Ch 1 · 2.1 / 3.1 Algebra · A Level Mathematics 9709
Topic 1 of 9 in Pure Mathematics 2 and 3 — A Level Mathematics 9709 — 68 lessons.
Introduction
This lesson is the algebra toolkit for the second year: modulus equations, dividing polynomials, and (for Paper 3) partial fractions and binomial series for any power. These tools reappear inside integration and differential equations, so fluency here pays off later. Sections marked (P2, P3) are syllabus 2.1 and 3.1; sections marked (P3 only) are in 3.1 but not in Paper 2.
Key ideas — The modulus function (P2, P3)
|x| is the size of x without its sign: |-5| = |5| = 5. The graph of y = |ax + b| is the line y = ax + b with the part below the x-axis reflected upwards, giving a V shape with its vertex on the x-axis at x = -b/a.
🧑🏫 Teacher's explanation
The modulus function |x| measures the 'distance' of a number from zero on a number line. Because distance cannot be negative, the output is always non-negative. Think of it like a car's odometer: whether you drive 5 km forward or 5 km in reverse, the odometer shows you have traveled a distance of 5 km.
📖 Definitions
🔍 Worked example — Easy
Solve |2x - 5| = 7.
🔍 Worked example — Standard
Solve the inequality |x - 3| < 5.
🔍 Worked example — Exam-style
Solve |3x + 2| = |x - 4|.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Polynomial division (P2, P3)
Dividing p(x) by a divisor d(x) gives a quotient q(x) and a remainder r(x) of lower degree than d(x):
🧑🏫 Teacher's explanation
Think of polynomial division like long division with numbers, but instead of dealing with digits, you are dealing with powers of x. When you divide 13 by 4, you get 3 with a remainder of 1. In polynomials, the 'size' of the remainder is determined by its degree. If you divide by a linear term like (x - 2), your remainder must be a constant (degree 0). If you divide by a quadratic like x^2 + 1, your remainder can be a linear expression ax + b (degree 1).
📖 Definitions
🔍 Worked example — Easy
Use the remainder theorem to find the remainder when p(x) = 2x^3 - 5x^2 + x + 7 is divided by (x - 2).
🔍 Worked example — Standard
Find the quotient and remainder when p(x) = x^4 - 3x^2 + 2x - 5 is divided by (x^2 - 1).
🔍 Worked example — Exam-style
Given that (x - 3) is a factor of p(x) = 2x^3 + ax^2 - 5x + b, and the remainder when p(x) is divided by (x + 1) is 12, find the values of a and b.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — The remainder and factor theorems (P2, P3)
Use them to find unknown coefficients (one equation per given fact), to factorise a cubic (find one root by trial, divide, then factorise the quadratic), and to solve polynomial equations.
🧑🏫 Teacher's explanation
Imagine you have a large number like 125 and you want to see if it is divisible by 5. You perform the division and see if there is anything left over. In algebra, polynomials behave just like integers. When we divide a polynomial p(x) by a linear term like (x - 2), we are essentially asking: "If I try to fit this linear piece into the larger polynomial, is there a leftover piece (the remainder) or does it fit perfectly (the factor)?"
📖 Definitions
🔍 Worked example — Exam-style
The polynomial p(x) = x^3 + ax^2 + bx + 6 has (x - 2) as a factor and leaves a remainder of 10 when divided by (x + 1). Find the values of a and b.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Partial fractions (P3 only)
A fraction with a factorised denominator splits into simpler fractions, provided the degree of the numerator is less than the degree of the denominator:
🧑🏫 Teacher's explanation
Imagine you have a complex machine made of several interconnected parts. Partial fractions is the mathematical process of taking that complex machine and 'deconstructing' it back into its individual, simpler components. Instead of looking at one giant, complicated fraction, we want to see the individual simple fractions that were added together to create it.
📖 Definitions
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Binomial expansion for any rational n (P3 only)
(1 + x)^n = 1 + nx + n(n-1)/2!x^2 + n(n-1)(n-2)/3!x^3 + dots |x| < 1
🧑🏫 Teacher's explanation
When n is a positive integer, the binomial expansion is a finite polynomial. However, when n is a fraction (like 1/2) or a negative integer (like -1), the sequence of terms never hits zero. This creates an infinite series. Think of it like a staircase that never ends; you can keep stepping down, but you never reach the floor.
