Edexcel International A Level Maths · P2 · Pure Mathematics 2
Proof and algebraic division
28 free, high-quality past-paper-style questions on Proof and algebraic division for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 233 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P2-1-B-01P2 · ASEasy5 marks
In this question, m and n are integers, and you may use the fact that an integer is even if it can be written as 2 × (an integer) and odd if it can be written as 2 × (an integer) + 1.
(a)
Prove that the product of any two odd numbers is odd.
(3)
(b)
Prove that the sum of any even number and any odd number is odd.
(2)
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Answer
Mark
Notes
Part (a)
let the odd numbers be 2m + 1 and 2n + 1 (different letters, so any two odd numbers are covered)
B1
(2m + 1)(2n + 1) = 4mn + 2m + 2n + 1
M1
= 2(2mn + m + n) + 1, which is 2 × an integer + 1, so the product is odd
A1
Part (b)
2m + (2n + 1) = 2(m + n) + 1
M1
m + n is an integer, so the sum is 2 × an integer + 1, which is odd
A1
Common error: Writing both odd numbers as 2n + 1. That only proves the result for an odd number multiplied by itself — use two different letters.
Each of the following statements is false. For each one, give a counterexample and show clearly why it disproves the statement.
(a)
For every real number x, x2 ≥ x.
(2)
(b)
The sum of any two irrational numbers is irrational.
(2)
(c)
If p is a prime number, then 2p + 1 is also prime.
(2)
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Answer
Mark
Notes
Part (a)
for example x = ½: x2 = ¼
M1
¼ < ½, so x2 < x and the statement is false
A1
Part (b)
for example √3 and −√3 (or 2 + √5 and 2 − √5), which are both irrational
M1
their sum is 0 (or 4), which is rational, so the statement is false
A1
Part (c)
for example p = 7 (prime): 2 × 7 + 1 = 15
M1
15 = 3 × 5 is not prime, so the statement is false (accept p = 13, 17, 19, … with a correct factorisation)
A1
Exam tip: A counterexample needs three things: the specific value, the calculation showing the statement fails, and a sentence saying the statement is false.
Figure 1 shows part of the curve with equation y = f(x), where f(x) = x3 + ax2 − 10x + 8 and a is a constant. The curve crosses the x-axis at the point A(−4, 0) and at the points P and Q.
(a)
Use the point A to show that a = 1.
(2)
(b)
Factorise f(x) completely.
(3)
(c)
Write down the x-coordinates of P and Q.
(1)
(d)
Find the remainder when f(x) is divided by (x + 1).
(1)
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Answer
Mark
Notes
Part (a)
f(−4) = 0: (−4)3 + a(−4)2 − 10(−4) + 8 = 0
M1
−64 + 16a + 40 + 8 = 16a − 16 = 0, so a = 1
A1
Part (b)
(x + 4) is a factor; divides x3 + x2 − 10x + 8 by (x + 4) or compares coefficients
M1
quadratic factor x2 − 3x + 2
A1
f(x) = (x + 4)(x − 1)(x − 2)
A1
Part (c)
P: x = 1 and Q: x = 2 (ft their factors)
B1
Part (d)
f(−1) = −1 + 1 + 10 + 8 = 18
B1
Common error: Reading A(−4, 0) as a sign that (x − 4) is a factor. A root at x = −4 gives the factor (x + 4), because x + 4 = 0 when x = −4.
Figure 1 shows part of the curve with equation y = g(x), where g(x) = x3 − 4x2 + x + 6. The point D(4, 10) lies on the curve, and the curve crosses the x-axis at three points whose x-coordinates are integers.
(a)
Using the point D and the remainder theorem, write down the remainder when g(x) is divided by (x − 4).
(2)
(b)
Use algebraic long division to divide g(x) by (x − 4). State the quotient and confirm the remainder.
(3)
(c)
Use the factor theorem to show that (x − 3) is a factor of g(x). Hence, using Figure 1, write g(x) as a product of three linear factors.
