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Chapter 06 – Polynomials Q1
x −2 − y −2
x −1 − y −1
( x −1 ) 2 − ( y −1 ) 2
=
x −1 − y −1
2ab − a 2 − b 2
= −(a 2 − 2ab + b 2 )
= −( a − b ) 2
= x −1 + y −1
Option A
Option E
Chapter 06 – Polynomials Q3
Chapter 06 – 1Polynomials Q4
x+
1
1
x
+
+
x +1 x −1 x − 1
x
x − 1 + x + 1 x2 + 1
=
+ 2
x2 −1
x −1
x2 + 2 x + 1
=
x2 −1
( x + 1) 2
x +1
=
=
( x − 1)( x + 1) x − 1
x+ y
1
1
=
+
a+b a −b
a −b+ a+b
=
a 2 − b2
2a
= 2 2
a −b
Option D
Option C
Chapter 06
– Polynomials Q5
−1
x y

 y + x +2


 x−y 
 y x 


−1
 x 2 + y 2 + 2 xy 
=

2
2
 x −y

−1
 ( x + y)2 
x− y
=
 =
(
)(
)
x
+
y
x
−
y
x
+y


Option A
Chapter 06 – Polynomials Q2
Chapter 06 – Polynomials Q6
b3
c4
second expression
third expression
a3 b2
a 4 b6
c2
c
H.C.F.
L.C.M.
a2
a4
first
expression
=
=
a2
b2
b6
c
c4
Option D
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Chapter 06 – Polynomials Q7
2a
1
+
2
2
a − 4b
2b − a
2a
2b + a
= 2
−
a − 4b 2 a 2 − 4b
a − 2b
= 2
a − 4b 2
1
=
a + 2b
Option A
Chapter 06 – Polynomials Q9
6
5
−
x − 9 x2 + x − 6
6
5
=
−
( x − 3)( x + 3) ( x − 2)( x + 3)
6( x − 2) − 5( x − 3)
=
( x − 3)( x − 2)( x + 3)
x+3
1
=
=
( x − 3)( x − 2)( x + 3) ( x − 3)( x − 2)
Option A
2
Chapter 06 – Polynomials Q11
a 3 − 1 = ( a − 1)(a 2 + a + 1)
a 4 − 1 = ( a 2 − 1)( a 2 + 1) = ( a − 1)(a + 1)( a 2 + 1)
H.C.F. = a − 1
Option C
Chapter 06 – Polynomials Q8
( a −2 − 3b −1 ) −1−1
 1 3
= 2 − 
a b
 b − 3a 2 
= 2 
 a2 b 
ab
=
b − 3a 2
−1
Option D
Chapter 06 – Polynomials Q10
1 1
1 1 1
1
1
+ 3 ( + )( 2 −
+ 2)
3
a b = a b a
ab b
1 1
1 1
+
+
a b
a b
1
1
1
= 2−
+ 2
a
ab b
Option C
Chapter 06 – Polynomials Q12
f ( x ) = (2 x + 1)Q ( x ) + R
−1
−1
f ( ) = 0Q ( ) + R = R
2
2
Option E
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Chapter 06 – Polynomials Q13
4
3
−
( x − 2)( x + 1) x 2 − 1
4
3
=
−
( x − 2)( x + 1) ( x − 1)( x + 1)
4( x − 1) − 3( x − 2)
=
( x − 2)( x + 1)( x − 1)
x+2
=
( x − 2)( x + 1)( x − 1)
Option B
Chapter 06 – Polynomials Q15
2
1
4x
−
− 2
1 + x 1 − x x −1
2
1
4x
=
−
+
1 + x 1 − x (1 − x)(1 + x)
2(1 − x ) − (1 + x ) + 4 x
=
(1 − x )(1 + x )
1+ x
1
=
=
(1 − x)(1 + x) 1 − x
Option A
Chapter 06 – Polynomials Q17
( x − y ) −1 ( x −2 − y −2 )
1  1
1 
=
 2− 2
x− y x
y 
1
y 2 − x2
=
× 2 2
x− y
x y
1
( y − x )( y + x) − y − x −1
1
=
×
= 2 2 = 2 − 2
2 2
x− y
x y
x y
x y xy
Option B
Chapter 06 – Polynomials Q14
f ( x ) = x 2 + ax + b
f ( −2) = ( −2) 2 + a ( −2) + b = 0
2a − b = 4
2a − b + 3 = 4 + 3 = 7
Option E
Chapter 06 – Polynomials Q16
2
2
b a b −a
−
a b = ab
b−a
1 1
−
ab
a b
b 2 − a 2 (b − a )(b + a )
=
=
b−a
b−a
= a+b
Option A
Chapter 06 – Polynomials Q18
2a 2 − 2b 2 = 2( a − b)( a + b)
a 3 − 2a 2b + ab 2 = a (a − b) 2
L.C.M. = 2a( a − b) 2 ( a + b)
Option D
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Chapter 06 – Polynomials Q19
f ( x) = 3 x3 − ax 2 + 5 x − 3b
f ( −3) = 3( −3)3 − a (−3) 2 + 5(−3) − 3b = 0
9a + 3b = −96
3a + b = −32
Option A
Chapter 06 – Polynomials Q21