📖 Definitions
🔍 Worked example — Standard
Find the first three terms of the expansion of (4+x)^-1/2 and state the range of values for x.
🔍 Worked example — Exam-style
Find the coefficient of x^2 in the expansion of (1-3x)^-3 and state the range of values for x.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Worked examples — Example 1: modulus equation and inequality
Solve |3x - 2| = |2x + 7|. Square: 9x^2 - 12x + 4 = 4x^2 + 28x + 49 ⇒ x^2 - 8x - 9 = 0, so x = 9 or x = -1.
🔍 Worked example — Easy
Solve |3x - 2| = |2x + 7|
🔍 Worked example — Standard
Solve |x + 1| > 2x + 5
🔍 Worked example — Exam-style
Solve |2x + 5| < |x + 1|
⚠️ Common errors
Worked examples — Example 2: unknown coefficients
p(x) = 2x^3 + ax^2 + bx - 6. (x - 2) is a factor, and the remainder on division by (x + 1) is -12.
🔍 Worked example — Easy
Given f(x) = x^3 + ax + b, find a and b if (x - 1) is a factor and the remainder when f(x) is divided by (x + 1) is 6.
🔍 Worked example — Standard
p(x) = 2x^3 + ax^2 + bx - 6. (x - 2) is a factor, and the remainder on division by (x + 1) is -12. Find a and b.
🔍 Worked example — Exam-style
The polynomial f(x) = x^3 + kx^2 + 2x + m has a factor (x + 2). When f(x) is divided by (x - 1), the remainder is 12. Find the values of k and m.
⚠️ Common errors
Worked examples — Example 3: dividing by a quadratic
(x^4 + 2x^3 - x + 3) ÷ (x^2 + 1): quotient x^2 + 2x - 1, remainder -3x + 4. Check: (x^2 + 1)(x^2 + 2x - 1) - 3x + 4 = x^4 + 2x^3 - x + 3 ✓.
🔍 Worked example — Exam-style
Find the remainder when 2x^4 - 3x^3 + x^2 - 5 is divided by x^2 - x + 1.
⚠️ Common errors
Worked examples — Example 4: partial fractions (P3)
3x + 1/(x + 1)(x - 1)^2 ≡ A/x + 1 + B/x - 1 + C/(x - 1)^2, so 3x + 1 ≡ A(x-1)^2 + B(x+1)(x-1) + C(x+1).
🔍 Worked example — Exam-style
Express x^2 + 1/(x + 2)(x^2 + 1) in partial fractions (Note: simplify first).
⚠️ Common errors
Worked examples — Example 5: binomial series (P3)
Expand (1 - 2x)^-1/2 up to x^3: 1 + (-tfrac12 )(-2x) + (-tfrac12 ) (-tfrac32 )/2(-2x)^2 + (-tfrac12 ) (-tfrac32 ) (-tfrac52 )/6(-2x)^3 = 1 + x + tfrac32x^2 + tfrac52x^3, valid for |x| < tfrac12.
🔍 Worked example — Easy
Expand (1 + x)^n for n = -2 up to the term in x^2.
🔍 Worked example — Standard
Expand (1 - 2x)^-1/2 up to the term in x^3.
🔍 Worked example — Exam-style
Find the range of values of x for which the expansion of (1 + 3x)^-2 is valid.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) -2 < x < 3; squaring: 8x^2 - 28x + 12 = 0 ⇒ 2x^2 - 7x + 3 = 0, so x = 3 or x = tfrac12. (2) quotient x^2 - x - 1, remainder 2. (3) p (tfrac12 ) = tfrac14 + tfrac34 - 4 + 3 = 0; (2x - 1)(x^2 + 2x - 3) = (2x - 1)(x + 3)(x - 1). (4) 1/x + 1 + 2 - x/x^2 + 3. (5) 2 (1 + tfrac x4 )^1/2 = 2 + x/4 - x^2/64, valid for |x| < 4.