(2)
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Answer
Mark
Notes
Part (a)
by the remainder theorem the remainder is g(4)
M1
D(4, 10) lies on the curve, so g(4) = 10: remainder = 10
A1
Part (b)
first term x2: (x3 − 4x2) − (x3 − 4x2) = 0, bring down x + 6
remainders are 0, 1 and 1, so in every case the remainder is 0 or 1
A1
Part (b)
by (a), n2 + 1 leaves a remainder of 0 + 1 = 1 or 1 + 1 = 2 when divided by 3
M1
the remainder is never 0, so n2 + 1 is never a multiple of 3
A1
Part (c)
n = 1 gives 1 + 2 = 3, which is a multiple of 3, so the claim is false (any n not a multiple of 3 works)
B1
Exam tip: Say why your cases cover everything (“every integer is 3k, 3k + 1 or 3k + 2”) and finish with a conclusion that brings all the cases together.
The polynomial q(x) = x3 + ax2 + bx + 10 has unknown constants a and b. Dividing q(x) by (x − 2) leaves no remainder, while dividing q(x) by (x + 1) leaves a remainder of 24.
(a)
Find the values of a and b.
(5)
(b)
Hence factorise q(x) completely.
(3)
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Answer
Mark
Notes
Part (a)
q(2) = 0: 8 + 4a + 2b + 10 = 0
M1
q(−1) = 24: −1 + a − b + 10 = 24
M1
2a + b = −9 and a − b = 15
A1
solves simultaneously, e.g. adding: 3a = 6
M1
a = 2, b = −13
A1
Part (b)
divides by (x − 2) to get a quadratic factor
M1
x2 + 4x − 5
A1
q(x) = (x − 2)(x − 1)(x + 5)
A1
Exam tip: Turn each sentence of information into one equation before solving: “no remainder” means q(2) = 0, “remainder 24” means q(−1) = 24.
Figure 1 shows a large square made from four congruent right-angled triangles and a smaller square. Each triangle has shorter sides of lengths a and b and a hypotenuse of length c, where a, b and c are positive.
(a)
By finding the area of the large square in two different ways, prove that a2 + b2 = c2.
(4)
(b)
Hence prove that a + b ≤ √2 c.
(3)
(c)
State the condition on a and b for which a + b = √2 c, and describe the triangles in Figure 1 in this case.
(1)
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Answer
Mark
Notes
Part (a)
the large square has side a + b, so its area is (a + b)2
B1
it is also the inner square plus four triangles: c2 + 4 × ½ab
M1
a2 + 2ab + b2 = c2 + 2ab
M1
subtracting 2ab: a2 + b2 = c2
A1
Part (b)
(a − b)2 ≥ 0, so a2 + b2 ≥ 2ab
M1
(a + b)2 = a2 + b2 + 2ab ≤ 2(a2 + b2) = 2c2
M1
a + b > 0 and c > 0, so taking square roots: a + b ≤ √2 c
A1
Part (c)
a = b: the four triangles are then isosceles right-angled triangles
B1
Exam tip: Taking square roots keeps the direction of an inequality only when both sides are non-negative. Say that a + b > 0 and c > 0 before you do it.
Figure 1 shows a cuboid. The lengths of two of its edges, in centimetres, are shown in terms of x, and the third edge has length d cm. The volume of the cuboid is V cm3, where V = 2x3 + 9x2 + 7x − 6.
(a)
Use algebraic division to show that d = x + 2.
(3)
(b)
Explain why x > ½.
(1)
(c)
Given that V = 150, show that x = 3 is a solution of 2x3 + 9x2 + 7x − 156 = 0, and prove that it is the only real solution.
(3)
(d)
Find the dimensions of the cuboid when its volume is 150 cm³.
(1)
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Answer
Mark
Notes
Part (a)
divides 2x3 + 9x2 + 7x − 6 by (x + 3) (or by (2x − 1))
M1
quotient 2x2 + 3x − 2 (or x2 + 5x + 6)
A1
2x2 + 3x − 2 = (2x − 1)(x + 2), so V = (x + 3)(2x − 1)(x + 2) and d = x + 2
A1
Part (b)
every edge length must be positive; 2x − 1 > 0 gives x > ½ (and then x + 3 and x + 2 are positive too)
B1
Part (c)
2(27) + 9(9) + 7(3) − 156 = 54 + 81 + 21 − 156 = 0, so (x − 3) is a factor
B1
2x3 + 9x2 + 7x − 156 = (x − 3)(2x2 + 15x + 52)
M1
discriminant 152 − 4(2)(52) = 225 − 416 = −191 < 0, so the quadratic factor has no real roots and x = 3 is the only solution
A1
Part (d)
6 cm, 5 cm and 5 cm
B1
Common error: Stopping as soon as x = 3 works. To show it is the only value you must deal with the quadratic factor, here by showing that its discriminant is negative.
