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CHAPTER 2

POLYNOMIALS

(A) Main Concepts and Results

Meaning of a Polynomial
Degree of a polynomial
Coefficients
Monomials, Binomials etc.
Constant, Linear, Quadratic Polynomials etc.
Value of a polynomial for a given value of the variable
Zeroes of a polynomial
Remainder theorem
Factor theorem
Factorisation of a quadratic polynomial by splitting the middle term
Factorisation of algebraic expressions by using the Factor theorem
Algebraic identities –
(x + y) 2 = x2 + 2xy + y2
(x – y) 2 = x2 – 2xy + y2
x2 – y2 = (x + y) (x – y)
(x + a) (x + b) = x2 + (a + b) x + ab
(x + y + z) 2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
(x + y) 3 = x3 + 3x2 y + 3xy2 + y3 = x3 + y3 + 3xy (x + y)
(x – y) 3 = x3 – 3x2 y + 3xy2 – y3 = x3 – y3 – 3xy (x – y)

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14 EXEMPLAR PROBLEMS

x3 + y3 = (x + y) (x2 – xy + y2)
x3 – y3 = (x – y) (x2 + xy + y2)
x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z 2 – xy – yz – zx)

(B) Multiple Choice Questions

Sample Question 1 : If x2 + kx + 6 = (x + 2) (x + 3) for all x, then the value of k is
(A) 1 (B) –1 (C) 5 (D) 3
Solution : Answer (C)

EXERCISE 2.1
Write the correct answer in each of the following :
1. Which one of the following is a polynomial?
x2 2
(A) – (B) 2 x −1
2 x2
3
3x 2 x −1
(C) x2 + (D)
x +1
x
2. 2 is a polynomial of degree

1
(A) 2 (B) 0 (C) 1 (D)
2
3. Degree of the polynomial 4x4 + 0x3 + 0x5 + 5x + 7 is
(A) 4 (B) 5 (C) 3 (D) 7
4. Degree of the zero polynomial is
(A) 0 (B) 1 (C) Any natural number
(D) Not defined

( )
5. If p ( x ) = x2 – 2 2 x + 1 , then p 2 2 is equal to

(A) 0 (B) 1 (C)

4 2 (D) 8 2 +1
6. The value of the polynomial 5x – 4x2 + 3, when x = –1 is
(A) – 6 (B) 6 (C) 2 (D) –2

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POLYNOMIALS 15

7. If p(x) = x + 3, then p(x) + p(–x) is equal to

(A) 3 (B) 2x (C) 0 (D) 6
8. Zero of the zero polynomial is
(A) 0 (B) 1
(C) Any real number (D) Not defined
9. Zero of the polynomial p(x) = 2x + 5 is
2 5 2 5
(A) – (B) – (C) (D)
5 2 5 2
10. One of the zeroes of the polynomial 2x2 + 7x –4 is

1 1
(A) 2 (B) (C) – (D) –2
2 2
11. If x51 + 51 is divided by x + 1, the remainder is
(A) 0 (B) 1 (C) 49 (D) 50
2
12. If x + 1 is a factor of the polynomial 2x + kx, then the value of k is
(A) –3 (B) 4 (C) 2 (D) –2
13. x + 1 is a factor of the polynomial
(A) x3 + x2 – x + 1 (B) x3 + x2 + x + 1
(C) x4 + x3 + x2 + 1 (D) x4 + 3x3 + 3x2 + x + 1
14. One of the factors of (25x2 – 1) + (1 + 5x)2 is
(A) 5 + x (B) 5 – x (C) 5x – 1 (D) 10x
15. The value of 249 – 248 is
2 2

(A) 1 2 (B) 477 (C) 487 (D) 497

2
16. The factorisation of 4x + 8x + 3 is
(A) (x + 1) (x + 3) (B) (2x + 1) (2x + 3)
(C) (2x + 2) (2x + 5) (D) (2x –1) (2x –3)
17. Which of the following is a factor of (x + y)3 – (x3 + y3)?
(A) x2 + y2 + 2xy (B) x2 + y2 – xy (C) xy 2 (D) 3xy
18. The coefficient of x in the expansion of (x + 3)3 is
(A) 1 (B) 9 (C) 18 (D) 27
x y
19. If y + x = –1 ( x , y ≠ 0 ) , the value of x3 – y3 is

