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Revision Question Bank
Polynomials
1.
Find the remainder when 4x3 –3 x2 +2x –4 is divided by
(a) x –1
(b) x + 2
(c) x +
1
2
Solution:
We know by remainder theorem, that when polynomial p(x) is divided by another linear
polynomial (x – a), then remainder is given by p(a).
(a) Let p(x) = 4x3 – 3x2 + 2x – 4
 remainder is p(1)
i.e., p(1) = 4(1)3 –(1)3 + 2(1) – 4
=4–3+2–4
=–1
(b) p(–2) = 4(–2)3 – 3(–2)2 + 2(–2) – 4
= –32 – 12 – 4 – 4
= –44 – 8
= – 52
3
2
 1
 1
 1 
 1 
(c) p     4     3   2   4
 2
 2
 2 
 2 
1
1
1
= 4   3   2   4
8
4
2
1 3
=   1 4
2 4
1 3
=   5
2 4
=
2  3  20
4
=
5  20
25

4
4
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2.
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Find the value of k, if (x – 2) is a factor of the polynomial p(x)= kx2 – 2 x +1.
Solution:
If (x – 2)is a factor of the polynomial.
 k 2  2 2  1  0
2
 4k  2 2  1
 k
3.
2 2 1
4
Evaluate (103)3 by using suitable identities.
Solution:
(103)3 = (100 + 3)3
= (100)3 + (3)3 + 3 × 100 × 3 (100 + 3)
(using identity (a + b)3 = a3 + b3 + 3ab (a + b))
= 1000000 + 27 + 900 × 103
= 1092727
4.
Factorize: 3u3 –4u2 –12u +16.
Solution:
3u3 – 4u2 – 12u + 16
= u2(3u – 4) – 4(3u – 4)
= (3u – 4)( (u2 – 4)
= (3u – 4) (u – 2) (u + 2)
(using a2 – b2 = (a + b) (a – b))
5.
Using a suitable identity and without multiplying directly, find the value of (0.98 × 1.02).
Solution:
(0.98 × 1.02) = (1.00– 0.02) (1.00 + 0.02)
= (1.00)2 – (0.0)2
[
(a – b) (a + b) = a2 – b2]
= 1 – 0.0004
= 0.9996
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6.
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For what value of m is 2x3 + mx2 + 11x + m + 3 exactly divisible by (2x–1)?
Solution:
Let p(x) = 2x3 + mx2 + 11x + m + 3.
We know that p(x) will be exactly divisible by (2x – 1) if remainder is zero
1
i.e. if p   = 0
2
3
2
1
1
1
1
 p    2   m    11   m  3  0
2
 2
 2
 2
1
1
1
 2   m   11   m  3  0
8
4
2

1 m 11
 
 m30
4 4 2

1  22  12 m  4m

0
4
4

35 5m

0
4
4
 35 + 5m = 0
 5m = – 35
 m=
35
5
 m=–7
7.
If x +
1
1
=7, then find the value of x3  3 .
x
x
Solution:
1
We have, x   7
x
We know that,
3
1
1
1