x
+
y
y
x
 x+ y
=
 xy 


( x + y )2
=
xy



Chapter 06 – Polynomials Q20
x2
1 2
− 3 y2
(x − 9 y2 )
3
3
=
3
3
( x − 3 y)
( x − 3 y)
2
12
( x − 3 y )( x + 3 y )
2
=3
= ( x + 3 y)
3
9
( x − 3 y)
2
Option E
Chapter 06 – Polynomials Q22
2
2
Option A
Chapter 06 – Polynomials Q23
12a 2 b = 2 2 × 3a 2b
18ab 3c = 2 × 32 ab 3c
L.C.M. = 22 × 32 a 2b3c = 36a 2b3c
Option D
Chapter 06 – Polynomials Q24
2
2 x 3 + 3 x 2 − 11x − 6
= ( x − 2)(2 x 2 + 7 x + 3)
= ( x − 2)( x + 3)(2 x + 1)
Option C
2
 x + 1   x −1 

 −

 x   x 
 x + 1 x −1   x + 1 x −1 
=
−
+


x  x
x 
 x
2
= ×2
x
4
=
x
Option B
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Chapter 06 – Polynomials Q25
f ( −1) = ( −1) 2 + p ( −1) + q = 4
−p+q =3
2 p − 2q + 1 = −2(− p + q ) + 1
Chapter 06 – Polynomials Q26
( x 2 − 3 x)(4 x 2 − 1) = x( x − 3)(2 x − 1)(2 x + 1)
(2 x − 1)( x 2 − 6 x + 9) = ( x − 3) 2 (2 x − 1)
H.C.F. = ( x − 3)(2 x − 1)
= −2(3) + 1 = −5
Option B
Chapter 06 – Polynomials Q27
x 2 − 2 x x 2 − 2 x − 15
×
x3 − 25 x x 2 + x − 6
x( x − 2)
( x − 5)( x + 3)
=
×
x( x − 5)( x + 5) ( x − 2)( x + 3)
1
=
x+5
Option C
Chapter 06 – Polynomials Q29
1
1
+ 2
2
2x − x
x + x−6
1
1
=
+
x(2 − x) ( x − 2)( x + 3)
( x + 3)
x
=
−
x(2 − x )( x + 3) x (2 − x)( x + 3)
3
=
x(2 − x)( x + 3)
Option A
Option C
Chapter 06 – Polynomials Q28
f (1) = a + b + c = 10⋯ (1)
f (−1) = a − b + c = 6⋯ (2)
(1) − (2) : 2b = 4
b=2
Option C
Chapter 06 – Polynomials Q30
x 3 − 125 = ( x − 5)( x 2 + 5 x + 25)
4 x 2 − 9 y 2 = (2 x − 3 y )(2 x + 3 y )
x 3 + 125 = ( x + 5)( x 2 − 5 x + 25)
3 x 2 + 6 xy + 3 y 2 = 3( x + y ) 2
Option D