Figure 1 shows the curve with equation y = f(x), where f(x) = x3 + ax2 + bx + c and a, b and c are constants. The curve passes through the labelled points (−2, 0), (1, 0) and (0, −8), and crosses the x-axis again at R.
(a)
Write down the value of c.
(1)
(b)
Find the value of a and the value of b.
(4)
(c)
Find the x-coordinate of R.
(2)
(d)
Find the remainder when f(x) is divided by (x − 2).
Exam tip: Every labelled point on a curve gives one equation. A point on the y-axis gives the constant term straight away, which leaves fewer unknowns for the other points.
Figure 1 shows the integers from 1 to 36 written in six columns. The column headings describe the numbers in each column, where k is an integer (k = 0 in the first row). The prime numbers are circled.
(a)
Using Figure 1, state the two columns that contain the primes greater than 3.
(1)
(b)
Prove, by exhaustion, that every prime number greater than 3 is of the form 6k + 1 or 6k + 5, where k is an integer.
(4)
(c)
Show that the statement “every number of the form 6k + 1, where k ≥ 1, is prime” is false.
(2)
(d)
Prove that, if p is a prime number greater than 3, then p2 − 1 is a multiple of 12.
(2)
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Answer
Mark
Notes
Part (a)
6k + 1 and 6k + 5
B1
Part (b)
every integer can be written as 6k, 6k + 1, 6k + 2, 6k + 3, 6k + 4 or 6k + 5
B1
6k = 2(3k), 6k + 2 = 2(3k + 1) and 6k + 4 = 2(3k + 2) are multiples of 2
M1
6k + 3 = 3(2k + 1) is a multiple of 3
M1
a multiple of 2 or of 3 that is greater than 3 is not prime, so a prime greater than 3 must be of the form 6k + 1 or 6k + 5
A1
Part (c)
25 = 6(4) + 1 is of this form
M1
25 = 5 × 5 is not prime, so the statement is false (accept 49 = 6(8) + 1 = 7 × 7, etc.)
p = 6k + 5: (6k + 5)2 − 1 = 36k2 + 60k + 24 = 12(3k2 + 5k + 2); in both cases p2 − 1 is 12 × an integer
A1
Common error: Saying “6k + 2 is even, so it is not prime” without care: when k = 0 it equals 2, which is prime. The argument works because the claim is only about primes greater than 3.
Figure 1 shows the curve C with equation y = 2x3 − 3x2 − 7x + 14 and the straight line L with equation y = x + 2. The line meets the curve at the points A(−2, 0) and B(2, 4), and at a third point P.
(a)
Show that the x-coordinates of the points where L meets C satisfy 2x3 − 3x2 − 8x + 12 = 0.
(2)
(b)
Use Figure 1 to write down one linear factor of 2x3 − 3x2 − 8x + 12, and use the factor theorem to confirm it.
(2)
(c)
Hence factorise 2x3 − 3x2 − 8x + 12 completely, and find the exact coordinates of P.
(4)
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Answer
Mark
Notes
Part (a)
2x3 − 3x2 − 7x + 14 = x + 2
M1
2x3 − 3x2 − 8x + 12 = 0
A1
Part (b)
A gives (x + 2) (or B gives (x − 2))
B1
x = −2: −16 − 12 + 16 + 12 = 0 (or x = 2: 16 − 12 − 16 + 12 = 0), so it is a factor
B1
Part (c)
divides by (x + 2) (or by (x − 2))
M1
2x2 − 7x + 6 (or 2x2 + x − 6)
A1
(x + 2)(x − 2)(2x − 3)
A1
P: x = 3/2, y = 3/2 + 2 = 7/2, so P is (3/2, 7/2)
A1
Exam tip: The points where a line meets a curve come from solving their equations simultaneously. Any intersection you can already see gives you a factor of the resulting cubic.
The triangular numbers Tn are given by Tn = ½n(n + 1) for positive integers n. Figure 1 shows dot patterns for T1 to T4, and a 5 × 5 square of dots split into two triangular patterns.
(a)
Explain how the 5 × 5 square in Figure 1 illustrates that T4 + T5 = 52.
