Ex. 1Multiple choice
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
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Solution : Answer (C)
Chapter 2 – Polynomials
Class 9 Mathematics · 74 questions · 40 with answers
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Ex. 1True / FalseExercise 2.1
Write whether the following statements are True or False. Justify your answer.
Ex. 1Multiple choiceExercise 2.2
- (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
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Solution : (i) p(x) will be a multiple of g(x) if g(x) divides p(x). Now, g(x) = 2 – 3x = 0 gives x = 2 2 2 Remainder = p = − +1 3 3 3
Ex. 2Short answerExercise 2.2
Find the value of a, if x – a is a factor of x3 – ax2 + 2x + a – 1.
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Solution : Let p(x) = x3 – ax2 + 2x + a – 1 Since x – a is a factor of p(x), so p(a) = 0. i.e., a3 – a(a)2 + 2a + a – 1 = 0 a3 – a3 + 2a + a – 1 = 0 3a = 1 Therefore, a =
Ex. 3Multiple choiceExercise 2.2
- (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.
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Solution : We know that x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx). If x + y + z = 0, then x3 + y3 + z3 – 3xyz = 0 or x3 + y3 + z3 = 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).
Ex. 1Short answerExercise 2.3
If x + y = 12 and xy = 27, find the value of x3 + y3.
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Solution : x3 + y3 = (x + y) (x2 – xy + y2) = (x + y) [(x + y)2 – 3xy] = 12 [122 – 3 × 27] = 12 × 63 = 756 POLYNOMIALS 23 Alternative Solution : x3 + y3 = (x + y)3 – 3xy (x + y) = 123 – 3 × 27 × 12 = 12 [122 – 3 × 27] = 12 × 63 = 756
Questions
Which one of the following is a polynomial? x2 2
- (A)–
- (B)2 x −1
x2 3x 2 x −1 (C) x + (D) x x +1 2. 2 is a polynomial of degree (A) 2 (B) 0 (C) 1 (D)
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(B)
Degree of the polynomial 4x4 + 0x3 + 0x5 + 5x + 7 is
- (A)4
- (B)5
- (C)3
- (D)7
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(A) 4
Degree of the zero polynomial is
- (A)0
- (B)1
- (C)Any natural number
- (D)Not defined ( )
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(D) Not defined ( )
If p ( x ) = x 2 – 2 2 x + 1 , then p 2 2 is equal to
- (A)0
- (B)1
- (C)4 2
- (D)8 2 +1
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(B) 1
The value of the polynomial 5x – 4x + 3, when x = –1 is
- (A)– 6
- (B)6
- (C)2
- (D)–2 POLYNOMIALS 15
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(A) – 6
If p(x) = x + 3, then p(x) + p(–x) is equal to
- (A)3
- (B)2x
- (C)0
- (D)6
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(D) 6
Zero of the zero polynomial is
- (A)0
- (B)1
- (C)Any real number
- (D)Not defined
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(C) Any real number
Zero of the polynomial p(x) = 2x + 5 is 2 5 2 5
- (A)–
- (B)–
- (C)
- (D)5 2 5 2
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(B) –
One of the zeroes of the polynomial 2x2 + 7x –4 is 1 1
- (A)2
- (B)
- (C)–
- (D)–2 2 2
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(B)
If x51 + 51 is divided by x + 1, the remainder is
- (A)0
- (B)1
- (C)49
- (D)50
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(D) 50
If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is
- (A)–3
- (B)4
- (C)2
- (D)–2
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(C) 2
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
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(B) x3 + x2 + x + 1
One of the factors of (25x2 – 1) + (1 + 5x)2 is
- (A)5 + x
- (B)5 – x
- (C)5x – 1
- (D)10x 2 2
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(D) 10x 2 2
The value of 249 – 248 is
- (A)12
- (B)477
- (C)487
- (D)497
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(D) 497
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)
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(B) (2x + 1) (2x + 3)
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
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(D) 3xy
The coefficient of x in the expansion of (x + 3) is
- (A)1
- (B)9
- (C)18
- (D)27
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(D) 27
If y + x = –1 ( x, y ≠ 0) , the value of x3 – y3 is
- (A)1
- (B)–1
- (C)0
- (D)1 1
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(C) 0
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
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(C)
If a + b + c = 0, then a3 + b3 + c3 is equal to
- (A)0
- (B)abc
- (C)3abc
- (D)2abc
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(C) 3abc
6 x + x2
- (i)x2 +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. 6 x + x2 (ii) True, because = 6 + x , which is a polynomial.