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Chapter 06 – Polynomials Q31
H.C.F. of 24ab 2 c 3 and 12a 2bc 4 is 12abc 3
H.C.F. of 24ab 2 c 3 and 30 a 2 bc 3 is 6abc 3
H.C.F. of 24ab 2 c 3 and 32 a 2 bc 5 is 8abc 3
H.C.F. of 24ab 2 c 3 and 40ab 2 c 3 is 8ab 2 c 3
H.C.F. of 24ab 2 c 3 and 48a 3bc 5 is 24abc 3
Option C
Chapter 06 – Polynomials Q33
g ( −31 ) = f ( −31 )
27( −31 )3 − 18( −31 ) + 4 = a ( −31 ) 2 − 5
a
−1 + 6 + 4 = − 5
9
a = 126
Option E
Chapter 06 – Polynomials Q35
x+ y−x+ y
x− y
x+ y
x+ y
=
x− y−x− y
x+ y
1−
x− y
x− y
2y
x− y
=
×
x + y −2 y
y−x
=
x+ y
Option A
1−
Chapter 06 – Polynomials Q32
27 x 3 − 8
3x − 2
(3 x − 2)  (3 x ) 2 + (3 x )(2) + (2) 2 
=
3x − 2
= 9 x2 + 6x + 4
Option E
Chapter 06 – Polynomials Q34
(1 − x 2 ) n + (1 − x ) n
(1 − x ) 2 n
(1 − x ) n (1 + x ) n + (1 − x ) n
=
(1 − x ) 2 n
n
(1 − x )  (1 + x ) n + 1
=
(1 − x ) 2 n
(1 + x ) n + 1
=
(1 − x ) n
Option A
Chapter 06 – Polynomials Q36
a 3 + 8 a −3
3
2
= a3 +  
a
2 
2
2 

=  a +   a 2 − a( ) + ( )2 
a
a
a



2 2
4
= ( a + )( a − 2 + 2 )
a
a
Option E
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Chapter 06 – Polynomials Q37
f ( k ) = 3k 3 − 4k + k = 3k 3 − 3k = 0
The required remainder
= f (− k ) = 3( − k )3 − 4( − k ) + k = −3k 3 + 5k
= −(3k 3 − 3k ) + 2k = 2k
Option A
Chapter 06 – Polynomials Q39
1
1
−
2
1− x
(1 + x ) 2
1
1
=
−
(1 − x )(1 + x ) (1 + x ) 2
1+ x
1− x
=
−
(1 − x )(1 + x ) 2 (1 − x )(1 + x ) 2
2x
=
(1 − x )(1 + x ) 2
Option E
Chapter 06 – Polynomials Q41
1
+
x3
1
+
x
 1 1  1 2 1 1
1 2
1
 x + y   ( x ) − ( x )( y ) + ( y ) 
3
y