(1)
(b)
Prove that Tn + Tn + 1 = (n + 1)2 for every positive integer n.
(3)
(c)
Prove that 8Tn + 1 is a perfect square for every positive integer n.
(2)
(d)
Disprove the claim “no triangular number greater than 1 is a perfect square”.
(2)
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Answer
Mark
Notes
Part (a)
the 25 dots split into a triangle of 10 = T4 dots and a triangle of 15 = T5 dots, so 10 + 15 = 25 = 52
B1
Part (b)
Tn + 1 = ½(n + 1)(n + 2)
B1
Tn + Tn + 1 = ½(n + 1)[n + (n + 2)]
M1
= ½(n + 1)(2n + 2) = (n + 1)2
A1
Part (c)
8Tn + 1 = 4n(n + 1) + 1 = 4n2 + 4n + 1
M1
= (2n + 1)2, the square of an integer
A1
Part (d)
T8 = ½ × 8 × 9 = 36
M1
36 = 62 is a perfect square greater than 1, so the claim is false (accept T49 = 1225 = 352)
A1
Common error: Checking a few values of n and calling it a proof. Work with a general n and simplify exactly to the target form.
f(x) = 2x3 + ax2 + bx + 15, where a and b are integers. Given that (2x − 3) is a factor of f(x), and that −49 is the remainder when f(x) is divided by (x + 2),
(a)
find the value of a and the value of b.
(5)
(b)
Factorise f(x) completely.
(3)
(c)
Hence solve the equation 2y6 − 11y4 + 2y2 + 15 = 0, giving all real solutions in exact form.
(3)
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Answer
Mark
Notes
Part (a)
f(3/2) = 0: 2(27/8) + a(9/4) + b(3/2) + 15 = 0
M1
f(−2) = −49: −16 + 4a − 2b + 15 = −49
M1
3a + 2b = −29 and 2a − b = −24 (or equivalent simplified pair)
A1
solves simultaneously, e.g. b = 2a + 24 ⇒ 7a = −77
M1
a = −11, b = 2
A1
Part (b)
divides 2x3 − 11x2 + 2x + 15 by (2x − 3) or compares coefficients
M1
quadratic factor x2 − 4x − 5
A1
f(x) = (2x − 3)(x + 1)(x − 5)
A1
Part (c)
the equation is f(y2) = 0, so y2 = 3/2, −1 or 5
M1
y2 = −1 has no real solutions
B1
y = ±√(3/2) = ±√6/2, y = ±√5
A1
Exam tip: In a “hence” part, look for the earlier expression hiding inside the new one: here every power of y is even, so put x = y² and use the factors you already have.
In this question n is an integer. Three claims are made about expressions of the form nr − ns.
(a)
Prove that n3 − n is a multiple of 6 for every integer n.
(4)
(b)
Show that the claim “n4 − n is a multiple of 4 for every integer n” is false.
(2)
(c)
Prove, by considering the cases n even and n odd, that n4 − n2 is a multiple of 4 for every integer n.
(4)
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Answer
Mark
Notes
Part (a)
n3 − n = n(n2 − 1) = (n − 1)n(n + 1)
M1
this is the product of three consecutive integers
A1
at least one of any two consecutive integers is even, so the product is a multiple of 2
B1
exactly one of any three consecutive integers is a multiple of 3, so the product is a multiple of 3; being a multiple of both 2 and 3, it is a multiple of 6
B1
Part (b)
n = 2: 24 − 2 = 14
M1
14 is not a multiple of 4, so the claim is false (accept other valid values, e.g. n = 3 gives 78)
A1
Part (c)
n4 − n2 = n2(n2 − 1)
M1
n even, n = 2k: n2 = 4k2, so n2(n2 − 1) = 4k2(4k2 − 1), a multiple of 4
A1
n odd, n = 2k + 1: n2 − 1 = 4k2 + 4k = 4(k2 + k), so the product is a multiple of 4
A1
every integer is even or odd, so n4 − n2 is a multiple of 4 for all integers n
A1
Common error: Stopping at “a product of three consecutive integers”. You must also say why such a product is divisible by 2 and by 3 to reach 6.
This question develops some inequalities step by step. Each proof must start from a statement that is known to be true.
(a)
Prove that u2 + v2 ≥ 2uv for all real numbers u and v.
(2)
(b)
Hence prove that a/b + b/a ≥ 2 for all positive real numbers a and b.
