Algebra: AS Level Mathematics questions with answers
Free Algebra questions for AS Level Mathematics (9709), part of Pure Mathematics 2. Each question has the answer and the marking points an examiner would look for.
The polynomial p(x) is defined by p(x) = 2x3 + ax2 + bx + 6, where a and b are constants. It is given that (x − 2) is a factor of p(x) and that the remainder is 18 when p(x) is divided by (x + 1).
(a)
Find the values of a and b.
[5]
(b)
Hence factorise p(x) completely.
[2]
(c)
Hence solve p(3y) = 0, giving your answers correct to 3 significant figures where appropriate.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
a = 1, b = −13
p(2) = 0 substituted M1 p(−1) = 18 substituted M1 two correct linear equations A1 solve simultaneously M1 a = 1, b = −13 A1
5
(b)
(x − 2)(2x − 1)(x + 3)
divide by (x − 2) to obtain the quadratic factor M1 complete factorisation A1
2
(c)
y = 0.631, y = −0.631 (3y > 0 so negative root rejected)
3y = positive root(s) M1 logarithms used correctly M1 0.631, −0.631, and negative root rejected A1
3
Question 2Structured · 4 marksA* challenge
Solve the inequality |2x + 4| > |x + 6|.
[4]
Show answers and mark scheme
Part
Answer and marking points
Marks
x < −103 or x > 2
square both sides or consider non-modular equations M1 critical values −103 and 2 A1 correct method for the inequality (outer region) M1 x < −103, x > 2 A1
4
Question 3Structured · 5 marks
The polynomial p(x) is defined by p(x) = 2x3 + ax2 − 6x + 9, where a is a constant. It is given that (x − 1) is a factor of p(x).
The polynomial p(x) is defined by p(x) = 2x3 + ax2 + bx + 18, where a and b are constants. It is given that (x − 2) is a factor of p(x) and that the remainder is 12 when p(x) is divided by (x + 1).
(a)
Find the values of a and b.
[5]
(b)
Hence factorise p(x) completely.
[2]
(c)
Hence solve p(3y) = 0, giving your answers correct to 3 significant figures where appropriate.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
a = −7, b = −3
p(2) = 0 substituted M1 p(−1) = 12 substituted M1 two correct linear equations A1 solve simultaneously M1 a = −7, b = −3 A1
5
(b)
(x − 2)(2x + 3)(x − 3)
divide by (x − 2) to obtain the quadratic factor M1 complete factorisation A1
2
(c)
y = 0.631, y = 1 (3y > 0 so negative root rejected)
3y = positive root(s) M1 logarithms used correctly M1 0.631, 1, and negative root rejected A1
3
Question 5Structured · 4 marksA* challenge
Solve the inequality |2x + 2| > |x − 3|.
[4]
Show answers and mark scheme
Part
Answer and marking points
Marks
x < −5 or x > 13
square both sides or consider non-modular equations M1 critical values −5 and 13A1 correct method for the inequality (outer region) M1 x < −5, x > 13A1
4
Question 6Structured · 4 marks
(a)
Solve the inequality |4x − 6| < |x + 6|.
[4]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
0 < x < 4
square both sides or consider cases M1 (4x − 6)2 < (x + 6)2A1 critical values 0 and 4 A1 0 < x < 4 A1
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in