=
1
1 1
+
y
x y
1
1
1
= 2− + 2
x
xy y
Option E
Chapter 06 – Polynomials Q38
first
expression
second expression
third
expression
H.C.F.
=
L.C.M.
=
x2
x3
x
y3
y
y5
z3
z2
z4
x
x3
y
y5
z2
z4
Option D
Chapter 06 – Polynomials Q40
x 3 − 4 x 2 + x + 6 = ( x + 1)( x 2 − 5 x + 6)
= ( x + 1)( x − 2)( x − 3)
Option A
Chapter 06 – Polynomials Q42
The L.C.M. of x, 2 x 2 ,3 x 3 , 4 x 4 , 5 x 5
= 60x 5
Option C
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Chapter 06 – Polynomials Q43
Chapter 06 – Polynomials Q44
1 1
+
a b
b
a
=
+
ab ab
a+b
=
ab
2
4 (a + b) − 9 ( a − b)
= [2 ( a + b )]2 − [3 ( a − b )]2
= [ 2( a + b) + 3( a − b) ][ 2( a + b) − 3(a − b) ]
= (5a − b )(5b − a )
Option A
Option A
Chapter 06 – Polynomials Q45
L.C.M. of
L.C.M. of
P and Q
X , Y and Z
L.C.M. of
P, Q, X , Y , Z
b3
b3
c2
c
a2
b3
c2
60
)(
)
3x + 1 x2 + 3x + 1
2
)(
+ 1) − 3 x ( x 2 + 1) + 3 x
2
f ( x ) = ( x − 1)Q ( x )
f ( x − 1) = ( x − 1 − 1)Q ( x − 1)
f ( x − 1) = ( x − 2)Q ( x − 1)
which is divisible by x − 2
Option A
Chapter 06 – Polynomials Q47
2
Chapter 06 – Polynomials Q46
12 a
30 a 2
Option B
(x −
= ((x
2
2
= ( x + 1) − ( 3 x )
2
= x 4 + 2 x 2 + 1 − 3x 2
)
Chapter 06 – Polynomials Q48
Let f ( x ) = x 2 − 2 x + k
f ( −1) = ( −1) 2 − 2( −1) + k = 0
k = −3
f ( −3) = ( −3) 2 − 2(−3) − 3 = 12
= x4 − x2 + 1
Option B
Option C
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Chapter 06 – Polynomials Q49
a b2
c
a b
H.C.F. a b
2
L.C.M. a b
c3
c
c3
Chapter 06 – Polynomials Q50
a 4 + a 2b 2 + b 4
= a 4 + 2a 2b 2 + b 4 − a 2b 2
= (a 2 + b 2 ) 2 − ( ab) 2
= [( a 2 + b 2 ) − (ab )][( a 2 + b 2 ) + ( ab)]
= (a 2 − ab + b 2 )( a 2 + ab + b 2 )
Option B
Option D
Chapter 06 – Polynomials Q51
( x − 1) 2
x2 −1
x3 − 1
=
( x − 1) 2
=
( x − 1)( x + 1)
= ( x − 1)( x 2 + x + 1)
L.C.M. ( x − 1) 2 ( x + 1)( x 2 + x + 1)
Option C
Chapter 06 – Polynomials Q53
2y − x
2 1
−
xy
x y
=
2
4y x
4 y − x2
−
x y
xy
2y − x
=
(2 y − x )(2 y + x )
1
=
2y + x
Option D
Chapter 06 – Polynomials Q52
a 2 − 2ab + b 2 − a + b
= ( a 2 − 2ab + b 2 ) − ( a − b)
= (a − b) 2 − ( a − b)
= (a − b)( a − b − 1)
Option A
Chapter 06 – Polynomials Q54
P ( x ) = (5 x − 2)Q ( x ) + R
P ( x ) = (2 − 5 x )[ −Q ( x )] + R
Option A
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Chapter 06 – Polynomials Q55
f ( −1) = ( −1)99 + 99( −1) + k = 0
k = 100
x3 − x
=
x ( x − 1)( x + 1)
4
x − 1 = ( x − 1)( x + 1)( x 2 + 1)
L.C.M. x ( x − 1)( x + 1)( x 2 + 1)
Option D
Option C
Chapter 06 – Polynomials Q57
2a n +1 − 7 a n − 30a n −1
= a n −1 (2a 2 − 7 a − 30)
= a n −1 ( a − 6)(2a + 5)
Option E
Chapter 06 – Polynomials Q59
4 x 2 yz
6 xy 3
L.C.M. 12 x 2 y 3 z
Option C
Chapter 06 – Polynomials Q56
Chapter 06 – Polynomials Q58
y−x y−x
x
 y 
×
 − 1 1 − 
x
y
y
 x 
=
2
2
x y
x −y
−
y x
xy
( y − x)2
=
( x − y )( x + y )
x− y
=
x+ y
Option A