(3)
(c)
Show, by means of a counterexample, that the inequality in part (b) is not true for all non-zero real numbers a and b.
(2)
(d)
Prove that x + 9/x ≥ 6 for all x > 0, and state the value of x for which equality holds.
(4)
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Answer
Mark
Notes
Part (a)
(u − v)2 ≥ 0 for all real u, v
M1
u2 − 2uv + v2 ≥ 0, so u2 + v2 ≥ 2uv
A1
Part (b)
by (a) with u = a, v = b: a2 + b2 ≥ 2ab
M1
a, b > 0 so ab > 0, and dividing by ab keeps the direction of the inequality
M1
a/b + b/a ≥ 2
A1
Part (c)
e.g. a = 1, b = −1: a/b + b/a = −1 + (−1) = −2
M1
−2 < 2, so the inequality fails for these non-zero values
A1
Part (d)
(x − 3)2 ≥ 0 for all real x
M1
x2 − 6x + 9 ≥ 0
A1
x > 0, so dividing by x keeps the inequality: x − 6 + 9/x ≥ 0, i.e. x + 9/x ≥ 6
A1
equality when (x − 3)2 = 0, i.e. x = 3
B1
Exam tip: If you found an inequality proof by working backwards from the answer, rewrite it forwards from a true statement such as (x − 3)² ≥ 0, and say where you used positivity.
A cubic function is given by f(x) = 2x3 + px2 + qx + 12, with constants p and q. The graph of y = f(x) passes through the point (2, 0). When f(x) is divided by (2x − 1), the remainder is 9.
(a)
Find the values of p and q.
(5)
(b)
Show that f(x) = (x − 2)2(2x + 3).
(3)
(c)
Write down the coordinates of the points where the graph of y = f(x) meets the x-axis, and explain why the graph touches the x-axis, without crossing it, at one of these points.
(2)
(d)
Hence find the set of values of x for which f(x) > 0.
(2)
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Answer
Mark
Notes
Part (a)
f(2) = 0: 16 + 4p + 2q + 12 = 0, so 2p + q = −14
M1
f(½) = 9: ¼ + p/4 + q/2 + 12 = 9
M1
p + 2q = −13
A1
solves the pair simultaneously
M1
p = −5, q = −4
A1
Part (b)
(x − 2) is a factor as f(2) = 0; divides 2x3 − 5x2 − 4x + 12 by (x − 2)
M1
quadratic factor 2x2 − x − 6
A1
2x2 − x − 6 = (x − 2)(2x + 3), so f(x) = (x − 2)2(2x + 3)
A1
Part (c)
(2, 0) and (−3/2, 0)
B1
(x − 2) is a repeated factor and (x − 2)2 ≥ 0, so f does not change sign at x = 2: the graph touches the axis at (2, 0)
B1
Part (d)
(x − 2)2 > 0 for x ≠ 2, so need 2x + 3 > 0
M1
x > −3/2, x ≠ 2
A1
Model sketch
Figure: model sketch: the graph crosses at (−3/2, 0) and touches at (2, 0)
Common error: Forgetting to exclude x = 2 in the last part. At a repeated root f(x) = 0, so a strict inequality f(x) > 0 fails there.
f(x) = 3x3 + px2 + qx − 10, where p and q are constants. Figure 1 shows part of the curve with equation y = f(x). The curve passes through the labelled points A(2, 0) and B(1, −16). The dashed line y = −10 meets the curve at the three points P, Q and R.
(a)
Find the value of p and the value of q.
(5)
(b)
Factorise f(x) completely.
(3)
(c)
Find the remainder when f(x) is divided by (3x − 1).
(2)
(d)
Find the exact x-coordinates of P, Q and R.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
A: 24 + 4p + 2q − 10 = 0
M1
B: 3 + p + q − 10 = −16
M1
2p + q = −7 and p + q = −9
A1
solves simultaneously, e.g. subtracting: p = 2
M1
p = 2, q = −11
A1
Part (b)
(x − 2) is a factor because A lies on the curve; divides 3x3 + 2x2 − 11x − 10 by (x − 2)
x = (−2 ± √136)/6 = (−1 ± √34)/3; P is at (−1 − √34)/3 and R at (−1 + √34)/3
A1
Exam tip: To find where y = f(x) meets the horizontal line y = k, solve f(x) = k. Here the constant terms cancel, so x is a common factor and one solution is x = 0.