3
x


x


3
x



x 
x3
x 


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=
7 
3
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 3  7  342  21
= 322
8.
If a + b=10 and a2 + b2 = 58, then find the value of a3 + b3.
Solution:
We have,
a + b = 10 and a2 + b2 = 58
We know that,
(a + b)2 = a2 + b2 + 2ab
 (10)2 = 58 + 2ab
 100 = 58 + 2ab
 2ab = 100 – 58
 2ab = 42
 ab 
42
 21
2
Now, we know that,
a3 + b3 = (a + b)3 – 3ab(a + b)
= (10)3 – 3 × 21 ×10
= 1000 – 630 = 379
9.
The polynomial p(x)=4x3 – 2x2 + px + 5and q(x)=x3 + 6x2 + p, leave the remainders a and b
respectively, when divided by (x+2). Find the value of p, if a + b=0.
Solution:
We have, p(x) = 4x3 – 2x2 + px + 5
and
q(x) = x3 + 6x2 + p
When p(x) and q(x) divided by (x + 2), then leave the remainders a and b, respectively.
p(–2) = 4(–2)3 – 2 – (2)2 + p(–2) + 5
 a = – 32 – 8 – 2p + 5
 a = – 35 – 2p
and
p(–2) = (–2)3 + 6(–2)2 + p
 b = – 8 + 24 + p
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 b = 16 + p
Given, a + b = 0
 –2p – 35 + p  +16 = 0
 –p – 19 = 0
 –p = 19
 p = – 19
10. Factorize:(3a –5b)3 + (4c –3a)3 + (5b – 4c)3.
Solution:
Given expression is
(3a – 5b)3 + (4c – 3a)3 + (5b – 4c)3
Now, (3a – 5b) + (4c – 3a) + (5b – 4c)
= 3a – 3a – 5b + 4c – 4c
=0
 (3a – 5b)3 + (4c – 3a)3 + (5b – 4c)3
= 3(3a – 5b) (4c – 3a)(5b – 4c)
[
if a + b + c = 0, then a3 + b3 + c3 = 3abc]
Chapter Test {Polynomials}
M:M: 40
1.
Factorize: (i) 36a3b – 60a2bc
M: Time: 40Min.
(ii) 5x2 – 20xy
[4]
Solution:
(i) 36a3b – 60a2bc = 12a2b(3a – 5c)
(ii) 5x2 – 20xy = 5x(x – 4y)
2.
Find the value of 64x3 – 125y3, if
4x – 5y = 16
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and xy = 12
[4]
Solution:
Given, 4x – 5y =16
On cubing both sides, we get
(4x-5y)3=(16)3
 (4x)3 – (5y)3 – 3 × 4 × 5y(4x – 5y) = (16)3
[using (a – b)3 = a3 – b3 – 3ab(a – b)]
 64x3 – 125y3 – 60xy(4x – 5y) = 4096
 64x3 –125y3 – 60 ×12 × 16 = 4096
[
xy = 12 and 4x – 5y = 16, given]
 64x3 –125y3 –11520 = 4096
 64x3 –12573 =4096+11520
Hence,
3.
64x3 –125y3 =15616
Find the zeroes of the polynomial f(x) = 2x3 + 3x2 – 11x – 6.
[4]
Solution:
Let
f(x) = x40 + x51 +5
If f(x) is divisible by (x + 1), then remainder is f(–1).
Remainder = f(–1)
= (-1)40 +(–1)51 +5
=1– 1 + 5 = 5
4.
If x40 + x51 + 5 is divisible by x + 1, then find the remainder.
[4]
Solution:
Consider f(x) = ax2 + 5x + b
1

(x – 3) and  x   are factors of f(x).
3

1
 f(3) = 0 and f   = 0
3
 a × 32 +5 × 3 + b = 0
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2
1
1
a  5   b  0
3
3
and
 9a + 15 + b
and
=0
a 5
 b0
9 3
 9a + b+15 = 0
and
a +9b + 15 = 0
 9a + b = –15
… (i)
and
...(ii)
a + 9b =–15
From Eqs. (i) and (ii), we get
9a + b = a + 96
 8a = 8b
 a=b
5.
1

If (x – 3) and  x   are both factors of ax2 + 5x + b, then show that a = b.
3

[4]
Solution:
Given, f(x)=x2 –5x + 1
...(i)
Put x = 2 in Eq. (i),
f(2) = (2)2
– 5 × 2+1
= 4 – 10 + 1
= 5 – 10 =
Put x =
–5
– 1 in Eq. (i),
f(–1) = (–1)2–5 ×
(–1) + 1
=1+ 5 + 1 = 7
Put x =
1
in Eq. (i),
3
2
1
1 1
f      5  1
3
3 3
=
1 5
1  15  9
5
 1 