Chapter 06 – Polynomials Q60
Let f ( x ) = x 3 − x 2 + 1
The required remainder
−1
= f( )
23
2
 −1   −1 
=   −   +1
 2   2 
5
=
8
Option B
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Chapter 06 – Polynomials Q61
ab − ac = a (b − c )
a (b − c ) − b + c = ( a − 1)(b − c )
a (b − c ) − b − c cannot be factorized
Option B
Chapter 06 – Polynomials Q63
9 − a 2 − b 2 + 2ab
= 9 − ( a 2 − 2ab + b 2 )
= 32 − ( a − b ) 2
= (3 − a + b )(3 + a − b)
Chapter 06 – Polynomials Q62
1
1
3x − 1
+
+
x −1 x + 1 1 − x2
−1
1
3x − 1
=
+
+
1 − x 1 + x 1 − x2
−1 − x + 1 − x + 3 x − 1
=
1 − x2
x −1
−1
=
=
1 − x2 1 + x
Option C
Chapter 06 – Polynomials Q64
Let f ( x ) = 2 x 2 + x + m
f (2) = 0
2(2) 2 + 2 + m = 0
m = −10
f ( x ) = 2 x 2 + x − 10 = (2 x + 5)( x − 2)
Option D
Chapter 06 – Polynomials Q65
4
3
−
x2 − 4 x2 − x − 2
4
3
=
−
( x − 2)( x + 2) ( x + 1)( x − 2)
4( x + 1) − 3( x + 2)
=
( x − 2)( x + 2)( x + 1)
x−2
1
=
=
( x − 2)( x + 1)( x + 2) ( x + 1)( x + 2)
Option A
Option E
Chapter 06 – Polynomials Q66
f ( x) = 2 x 3 − x 2 − 7 x + 6
f ( x ) = ( x − 1)(2 x 2 + x − 6) ∵ f (1) = 0
f ( x ) = ( x − 1)( x + 2)(2 x − 3) ∵ f ( −2) = 0
Option D
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Chapter 06 – Polynomials Q67
x2 − y2 + 2 x + 1
= x2 + 2x + 1 − y2
= ( x + 1) 2 − y 2
= ( x + 1 − y )( x + 1 + y )
Option B
Chapter 06 – Polynomials Q69
x2 − y2 − x + y
= (x2 − y 2 ) − ( x − y)
Chapter 06 – Polynomials Q68
2
3
− 2
x −1 x − x − 2
2
3
=
−
( x − 1)( x + 1) ( x + 1)( x − 2)
2( x − 2) − 3( x − 1)
=
( x − 1)( x + 1)( x − 2)
−x −1
−1
=
=
( x − 1)( x + 1)( x − 2) ( x − 1)( x − 2)
2
Option A
Chapter 06 – Polynomials Q70
F (1) = 0, 13 − 4(1) 2 + a (1) + b = 0, a + b = 3⋯ (1)
F ( −1) = 12, (−1)3 − 4(−1) 2 + a (−1) + b = 12, a − b = −17 ⋯ (2)
= ( x − y )( x + y ) − ( x − y )
(1) + (2), 2a = −14, a = −7
= ( x − y )( x + y − 1)
(1) − (2), 2b = 20, b = 10
Option B
Chapter 06 – Polynomials Q71
2
x −1
− 2
2
x −1 x − 2x − 3
2
x −1
=
−
( x − 1)( x + 1) ( x + 1)( x − 3)
2( x − 3) − ( x − 1) 2
=
( x − 1)( x + 1)( x − 3)
−x2 + 4x − 7
=
( x − 1)( x + 1)( x − 3)
Option E
Option E
Chapter 06 – Polynomials Q72
x 2 − x − xy + y
= x 2 − x − ( xy − y )
= x ( x − 1) − y ( x − 1)
= ( x − y )( x − 1)
Option A
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Chapter 06 – Polynomials Q73
f ( x) = x3 − 2 x 2 − 5 x + 6
= ( x − 1)( x 2 − x − 6) ∵ f (1) = 0
= ( x − 1)( x + 2)( x − 3)
Option D
Chapter 06 – Polynomials Q75
(2 x 2 − 3 x + 1)(2 − 3 x )
= 4 x2 − 6 x + 2 − 6 x3 + 9 x 2 − 3x
= −6 x 3 + 13 x 2 − 9 x + 2
Chapter 06 – Polynomials Q74
a
b
2ab
+
+ 2
a + b b − a a − b2
a
b
2ab
=
−
+
a + b a − b a2 − b2
a ( a − b) − b( a + b) + 2ab
=
a2 − b2
2
2
a −b
= 2
=1
a − b2
Option E
Chapter 06 – Polynomials Q76
The required remainder
  −1     −1  
 −1 
=  2   − 1   + 1 + 2   + 1
2
2