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16 EXEMPLAR PROBLEMS

1
(A) 1 (B) –1 (C) 0 (D)
2

 1  1
20. If 49x2 – b =  7 x +   7 x –  , then the value of b is
 2  2

1 1 1
(A) 0 (B) (C) (D)
2 4 2
21. If a + b + c = 0, then a3 + b3 + c3 is equal to
(A) 0 (B) abc (C) 3abc (D) 2abc

(C) Short Answer Questions with Reasoning

Sample Question 1 : Write whether the following statements are True or False.
Justify your answer.
3
1
1 2 6 x + x2
(i) x + 1 is a polynomial (ii) is a polynomial, x ≠ 0
5 x
Solution :
(i) False, because the exponent of the variable is not a whole number.
3
6 x+ x2
(ii) True, because = 6 + x , which is a polynomial.
x

EXERCISE 2.2
1. Which of the following expressions are polynomials? Justify your answer:
(i) 8 (ii) 3x2 – 2 x (iii) 1 – 5x

(iv)
1
+ 5x + 7 (v)
( x – 2 )( x – 4 ) (vi)
1
5 x –2 x x +1

1 3 2 2 1
(vii) a – a + 4a – 7 (viii)
7 3 2x

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POLYNOMIALS 17

2. Write whether the following statements are True or False. Justify your answer.
(i) A binomial can have atmost two terms
(ii) Every polynomial is a binomial
(iii) A binomial may have degree 5
(iv) Zero of a polynomial is always 0
(v) A polynomial cannot have more than one zero
(vi) The degree of the sum of two polynomials each of degree 5 is always 5.

(D) Short Answer Questions

Sample Question 1 :
(i) Check whether p(x) is a multiple of g(x) or not, where
p(x) = x3 – x + 1, g(x) = 2 – 3x
(ii) Check whether g(x) is a factor of p(x) or not, where

x 1
p(x) = 8x3 – 6x2 – 4x + 3, g(x) = −
3 4
Solution :
(i) p(x) will be a multiple of g(x) if g(x) divides p(x).
2
Now, g(x) = 2 – 3x = 0 gives x =
3
3
 2 2   2
Remainder = p   =   −   +1
 3 3   3

8 2 17
= − +1 =
27 3 27
Since remainder ≠ 0, so, p(x) is not a multiple of g(x).
x 1 3
(ii) g(x) = − = 0 gives x =
3 4 4
 3
g(x) will be a factor of p(x) if p   = 0 (Factor theorem)
 4
3 2
 3 3  3  3 
Now, p   = 8  − 6   − 4  + 3
 4 4  4  4 

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18 EXEMPLAR PROBLEMS

27 9
= 8× − 6× −3 +3 = 0
64 16

 3
Since, p   = 0, so, g(x) is a factor of p(x).
 4
Sample Question 2 : Find the value of a, if x – a is a factor of x3 – ax2 + 2x + a – 1.
Solution : Let p(x) = x3 – ax2 + 2x + a – 1
Since x – a is a factor of p(x), so p(a) = 0.
i.e., a 3 – a(a)2 + 2a + a – 1 = 0
a 3 – a3 + 2a + a – 1 = 0
3a = 1
1
Therefore, a =
3
Sample Question 3 : (i)Without actually calculating the cubes, find the value of
483 – 303 – 183.
(ii)Without finding the cubes, factorise (x – y) 3 + (y – z) 3 + (z – x) 3.
Solution : We know that x3 + y3 + z 3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx).
If x + y + z = 0, then x3 + y3 + z 3 – 3xyz = 0 or x3 + y3 + z 3 = 3xyz.
(i) We have to find the value of 483 – 303 – 183 = 483 + (–30)3 + (–18)3.
Here, 48 + (–30) + (–18) = 0
So, 483 + (–30)3 + (–18)3 = 3 × 48 × (–30) × (–18) = 77760
(ii) Here, (x – y) + (y – z) + (z – x) = 0
Therefore, (x – y)3 + (y – z) 3 + (z – x) 3 = 3(x – y) (y – z) (z – x).

