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QQQ – C2 Chapter 1 – Algebra & Functions
1.
f(x) = ax3 + bx2 − 4x − 3, where a and b are constants.
Given that (x – 1) is a factor of f(x),
(a) show that a + b = 7.
(2)
Given also that, when f(x) is divided by (x + 2), the remainder is 9,
(b) find the value of a and the value of b, showing each step in your working.
(4)
f(x) = 2x3 – 7x2 – 10x + 24.
2.
(a) Use the factor theorem to show that (x + 2) is a factor of f(x).
(2)
(b) Factorise f(x) completely.
(4)
(9+ Bronze, 10+ Silver, 11+ Gold, 12 Platinum)
QQQ – C2 Chapter 1 – Algebra & Functions
1.
f(x) = ax3 + bx2 − 4x − 3, where a and b are constants.
Given that (x – 1) is a factor of f(x),
(a) show that a + b = 7.
(2)
Given also that, when f(x) is divided by (x + 2), the remainder is 9,
(b) find the value of a and the value of b, showing each step in your working.
(4)
f(x) = 2x3 – 7x2 – 10x + 24.
2.
(a) Use the factor theorem to show that (x + 2) is a factor of f(x).
(2)
(b) Factorise f(x) completely.
(4)
(9+ Bronze, 10+ Silver, 11+ Gold, 12 Platinum)
QQQ – C2 Chapter 1 – Algebra & Functions - RETAKE
1.
f(x) = x3 + ax2 + bx + 3, where a and b are constants.
Given that when f (x) is divided by (x + 2) the remainder is 7,
(a) show that 2a − b = 6.
(2)
Given also that when f(x) is divided by (x −1) the remainder is 4,
(b) find the value of a and the value of b.
(4)
f(x) = 2x3 − 7x2 − 5x + 4
2.
(a) Find the remainder when f(x) is divided by (x – 1).
(2)
(b) Use the factor theorem to show that (x + 1) is a factor of f(x).
(2)
(c) Factorise f(x) completely.
(4)
(11+ Bronze, 12+ Silver, 13+ Gold, 15 Platinum)
Answers
f (1)  a  b  4  3 = 0
1. (a)
Attempt f(±1)
M1
Must be f(1) and = 0 needs to be seen
A1
or a + b – 7 = 0
a+b=7*
(2)
(b)
f (2)  a  2   b  2   4  2   3  9
Attempt f(±2) and uses f(±2) = 9
M1
8a  4b  8  3  9
Correct equation with exponents
of (–2) removed
A1
3
2
(–8a + 4b = 4)
Solves the given equation from part (a) and
their equation in a and b from part (b) as far
as a =... or b = ...
a = 2 and b = 5
M1
Both correct
A1
(4)
[6]
2 (a)
f  2   2.  2   7.  2   10.  2   24
M1
= 0 so (x+2) is a factor
A1
3
2
(2)
(b)
f ( x)  ( x  2)(2 x 2  11x  12)
M1 A1
f ( x)  ( x  2)(2 x  3)( x  4)
dM1 A1
(4)
6 marks
Answers for Retake
1.
(a)
f (2)  8  4a  2b  3  7
M1
so 2a – b = 6 *
(b)
A1
f (1)  1  a  b  3  4
(2)
M1 A1
Solve two linear equations to give a = 2 and b = –2
M1 A1
(4)
6 marks
f ( x)  2x3  7 x2  5x  4
2.
(a)
Remainder = f (1)  2  7  5  4   6
6
f (1)  2  1  7  1  5 1  4
3
(b)
(c)
2
and so ( x  1) is a factor.
f ( x)  ( x  1) (2 x 2  9 x  4)
 ( x  1)(2 x  1)( x  4)
Attempts f (1) or f (1).
M1
6
A1
Attempts f (1).
M1
f (1)  0 with no sign or substitution
errors and for conclusion.
A1
[2
[2
M1 A1
dM1 A
[