9 3
9
9
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= 5  7 
5
5
 12 
9
9
108  5 113

9
9
=
6.
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1
If f(x) = x2 – 5x + 1, then evaluate: f 2  f  1  f   .
3
[4]
Solution:
f(x) = x2 – 5x + 1
 f(2) = (2)2 – 5(2) + 1
= 4 – 10 + 1
= –6 + 1 = – 5
f(–1) = (–1)2 – 5(–1) + 1
=1+5+1
=7
2
1 1
1
f       5   1
3 3
3
=
1 5
15 9
 1 
9 3
9
=
14  9 5

9
9
5
5
1
= 12 
 f(2) – f(–1) + f   = –5 – 7 –
9
9
3
=
7.
108  5 113

9
9
Factorize: the following: 12(x2 + 7x)2 – 8(x2 + 7x) (2x – 1) – 15(2x – 1)2.
[4]
Solution:
12(x2 + 7x)2 – (8x2 + 7x) (2x – 1) – 15(2x – 1)2
let a = x2 + 7x, b = 2x – 1
Therefore, we have
12a2 – 8ab – 15b2
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2


 a a

= b 12      15
 b  b



2
2


 a

a
a
= b 12   18    10   15
 b
 b
 b



2
  a    a     a  
= b2 6   2   3  5 2   3
  b    b     b  
 6a
 2a

= b2   5   3 
 b
 b

b2  6a  5b 2a  3b
=
b2
= (6a + 5b) (2a – 3b)
= {6(x2 + 7x) + 5(2x – 1)} {2(x2 + 7x) –3(2x – 1)}
= (6x2 + 42x + 10x – 5) (2x2 + 14x – 6x + 3)
= (6x2 + 52x – 5) (2x2 + 8x + 3)
8.
Factorize: x4 + x3 –7x2 – x + 6.
[4]
Solution:
x4 + x3 –7x2 – x + 6
= x4 + x3 – (7x2 + x – 6)
= x4 + x3 – (7x2 + 7x – 6x – 6)
= x4 + x3 – {7x(x + 1) –6(x + 1)}
= x4 + x3 – (7x – 6) (x + 1)
= x3 (x + 1) – (7x – 6) (x + 1)
= (x + 1) (x3 – 7x + 6)
= (x + 1) (x3 + x2 – 7x + 6 – x2)
(adding and subtracting x2)
= (x + 1) {x3 – x2 + (x2 – 7x + 6)}
= (x + 1) {x3 – x2 + (x2 – 6x –1x + 6)}
= (x + 1) {x2 (x – 1) + (x –1) (x –6)}
= (x + 1) (x – 1) (x2 + x – 6) = (x + 1) (x – 1) (x + 3) (x – 2)
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9.
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Factorize: a8 – b8.
[4]
Solution:
=
a    b 
a  b a  b 
=
=
4
a 8 – b8 =
2
2
4
4
4
4
4
a
4
 b4
a
4
 b4 a2  b2 a2  b2
a    b  
2
2
2

2

    


2
2
=  a2  b2  a2  b2  a  b a  b


=
a
2
=
a
 b2  2ab a2  b2  2ab a2  b2  a  b a  b
2

2
 b2  2a2b2
 a  b a  ba  b
2
2



10. Simplify (a + b)3 – (a – b)3 – 6b(a2 – b2).
[4]
Solution:
(a + b)3 – (a – b)3 – 6b(a2 – b2)
= {(a + b) – (a – b)} {(a + b)2 + (a – b)2 + (a + b) (a – b)2 –6b(a2 – b2)
= (a + b – a + b) (a2 + b2 + 2ab + a2 + b2 – 2ab + a2 – b2)
= 2b(2a2 + 2b2 + a2 – b2) –6b(a2 – b2)
= 2b(3a2 + b2 –3a2 + 3b2)
= 2b(4b2)
= 8b3.
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