequalities edexcel igcse questions in this area reward clear, organised working.

Practice questions

Use these edexcel igcse further pure mathematics practice questions to consolidate the techniques above. Full solutions follow.

Divide 3x³ + x² - 12x + 4 by (3x - 2) and state the quotient and remainder.
f(x) = 2x³ - x² - 13x - 6. Show that (x + 2) is a factor, and hence factorise f(x) completely.
Solve x² + 2x - 8 > 0.
Solve simultaneously: y = x + 2 and x² + xy = 3.
Minimise C = 2x + 5y subject to x + 2y >= 8, 3x + y >= 9, x >= 0, y >= 0.

Solutions

Question 1: Dividing 3x³ + x² - 12x + 4 by (3x - 2): quotient is x² + x - 10/3. Using the remainder theorem: f(2/3) = 3(8/27) + (4/9) - 12(2/3) + 4 = 8/9 + 4/9 - 8 + 4 = 12/9 - 4 = 4/3 - 4 = -8/3. The remainder is -8/3.

Question 2: f(-2) = 2(-8) - (4) - 13(-2) - 6 = -16 - 4 + 26 - 6 = 0. So (x + 2) is a factor. Dividing: 2x³ - x² - 13x - 6 = (x + 2)(2x² - 5x - 3) = (x + 2)(2x + 1)(x - 3).

Question 3: x² + 2x - 8 = (x + 4)(x - 2) = 0 gives x = -4 and x = 2. Parabola opens upward, so positive outside the roots: x < -4 or x > 2.

Question 4: Substitute y = x + 2 into x² + xy = 3: x² + x(x + 2) = 3, so 2x² + 2x - 3 = 0. x = (-2 +/- \u221a(4 + 24))/4 = (-2 +/- \u221a28)/4 = (-2 +/- 2\u221a7)/4 = (-1 +/- \u221a7)/2. Then y = x + 2 for each value.

Question 5: Vertices: intersection of x + 2y = 8 and 3x + y = 9 gives (2, 3), C = 19. x + 2y = 8 and x = 0 gives (0, 4), C = 20. 3x + y = 9 and y = 0 gives (3, 0), C = 6. Minimum C = 6 at (3, 0).

The igcse 4PM1 identities and inequalities section rewards methodical algebraic technique. These edexcel igcse further pure mathematics notes should become your reference for the standard procedures: long division, factor theorem testing, discriminant analysis, and vertex enumeration for linear programming. Precision in each step is what separates full marks from partial credit.

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