EXERCISE 2.3

1. Classify the following polynomials as polynomials in one variable, two variables etc.
(i) x2 + x + 1 (ii) y3 – 5y
(iii) xy + yz + zx (iv) x2 – 2xy + y2 + 1

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POLYNOMIALS 19

2. Determine the degree of each of the following polynomials :

(i) 2x – 1 (ii) –10
3 5
(iii) x – 9x + 3x (iv) y3 (1 – y4)
3. For the polynomial

x3 + 2 x + 1 7 2
– x – x 6 , write
5 2
(i) the degree of the polynomial
(ii) the coefficient of x3
(iii) the coefficient of x6
(iv) the constant term
4. Write the coefficient of x2 in each of the following :
π
(i) x + x 2 –1 (ii) 3x – 5
6
(iii) (x –1) (3x –4) (iv) (2x –5) (2x2 – 3x + 1)
5. Classify the following as a constant, linear, quadratic and cubic polynomials :
(i) 2 – x2 + x3 (ii) 3x3 (iii) 5t – 7 (iv) 4 – 5y2
(v) 3 (vi) 2+x (vii) y3 – y (viii) 1 + x + x2
(ix) t2 (x) 2x – 1
6. Give an example of a polynomial, which is :
(i) monomial of degree 1
(ii) binomial of degree 20
(iii) trinomial of degree 2
7. Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when
x = –3.

1 
8. If p(x) = x2 – 4x + 3, evaluate : p(2) – p(–1) + p  
2 
9. Find p(0), p(1), p(–2) for the following polynomials :
(i) p(x) = 10x – 4x2 – 3 (ii) p(y) = (y + 2) (y – 2)
10. Verify whether the following are True or False :
(i) –3 is a zero of x – 3

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20 EXEMPLAR PROBLEMS

1
(ii) – is a zero of 3x + 1
3

–4
(iii) is a zero of 4 –5y
5
(iv) 0 and 2 are the zeroes of t2 – 2t
(v) –3 is a zero of y2 + y – 6
11. Find the zeroes of the polynomial in each of the following :
(i) p(x) = x – 4 (ii) g(x) = 3 – 6x
(iii) q(x) = 2x –7 (iv) h(y) = 2y
12. Find the zeroes of the polynomial :
p(x) = (x – 2)2 – (x + 2)2
13. By actual division, find the quotient and the remainder when the first polynomial is
divided by the second polynomial : x4 + 1; x –1
14. By Remainder Theorem find the remainder, when p(x) is divided by g(x), where
(i) p(x) = x3 – 2x2 – 4x – 1, g(x) = x + 1
(ii) p(x) = x3 – 3x2 + 4x + 50, g(x) = x – 3
(iii) p(x) = 4x3 – 12x2 + 14x – 3, g(x) = 2x – 1
3
(iv) p(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – x
2
15. Check whether p(x) is a multiple of g(x) or not :
(i) p(x) = x3 – 5x2 + 4x – 3, g(x) = x – 2
(ii) p(x) = 2x3 – 11x2 – 4x + 5, g(x) = 2x + 1
16. Show that :
(i) x + 3 is a factor of 69 + 11x – x2 + x3 .
(ii) 2x – 3 is a factor of x + 2x3 – 9x2 + 12 .
17. Determine which of the following polynomials has x – 2 a factor :
(i) 3x2 + 6x – 24 (ii) 4x2 + x – 2
18. Show that p – 1 is a factor of p 10 – 1 and also of p 11 – 1.
19. For what value of m is x3 – 2mx2 + 16 divisible by x + 2 ?
20. If x + 2a is a factor of x5 – 4a2 x3 + 2x + 2a + 3, find a.
21. Find the value of m so that 2x – 1 be a factor of 8x4 + 4x3 – 16x2 + 10x + m.

