Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 2: Pure Mathematics 2 · official syllabus
ALGEBRA LAB
Learn it by playing it
Four short missions on modulus graphs, modulus inequalities and the remainder and factor theorems. When a mission glows green, read why it worked, then take the next one.
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Mission 1 of 4
Fold the line into a V
The dashed V is y = |2x - 6|: the line 2x - 6 with everything below the axis folded upwards. Your solid V is y = |ax + b|. Slide the steepness a and the shift b until your V lies exactly on the dashed one.
Before you slide: where does y = |2x - 6| touch the x-axis?
a = 1, b = 0: vertex at x = 0 with arm slope 1 - target vertex x = 3, arm slope 2
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 and the graph of y = |ax + b| 2.1
The modulus |x| is the distance of x from 0 on the number line, so it is never negative: |3| = 3 and |-3| = 3. Algebraically, |x| = x when x >= 0 and |x| = -x when x < 0.
The graph of y = |ax + b| (a nonzero) is a V shape: the line y = ax + b where ax + b >= 0, reflected in the x-axis where ax + b < 0. The vertex touches the x-axis at x = -b/a.
For y = |2x - 6| the vertex is at x = 3. To the right of 3 the graph is y = 2x - 6; to the left it is the reflection y = -(2x - 6) = 6 - 2x. The gradient changes sign at the vertex.
To sketch, find the vertex x = -b/a, one point on each arm, and mark the y-intercept |b|. The arms of a modulus graph are straight lines, never curves.
Only linear inside functions are in this topic: graphs of y = |f(x)| and y = f(|x|) for non-linear f are not in the 9709 Paper 2 syllabus.
Modulus equations and |a| = |b| ⇔ a² = b² 2.1
Squaring removes a modulus sign because |u|² = u² for every real u. So |a| = |b| if and only if a² = b², which factorises as (a - b)(a + b) = 0, giving a = b or a = -b.
For |3x - 2| = |2x + 7|, square both sides: (3x - 2)² = (2x + 7)². Expanding gives 9x² - 12x + 4 = 4x² + 28x + 49, so 5x² - 40x - 45 = 0 and x² - 8x - 9 = 0, giving x = 9 or x = -1.
The same equation can be solved by cases: 3x - 2 = 2x + 7 gives x = 9, and 3x - 2 = -(2x + 7) gives x = -1. Squaring is usually faster; cases show where the answers come from.
Squaring an equation with only one modulus, such as |x + 1| = 2x, can introduce values that fail the original equation, because squaring also solves |x + 1| = -2x. Always check each answer in the unsquared equation: here x = 1 works but x = -1/3 does not.
Modulus inequalities and the number line 2.1
|x - a| is the distance of x from a, so |x - a| < b (with b > 0) says x stays within b of a: a - b < x < a + b. For example |x - 3| < 2 means 1 < x < 5.
The flipped inequality reverses the regions: |x - a| > b means x < a - b or x > a + b. Distance language decides which shape you get: inside the band for <, outside it for >.
Squaring also works on inequalities with nonnegative sides: 2x + 5 < |x + 1| gives (2x + 5)² < (x + 1)² only where 2x + 5 >= 0, so handle x < -5/2 separately or sketch both graphs. Expanding: 3x² + 18x + 24 < 0, that is x² + 6x + 8 < 0, so (x + 2)(x + 4) < 0 and -4 < x < -2.
Sketching y = |ax + b| against a line is a reliable check: read off where the V is above or below the line, then confirm the boundary values algebraically.
Write strict and non-strict inequalities carefully: |x - 3| <= 2 gives 1 <= x <= 5 with the endpoints included.
Polynomial division: quotient and remainder 2.1
Dividing a polynomial f(x) of degree at most 4 by a divisor d(x) gives f(x) = d(x)·q(x) + r(x), where the quotient q(x) and remainder r(x) are polynomials and the degree of r is less than the degree of d. The remainder may be zero.
Dividing by a linear divisor (x - c) leaves a constant remainder. Dividing by a quadratic leaves a remainder of the form px + q.
Long division works term by term: at each step divide the leading terms, multiply back, subtract, and bring down the next term. Writing missing powers with coefficient 0 keeps the columns aligned.
Example: dividing x³ - 2x² + x + 1 by (x - 2) gives quotient x² + 1 and remainder 3, so x³ - 2x² + x + 1 = (x - 2)(x² + 1) + 3.
Match coefficients as an alternative to long division: if x³ + x + 2 = (x - 1)(x² + ax + b) + c, expanding and comparing coefficients gives a = 1, b = 2, c = 4.
The remainder theorem 2.1
When a polynomial f(x) is divided by (x - c), the remainder is f(c). This follows from f(x) = (x - c)q(x) + r by substituting x = c, so no long division is needed to find a remainder.
For division by (ax + b) the remainder is f(-b/a), since ax + b = 0 at x = -b/a. Example: the remainder when x³ + 2x - 1 is divided by (2x + 1) is f(-1/2) = -1/8 - 1 - 1 = -17/8.
Remainder questions often hide the divisor: "leaves remainder 5 on division by (x - 2)" is the statement f(2) = 5.
Use the theorem to find unknown coefficients: if x³ + kx + 4 has remainder 10 on division by (x + 1), then f(-1) = -1 - k + 4 = 10, so k = -7.
The factor theorem and solving polynomial equations 2.1
The factor theorem is the zero-remainder case: (x - c) is a factor of f(x) if and only if f(c) = 0. More generally (ax + b) is a factor if and only if f(-b/a) = 0.
To factorise a cubic, test small integer candidates c from the factors of the constant term. For g(x) = x³ - x² - 4x + 4, testing c = 1 gives g(1) = 0, so (x - 1) is a factor.
Divide out the found factor to reduce the degree: g(x) = (x - 1)(x² - 4) = (x - 1)(x - 2)(x + 2), so the equation g(x) = 0 has roots x = 1, 2, -2.
To find an unknown coefficient from a factor statement, set the corresponding value to zero: if (x + 2) is a factor of x³ + kx² + x - 6, then f(-2) = -8 + 4k - 2 - 6 = 0, so k = 4.
A factor theorem zero must be exact. A graph crossing or a calculator approximation suggests a candidate; verify by substituting into the polynomial.
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