Figure 1 shows a calendar for a month of 31 days, written in rows of seven consecutive dates. A 2 × 2 square of dates and a 3 × 3 square of dates are outlined. For any square of dates, the corner difference is defined as (top-right date × bottom-left date) − (top-left date × bottom-right date). In parts (b), (c) and (e), assume that the square lies wholly inside such a grid.
(a)
Show that the corner difference of the outlined 2 × 2 square is 7.
(1)
(b)
Prove that the corner difference of every 2 × 2 square of dates is 7.
(3)
(c)
Prove that the corner difference of every 3 × 3 square of dates is 28.
(3)
(d)
The nine dates in the 3 × 3 square with top-left date 1 add up to 81 = 92. Use the 3 × 3 square outlined in Figure 1 to disprove the claim that the nine dates in every 3 × 3 square add up to a square number.
(2)
(e)
Prove that the corner difference of every k × k square of dates, where k ≥ 2, is 7(k − 1)2.
(3)
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Answer
Mark
Notes
Part (a)
10 × 16 − 9 × 17 = 160 − 153 = 7
B1
Part (b)
with top-left date n, the dates are n, n + 1, n + 7 and n + 8
Exam tip: Name one date n and write every other date in terms of it. Keeping (k − 1) as a single block in the last part makes the cancelling easy to see.
An open box is made from a rectangular sheet of card 16 cm by 10 cm. A square of side x cm is cut from each corner, as shown in Figure 1, and the four sides are then folded up along the dashed lines. The volume of the box is V cm3.
(a)
Show that V = 4x3 − 52x2 + 160x.
(2)
(b)
State the set of possible values of x.
(1)
(c)
When x = 2 the volume is 144 cm3. Show that the equation V = 144 can be written as x3 − 13x2 + 40x − 36 = 0, and use the factor theorem to show that (x − 2) is a factor of x3 − 13x2 + 40x − 36.
(2)
(d)
Factorise x3 − 13x2 + 40x − 36 completely.
(3)
(e)
Hence prove that V ≤ 144 for every possible value of x, and state the value of x for which V = 144.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
base (16 − 2x) cm by (10 − 2x) cm, height x cm: V = x(16 − 2x)(10 − 2x)
(x − 2)2 ≥ 0 and 9 − x > 0 for 0 < x < 5, so 144 − V ≥ 0, i.e. V ≤ 144
A1
equality only when (x − 2)2 = 0, i.e. x = 2
A1
Common error: Including x = 9 as a value that gives a volume of 144 cm³. The squares must fit on the 10 cm side, so x < 5 and x = 9 is impossible for the box.
f(x) = 6x3 + px2 + qx − 8, where p and q are constants. Figure 1 shows an incomplete grid used to divide f(x) by (2x + 1). The quotient is the sum of the three column headings, and when f(x) is divided by (2x + 1) the remainder is the constant R.
(a)
Use Figure 1 to find the quotient.
(3)
(b)
Complete the grid, and hence find the values of p, q and R.
(3)
(c)
Use the remainder theorem to verify your value of R.
(2)
(d)
Hence factorise f(x) + 10 completely.
(2)
(e)
Find the set of values of x for which f(x) > −10.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
second heading: −10x2 ÷ 2x = −5x
B1
third heading: +2 ÷ (+1) = +2
B1
quotient 3x2 − 5x + 2
B1
Part (b)
missing cells: 4x in the 2x row; 3x2 and −5x in the +1 row
M1
p = −10 + 3 = −7 and q = 4 − 5 = −1
A1
constant term: 2 + R = −8, so R = −10
A1
Part (c)
f(−½) = 6(−⅛) − 7(¼) − (−½) − 8
M1
= −3/4 − 7/4 + 2/4 − 32/4 = −10 ✓
A1
Part (d)
f(x) + 10 = (2x + 1)(3x2 − 5x + 2)
M1
= (2x + 1)(3x − 2)(x − 1)
A1
Part (e)
need (2x + 1)(3x − 2)(x − 1) > 0, with critical values −½, ⅔ and 1
M1
−½ < x < ⅔ or x > 1
A1
Exam tip: In a division grid, each missing heading comes from one known cell: divide that cell by its row heading. The remainder is whatever is needed to make the constant term match.
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