Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 3: Pure Mathematics 3 · official syllabus
ALGEBRA LAB
Learn it by playing it
Four short missions on trapping a modulus inequality between two bounds, reading a remainder straight off with the remainder theorem, splitting a fraction into partial fractions, and matching a binomial expansion's coefficient and validity band. When a mission glows green, read why it worked, then take the next one.
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Mission 1 of 4
Trap it between two lines
The V-graph is y = |x - 2| and the flat line is y = 3. The inequality |x - 2| < 3 asks: where does the V dip below the line? Slide L and U to the two x-values where they cross - that traps the solution interval.
Before you slide: |x - a| < b is equivalent to...
Band reads (0, 3) - both ends off. Crossings solve x - 2 = 3 and x - 2 = -3.
Study notes
Played the lab above? These notes are the reference version of what you just felt - use them to check exact formulas and the exam wording.
The modulus function and its graph 3.1
|x| is the size of x with its sign dropped: |3| = 3 and |-3| = 3. On the graph of y = |x| the left half of y = x is reflected in the x-axis, giving the V-shape with its point at the origin.
To sketch y = |ax + b|, find where the inside is zero (x = -b/a) - that is the point of the V - and reflect whatever part of the straight line y = ax + b lies below the axis.
Worked example: y = |x - 2| is the line y = x - 2 with everything below the axis flipped up, so the V has its point at (2, 0) and both arms have gradient +-1.
Read equations off the sketch: |x - 2| = 3 asks where the V meets the line y = 3, and the two crossings give the two solutions x = -1 and x = 5.
Modulus equations and inequalities 3.1
The key equivalence: |x - a| < b says x is within b of a, so a - b < x < a + b. The flipped version |x - a| > b gives the two outside rays, x < a - b or x > a + b. Getting these the right way round is the whole skill.
Worked example: |x - 2| < 3 becomes -3 < x - 2 < 3, so -1 < x < 5 - exactly the interval where the V sits below the line y = 3.
For an equation with a modulus on both sides, |a| = |b| means a^2 = b^2, so square both sides and solve. Worked example: |x - 2| = |2x + 1| gives (x - 2)^2 = (2x + 1)^2, which expands to 3x^2 + 8x - 3 = 0 and factors as (3x - 1)(x + 3) = 0, so x = 1/3 or x = -3.
Squaring can introduce no false solutions here because both sides are non-negative, but always solve the squared equation fully and present both answers - each sign combination gives one.
Polynomial division, the remainder theorem and the factor theorem 3.1
Dividing a polynomial of degree up to 4 by a linear or quadratic divisor gives f(x) = divisor x Q(x) + R(x), where the remainder has lower degree than the divisor (a constant for a linear divisor, linear for a quadratic). Find Q and R by long division or by comparing coefficients.
Remainder theorem: the remainder when f(x) is divided by (x - a) is f(a), because substituting x = a into f(x) = (x - a)Q(x) + r kills the quotient term. Worked example: for f(x) = 2x^3 + x^2 - 7x + 3 divided by (x - 2), the remainder is f(2) = 16 + 4 - 14 + 3 = 9 - no division needed.
For a divisor (ax + b), use the root x = -b/a: the remainder is f(-b/a). Worked example: dividing by (2x - 1), evaluate f(1/2).
Factor theorem: (x - a) is a factor of f(x) exactly when f(a) = 0 - the remainder is zero. Use it to find unknown coefficients: if (x + 1) is a factor of x^3 + px + 6, then f(-1) = -1 - p + 6 = 0, so p = 5.
Partial fractions 3.1
Split a rational function into simpler pieces: with distinct linear factors, (5x + 7)/((x + 1)(x + 2)) = A/(x + 1) + B/(x + 2). Multiply back up to the identity A(x + 2) + B(x + 1) = 5x + 7, which must hold for every x.
Find the constants by substituting the values that make factors vanish: x = -1 gives A = 2 directly, x = -2 gives B = 3. Comparing coefficients works too - equating the x terms and constants gives the simultaneous pair A + B = 5 and 2A + B = 7.
Denominators go up to three distinct linear factors (ax + b)(cx + d)(ex + f), a repeated factor (ax + b)(cx + d)^2 - which needs both a C/(cx + d) and a D/(cx + d)^2 term - and a quadratic factor (ax + b)(cx^2 + d), which takes a linear numerator (Cx + D)/(cx^2 + d).
The numerator degree must not exceed the denominator degree; if it does, divide first and split the proper remainder. Check the answer by adding the pieces back over a common denominator.
The binomial series for rational n 3.1
For any rational n, (1 + x)^n = 1 + nx + n(n - 1)/2! x^2 + n(n - 1)(n - 2)/3! x^3 + ..., valid only for |x| < 1 when n is not a positive integer. For a positive integer n the series stops after n + 1 terms and works for all x.
Worked example: (1 + 2x)^-1 = 1 - 2x + 4x^2 - 8x^3 + ..., substituting u = 2x into 1 - u + u^2 - u^3 + .... The validity condition |u| < 1 becomes |2x| < 1, so the expansion holds for |x| < 1/2.
Adapting to forms like (2 - 2x)^-1: factor out the constant first, (2 - 2x)^-1 = (1/2)(1 - x)^-1, then expand the bracket and multiply through. The validity set comes from the bracketed form.
Always state the validity set as part of the answer - the infinite series only converges inside it, and exam mark schemes carry a mark for saying so.
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