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POLYNOMIALS 21

22. If x + 1 is a factor of ax3 + x2 – 2x + 4a – 9, find the value of a.

23. Factorise :
(i) x2 + 9x + 18 (ii) 6x2 + 7x – 3
(iii) 2x2 – 7x – 15 (iv) 84 – 2r – 2r 2
24. Factorise :
(i) 2x3 – 3x2 – 17x + 30 (ii) x3 – 6x2 + 11x – 6
(iii) x3 + x2 – 4x – 4 (iv) 3x3 – x2 – 3x + 1
25. Using suitable identity, evaluate the following:
(i) 1033 (ii) 101 × 102 (iii) 9992
26. Factorise the following:
(i) 4x2 + 20x + 25
(ii) 9y2 – 66yz + 121z2
2 2
(iii)  1  1
 2x +  −  x − 
 3  2
27. Factorise the following :
(i) 9x2 – 12x + 3 (ii) 9x2 – 12x + 4
28. Expand the following :
(i) (4a – b + 2c) 2
(ii) (3a – 5b – c) 2
(iii) (– x + 2y – 3z) 2
29. Factorise the following :
(i) 9x2 + 4y2 + 16z2 + 12xy – 16yz – 24xz
(ii) 25x2 + 16y2 + 4z2 – 40xy + 16yz – 20xz
(iii) 16x2 + 4y2 + 9z 2 – 16xy – 12yz + 24 xz
30. If a + b + c = 9 and ab + bc + ca = 26, find a2 + b2 + c2.
31. Expand the following :
3 3
1 y  1 
(i) (3a – 2b)3 (ii)  +  (iii)  4 – 
x 3  3x 
32. Factorise the following :
(i) 1 – 64a3 – 12a + 48a2

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22 EXEMPLAR PROBLEMS

12 2 6 1
(ii) 8 p3 + p + p+
5 25 125
33. Find the following products :

 x 
2
x
(i)  + 2 y  – xy + 4 y 2  (ii) (x2 – 1) (x4 + x2 + 1)
2  4 
34. Factorise :
(i) 1 + 64x3 (ii) a3 – 2 2b 3
35. Find the following product :
(2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz)
36. Factorise :
(i) a3 – 8b 3 – 64c3 – 24abc (ii) 3 3 3
2 2 a + 8b – 27c + 18 2 abc.
37. Without actually calculating the cubes, find the value of :
3 3 3
1  1 5 
(i)   +  –  (ii) (0.2)3 – (0.3) 3 + (0.1)3
2  3 6 
38. Without finding the cubes, factorise
(x – 2y) 3 + (2y – 3z) 3 + (3z – x) 3
39. Find the value of
(i) x3 + y3 – 12xy + 64, when x + y = – 4
(ii) x3 – 8y3 – 36xy – 216, when x = 2y + 6
40. Give possible expressions for the length and breadth of the rectangle whose area is
given by 4a2 + 4a –3.

(E) Long Answer Questions

Sample Question 1 : If x + y = 12 and xy = 27, find the value of x3 + y3.
Solution :
x 3 + y3 = (x + y) (x2 – xy + y2)
= (x + y) [(x + y)2 – 3xy]
= 12 [122 – 3 × 27]
= 12 × 63 = 756

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POLYNOMIALS 23

Alternative Solution :
x3 + y3 = (x + y) 3 – 3xy (x + y)
= 123 – 3 × 27 × 12
= 12 [122 – 3 × 27]
= 12 × 63 = 756

EXERCISE 2.4
1. If the polynomials az3 + 4z2 + 3z – 4 and z3 – 4z + a leave the same remainder
when divided by z – 3, find the value of a.
2. The polynomial p(x) = x4 – 2x3 + 3x2 – ax + 3a – 7 when divided by x + 1 leaves
the remainder 19. Find the values of a. Also find the remainder when p(x) is
divided by x + 2.
1
3. If both x – 2 and x – are factors of px 2 + 5x + r, show that p = r.
2
4. Without actual division, prove that 2x4 – 5x3 + 2x2 – x + 2 is divisible by x2 – 3x + 2.
[ Hint: Factorise x2 – 3x + 2]
5. Simplify (2x – 5y) 3 – (2x + 5y) 3.
6. Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (– z + x – 2y).

a 2 b 2 c2
7. If a, b, c are all non-zero and a + b + c = 0, prove that + + = 3.
bc ca ab
8. If a + b + c = 5 and ab + bc + ca = 10, then prove that a 3 + b3 + c3 –3abc = – 25.
9. Prove that (a + b + c)3 – a3 – b3 – c 3 = 3(a + b ) (b + c) (c + a).

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