Shatin Pui Ying College
F.4 Mathematics
Revision Ex. Ch.3 Ans
1.
Find the remainder when the polynomial x3 + 2x2 – 3x + 1 is divided by
(a) x – 1,
(b) x + 2.
##
Let f(x) = x3 + 2x2 – 3x + 1. By the remainder theorem:
(a) Remainder  f (1)
 (1) 3  2(1) 2  3(1)  1
 1 2  3 1
1
(b) Remainder  f (2)
 (2) 3  2(2) 2  3(2)  1
 8  8  6  1
7
2. When x4 – 2x3 + x2 – kx + k is divided by x + 1, the remainder is 10. Find the value of k.
##
Let f(x) = x4 – 2x3 + x2 – kx + k.
By the remainder theorem, we have
f (1)  10
(1)  2(1)  (1)  k (1)  k  10
4
3
2
4  2k  10
2k  6
k 3
When x2 + 4x – 20 is divided by x – k, the remainder is k2. Find the value of k.
##
Let f(x) = x2 + 4x – 20.
By the remainder theorem, we have
f (k )  k 2
(k ) 2  4(k )  20  k 2
4k  20  0
4k  20
k 5
3.
Find the remainders when the polynomial 6x2 – 3x + 1 is divided by
(a) 2x – 3,
(b) 3x + 1.
##
Let f(x) = 6x2 – 3x + 1.
3
(a) Remainder  f  
2
2
3
3
 6   3   1
2
2
27 9

 1
2 2
 10
 1
(b) Remainder  f   
 3
2
 1
 1
 6    3    1
 3
 3
2
 11
3
8

3
4.
When ax2 + bx – 2 is divided by x + 1 and x – 1, the remainders are –1 and 3 respectively. Find the values
of a and b.
##
Let f(x) = ax2 + bx – 2.
When f(x) is divided by x + 1,
f (1)  1
a(1) 2  b(1)  2  1
a  b  2  1
a  b  1  (1)
When f(x) is divided by x – 1,
f (1)  3
a(1) 2  b(1)  2  3
ab23
ab5
 (2)
2a  6
(1) + (2),
a3
By substituting a = 3 into (2), we have
3b  5
b2
5.
##
Find the quotient and the remainder of (2x3 + 4x2 – 6) ÷ (x + 3).
2x 2  2x  6
x  3 2x3  4x 2  0x  6
2x3  6x 2
 2x 2  0x
 2x 2  6x
6x  6
6 x  18
 24
∴The quotient is 2x2 – 2x + 6 and the remainder is –24.
6. Find the quotient and the remainder of (1 – 3x3 + 2x2 + x) ÷ (x2 – 1 + 2x).
(1  3x 3  2 x 2  x)  ( x 2  1  2 x)  (3x 3  2 x 2  x  1)  ( x 2  2 x  1)
 3x  8
x 2  2 x  1  3x 3  2 x 2 
x 1
 3x 3  6 x 2  3x
8x 2  2x  1
8 x 2  16 x  8
 18 x  9
∴The quotient is –3x + 8 and the remainder is –18x + 9.
7.
It is given that f(x) = x3 – 2x2 + kx + 42 is divisible by x – 1.
(a) Find the value of k.
(b) Hence, factorize f(x) completely.
(a) ∵ f (x) is divisible by x – 1.
∴ By the converse of the factor theorem,
f (1)  0
(1)  2(1)  k (1)  42  0
k   41
3
2
(b) By (a), we have f ( x)  x 3  2 x 2  41x  42.
By long division,
x 2  x  42
x  1 x 3  2 x 2  41x  42
x3  x2
 x 2  41x
 x2  x
 42 x  42
 42 x  42
2
∴ f ( x)  ( x  1)( x  x  42)
 ( x  1)( x  6)( x  7)
8.
It is given that f(x) = 4x3 + kx2 – 25x – 6 is divisible by 4x + 1.
(a) Find the value of k.
(b) Hence, factorize f(x) completely.
(a) ∵ f (x) is divisible by 4x + 1.
∴ By the converse of the factor theorem,
 1
f    0
 4
3
2
 1
 1
 1
4    k     25    6  0
 4
 4
 4
1
k 25
  
6  0
16 16 4
 1  k  100  96  0
k  3
(b) By (a), we have f ( x)  4 x 3  3x 2  25x  6.
By long division,
x2  x  6
4 x  1 4 x 3  3 x 2  25 x  6
4x3  x 2
 4 x 2  25 x
 4x 2  x
 24 x  6
 24 x  6
2
∴ f ( x)  (4 x  1)( x  x  6)
 (4 x  1)( x  2)( x  3)
9.
It is given that g(x) = ax3 – 17x2 + bx + 15. When g(x) is divided by x – 1, the remainder is –4, and x – 3 is
a factor of g(x).
(a) Find the values of a and b.
(b) Hence, factorize g(x) completely.
(a) When g(x) is divided by x – 1,
g (1)  4
a(1) 3  17(1) 2  b(1)  15  4
a  b  2
∵ x – 3 is a factor of g(x).
g (3)  0
∴
a(3) 3  17(3) 2  b(3)  15  0
27 a  153  3b  15  0
27 a  3b  138
9a  b  46
(2) – (1),
8a  48
(1)
(2)
a6
By substituting a = 6 into (1),
(6)  b  2
b  8
Therefore, a = 6 and b = 8.
(b) By (a), we have g ( x)  6 x 3  17 x 2  8x  15.
By long division,
6x 2  x  5
x  3 6 x 3  17 x 2  8 x  15
6 x 3  18 x 2
x 2  8x
x 2  3x
 5 x  15
 5 x  15
2
∴ g ( x)  ( x  3)(6 x  x  5)
 ( x  3)( x  1)(6 x  5)