CAIE A-Level Mathematics A2 3.1 Algebra Questions
Practise Pure 3 algebra skills with modulus graphs, polynomial division, theorem use, partial fractions and binomial expansions.
- Syllabus
- 2028–2030
- Course
- Mathematics 9709
- Level
- A2
Practise Pure 3 algebra skills with modulus graphs, polynomial division, theorem use, partial fractions and binomial expansions.
Solve the inequality |2 x+3|>3|x+2|.
Square both sides:
(2x+3)2>9(x+2)2.
Then
4x2+12x+9>9x2+36x+36,
so
5x2+24x+27<0.
Factorising,
(5x+9)(x+3)<0.
The critical values are x=-3 and x=−59. The product is negative between them, hence
−3<x<−59.
B1 for the non-modular squared inequality or equivalent linear equations. M1 for solving the quadratic/equations. A1 for critical values -3 and −59. A1 for −3<x<−59, with strict inequalities.
Find the quotient and remainder when 8x3+4x2+2x+7 is divided by 4x2+1.
Commence division and reach quotient of the form 2x±1
Or by inspection 8x3+4x2+2x+7=(4x2+1)(2x±1)+r
Or by inspection 8x3+4x2+2x+7=(4x2+1)(2x±1)
+r
Obtain (quotient) 2 x+1
Obtain (remainder) 6
3
The polynomial x3+5x2+31x+75 is denoted by p(x).
Show that (x+3) is a factor of p(x).
Substitute x=-3 to obtain value of p(-3)
Obtain p(-3)=0 and hence given result
Alternative method for Question 10(a)
Divide p(x) by ( x+3 ) to obtain quotient x2±2x+…
Obtain quotient x2+2x+25, with zero remainder and hence given result
Express 2x2−5x−126x2−9x−16 in partial fractions.
State or imply the form A+2x+3B+x−4C2x+3Dx+E+x−4F and 2x+3P+x−4Qx+R are also valid.
Use a correct method for finding a constant
SC: If score B0, they can score M1 A1 for one correct constant.
B0 M1 A0 available if they substitute two values to form simultaneous equations but get an incorrect answer, or they substitute one value and make an arithmetic error.
Obtain one of A=3, B=-2 and C=4
SC: If the horizontal equation is correct apart from an incorrect
value for A, the other A marks may be available.
Obtain a second value
SC: If denominator factorised as (x+23)(x−4) can score a
Marking guidance:
maximum of B0 M1 A1 A1 A0 for a split involving 3 terms.
Obtain a third value
ISW
Statement of the final split is not required.
2
Alternative method for Question 2
Divide numerator by denominator
Obtain 3(+2x2−5x−12Px+Q)(3+(2x+3)(x−4)6x+20)
State or imply the form 2x2−5x−12Px+Q=2x+3D+x−4E
Must deal with the 3 separately or include it correctly on both sides in their split.
Obtain one of D=-2 and E=4
SC: If denominator factorised as (x+23)(x−4), then can score a
maximum of B0 M1 A1 A1 A0 for a split involving three terms.
Obtain a second value
ISW
Statement of the final split is not required.
Expand (3+x)(1−2x)21 in ascending powers of x, up to and including the term in x2, simplifying the coefficients.
State correct unsimplified first two terms of the expansion of (1−2x)21,
e.g. 1+21(−2x)
Symbolic coefficients are not sufficient.
1-x
State correct unsimplified term in x2, e.g. 2!21(21−1)(−2x)2
Symbolic coefficients are not sufficient.
−21x2
Obtain sufficient terms of the product of (3+x) and the expansion up to the term
in x2
Obtain final answer 3−2x−25x2