 2 



= −1
Option C
Chapter 06 – Polynomials Q77
2( a − b) 2 − a 2 + b 2
= 2(a − b) 2 − ( a 2 − b 2 )
= 2( a − b ) 2 − ( a − b)( a + b)
= (a − b) [ 2(a − b) − ( a + b) ]
= (a − b)( a − 3b)
Option A
Option A
Chapter 06 – Polynomials Q78
1− x
x −1
+
x + 4x − 5 x +1
1− x
x −1
=
+
( x + 5)( x − 1) x + 1
−( x − 1)( x + 1) + ( x − 1) 2 ( x + 5)
=
( x + 5)( x − 1)( x + 1)
( x − 1) [ −( x + 1) + ( x − 1)( x + 5) ]
=
( x − 1)( x + 5)( x + 1)
x 2 + 3x − 6
=
( x + 5)( x + 1)
Option A
2
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Chapter 06 – Polynomials Q79
f ( −1) = 0, ( −1)3 + 2( −1) 2 + a ( −1) + b = 0, a − b = 1⋯ (1)
f (2) = 0, 23 + 2(2) 2 + a (2) + b = 0, 2a + b = −16⋯ (2)
(1) + (2), 3a = −15, a = −5
b = −6
f ( x ) = x 3 + 2 x 2 − 5 x − 6 = ( x + 1)( x − 2)( x + 3)
Option D
Chapter 06 – Polynomials Q80
2x
1
x−
x
2x2
= 1− 2
x −1
− x2 −1
= 2
x −1
x2 + 1
=− 2
x −1
Option D
1−
Chapter 06 – Polynomials Q81
Chapter 06 – Polynomials Q82
(−2) 2 + a ( −2) + b = −4, 2a − b = 8⋯ (1)
(−1)3 + 2(−1) 2 + k = 0
a (2) 2 + b(2) + 1 = 9, 2a + b = 4⋯ (2)
k = −1
(1) + (2), 4a = 12, a = 3
f ( x) = x 3 + 2 x 2 − 1
The remainder
= f (1) = 13 + 2(1) 2 − 1 = 2
Option D
Chapter 06 – Polynomials Q83
10
2
−
2
x + x−6 x−2
10
2
=
−
( x − 2)( x + 3) x − 2
10 − 2( x + 3)
=
( x − 2)( x + 3)
−2( x − 2)
=
( x − 2)( x + 3)
−2
=
x+3
Option B
Option C
Chapter 06 – Polynomials Q84
210 xy 2
30 x 2 yz
L.C.M. 210x 2 y 2 z
Option C
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Chapter 06 – Polynomials Q85
27
x − 3
x
3
3
= x3 −  
x
2
3 