Which of the following expressions are polynomials? Justify your answer:
- (i)8
- (ii)3x 2 – 2 x
- (iii)1 – 5x + 5x + 7 ( x – 2 )( x – 4 ) 1
- (iv)(v) (vi) 5 x –2 x x +1
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(i) 8
(ii) 3x 2 – 2 x
(iv) (v) (vi) 5 x –2 x x +1
3 2 2 1 (vii) a – a + 4a – 7 (viii) 7 3 2x POLYNOMIALS 17
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Polynomials: (i), (ii), (iv), (vii) because the exponent of the variable after simplification in each of these is a whole number.
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.
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(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.
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 27 9 = 8× −6× −3+ 3 = 0 64 16 3 Since, p = 0, so, g(x) is a factor of p(x). 4
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 POLYNOMIALS 19
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(i) x2 + x + 1
(ii) y3 – 5y
(iii) xy + yz + zx
(iv) x2 – 2xy + y2 + 1 POLYNOMIALS 19
Determine the degree of each of the following polynomials :
- (i)2x – 1
- (ii)–10
- (iii)x3 – 9x + 3x5
- (iv)y3 (1 – y4)
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(i) 2x – 1
(ii) –10
(iii) x3 – 9x + 3x5
(iv) y3 (1 – y4)
For the polynomial x3 + 2 x + 1 7 2 – x – x 6 , write
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(i) 6 (ii) (iii) –1 (iv) 5 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 : (i) x + x 2 –1 (ii) 3x – 5 (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) 3x 3 (iii) 5t – 7 (iv) 4 – 5y2 (v) 3 (vi) 2+x (vii) y3 – y (viii) 1 + x + x2 (ix) t2 (x) 2x – 1
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(iii) the coefficient of x6
(iv) the constant term 4. Write the coefficient of x2 in each of the following : (i) x + x 2 –1 (ii) 3x – 5 (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) 3x 3 (iii) 5t – 7 (iv) 4 – 5y2 (v) 3 (vi) 2+x (vii) y3 – y (viii) 1 + x + x2 (ix) t2 (x) 2x – 1
(i) the degree of the polynomial
(ii) the coefficient of x3
Give an example of a polynomial, which is :
- (i)monomial of degree 1
- (ii)binomial of degree 20
- (iii)trinomial of degree 2
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(i) monomial of degree 1
(ii) binomial of degree 20
(iii) trinomial of degree 2
Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = –3. 1
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61, –143
If p(x) = x2 – 4x + 3, evaluate : p(2) – p(–1) + p 2
Find p(0), p(1), p(–2) for the following polynomials :
- (i)p(x) = 10x – 4x2 – 3
- (ii)p(y) = (y + 2) (y – 2)
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(i) p(x) = 10x – 4x2 – 3
(ii) p(y) = (y + 2) (y – 2)
Verify whether the following are True or False :
- (i)–3 is a zero of x – 3
- (ii)– is a zero of 3x + 1 –4
- (iii)is a zero of 4 –5y
- (iv)0 and 2 are the zeroes of t2 – 2t (v) –3 is a zero of y2 + y – 6
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(i) –3 is a zero of x – 3
(ii) – is a zero of 3x + 1 –4
(iii) is a zero of 4 –5y
(iv) 0 and 2 are the zeroes of t2 – 2t (v) –3 is a zero of y2 + y – 6
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
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(i) p(x) = x – 4
(ii) g(x) = 3 – 6x
(iii) q(x) = 2x –7
(iv) h(y) = 2y
Find the zeroes of the polynomial : p(x) = (x – 2)2 – (x + 2)2
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0
By actual division, find the quotient and the remainder when the first polynomial is divided by the second polynomial : x4 + 1; x –1
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x3 + x2 + x + 1, 2 3 −136
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
- (iv)p(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – x
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(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
(iv) p(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – x
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
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(i) p(x) = x3 – 5x2 + 4x – 3, g(x) = x – 2
(ii) p(x) = 2x3 – 11x2 – 4x + 5, g(x) = 2x + 1
Show that :
- (i)x + 3 is a factor of 69 + 11x – x2 + x3 .