3 3 
=  x −   x2 + x   +   
x  

 x   x  
3
9
= ( x + )( x 2 + 3 + 2 )
x
x
3
Option D
Chapter 06 – Polynomials Q87
2−b
=
2−b
4 − b2
=
(2 − b)(2 + b)
8 − b3
= (2 − b)(2 + 2b + b 2 )
Chapter 06 – Polynomials Q86
3 2
−
x y
4x 9 y
−
y
x
3x − 2 x
= 2
4x − 9 y2
3y − 2x
=
(2 x − 3 y )(2 x + 3 y )
−1
=
2x + 3y
Option D
Chapter 06 – Polynomials Q88
x 3 − 7 x + 6 = ( x 2 − 3 x + k )( x − a )
x 3 − 7 x + 6 = x 3 − ( a + 3) x 2 + ( k + 3a ) x − ka
2
L.C.M. (2 − b )(2 + b)(4 + 2b + b )
Option D
0 = −( a + 3), a = −3
6 = − ka, k = 2
Option B
Chapter 06 – Polynomials Q89
Chapter 06 – Polynomials Q90
(2 x − 3)( x 2 + 3 x − 2)
3
2
x 2 ( x + 1)( x + 2)
2
= 2 x + 6 x − 4 x − 3x − 9 x + 6
= 2 x 3 + 3 x 2 − 13 x + 6
Option D
x ( x + 1)3
H.C.F. x ( x + 1)
Option A
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Chapter 06 – Polynomials Q91
The remainder
= (−1) 2 k +1 + k ( −1) + k
= −1 − k + k
= −1
Option A
Chapter 06 – Polynomials Q93
pr + qr − ps − qs
Chapter 06 – Polynomials Q92
1
1
−
x +1 x −1
x − 1 − ( x + 1)
=
x2 − 1
−2
= 2
x −1
2
= 2
x −1
Option A
Chapter 06 – Polynomials Q94
(− k )3 + k ( − k ) 2 + 2k ( − k ) + 3k = k
= pr + qr − ( ps + qs )
− k 3 + k 3 − 2k 2 + 3k = k
= r ( p + q) − s( p + q)
2k 2 − 2k = 0
k = 1 or 0 (rejected)
= ( p + q )( r − s )
Option A
Chapter 06 – Polynomials Q95
1
1
−
n +3 3− n
1
1
=
+
n+3 n−3
n − 3 + ( n + 3)
=
n2 − 9
2n
= 2
n −9
Option B
Chapter 06 – Polynomials Q96
( x + x )( y + y + y )
= (2 x )(3 y )
= 6 xy
Option A
Option D
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Chapter 06 – Polynomials Q97
Chapter 06 – Polynomials Q98
f ( x ) = ( x − 1)Q ( x )
−k
1
−
1− k k −1
k
1
=
−
k −1 k −1
k −1
=
=1
k −1
f (2 x + 1) = [ (2 x + 1) − 1] Q (2 x + 1)
= 2 xQ (2 x + 1)
Option A
Option A
Chapter 06 – Polynomials Q99
Chapter 06 – Polynomials Q100
(2 x 2 − 3 x + 1) − 2( x 2 + 2 x − 1)
x 2 − y 2 = ( x − y )( x + y )
= 2 x 2 − 3x + 1 − 2 x 2 − 4 x + 2
x 2 + y 2 cannot be factorized
= −7 x + 3
Option B
Chapter 06 – Polynomials Q101
The required remainder
= (−1) 2009 + ( −1) 2008 + ( −1) 2007 + ⋯ + ( −1)1
= −1 + 1 − 1 + 1 − ⋯ − 1 = −1
x ( x + y ) − x − y = x ( x + y ) − ( x + y ) = ( x − 1)( x + y )
Option C
Chapter 06 – Polynomials Q102
1
1
+
2x − 3 2x + 3
2x + 3
2x − 3
= 2
+
4 x − 9 4 x2 − 9
4x
= 2
4x − 9
Option A
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Chapter 06 – Polynomials Q103
ab + ac − a 2 − bc
Chapter 06 – Polynomials Q104
(1)3 + 5(1) 2 + 3k (1) − k = 0
= ab − a 2 + ac − bc
1 + 5 + 3k − k = 0
= a (b − a ) + c ( a − b )
= c (a − b) − a ( a − b )
2 k = −6
k = −3
= (a − b )(c − a )
Option B
Option A
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