- (ii)2x – 3 is a factor of x + 2x3 – 9x2 + 12 .
Determine which of the following polynomials has x – 2 a factor :
- (i)3x2 + 6x – 24
- (ii)4x2 + x – 2
Show that p – 1 is a factor of p10 – 1 and also of p11 – 1.
For what value of m is x3 – 2mx2 + 16 divisible by x + 2 ?
If x + 2a is a factor of x5 – 4a2x3 + 2x + 2a + 3, find a.
Find the value of m so that 2x – 1 be a factor of 8x4 + 4x3 – 16x2 + 10x + m. POLYNOMIALS 21
If x + 1 is a factor of ax3 + x2 – 2x + 4a – 9, find the value of a.
Factorise :
- (i)x2 + 9x + 18
- (ii)6x2 + 7x – 3
- (iii)2x2 – 7x – 15
- (iv)84 – 2r – 2r2
Factorise :
- (i)2x3 – 3x2 – 17x + 30
- (ii)x3 – 6x2 + 11x – 6
- (iii)x3 + x2 – 4x – 4
- (iv)3x3 – x2 – 3x + 1
Using suitable identity, evaluate the following:
- (i)1033
- (ii)101 × 102
- (iii)9992
Factorise the following:
- (i)4x2 + 20x + 25
- (ii)9y2 – 66yz + 121z2 2 2 1 1
- (iii)2x + − x − 3 2
Factorise the following :
- (i)9x2 – 12x + 3
- (ii)9x2 – 12x + 4
Expand the following :
- (i)(4a – b + 2c)2
- (ii)(3a – 5b – c)2
- (iii)(– x + 2y – 3z)2
Factorise the following :
- (i)9x2 + 4y2 + 16z2 + 12xy – 16yz – 24xz
- (ii)25x2 + 16y2 + 4z2 – 40xy + 16yz – 20xz
- (iii)16x2 + 4y2 + 9z2 – 16xy – 12yz + 24 xz
If a + b + c = 9 and ab + bc + ca = 26, find a2 + b2 + c2.
Expand the following : 3 3 1 y 1
- (i)3 (3a – 2b)
- (ii)+
- (iii)4 – x 3 3x
Factorise the following :
- (i)1 – 64a3 – 12a + 48a2 12 2 6 1
- (ii)8 p3 + p + p+ 5 25 125
Find the following products : x x
- (i)+ 2 y – xy + 4 y 2
- (ii)(x2 – 1) (x4 + x2 + 1) 2 4
Factorise :
- (i)1 + 64x3
- (ii)a 3 – 2 2b 3
Find the following product : (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz)
Factorise :
- (i)a3 – 8b3 – 64c3 – 24abc
- (ii)3 3 3 2 2 a + 8b – 27c + 18 2 abc.
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
Without finding the cubes, factorise (x – 2y)3 + (2y – 3z)3 + (3z – x)3
Find the value of
- (i)x3 + y3 – 12xy + 64, when x + y = – 4
- (ii)x3 – 8y3 – 36xy – 216, when x = 2y + 6
Give possible expressions for the length and breadth of the rectangle whose area is given by 4a2 + 4a –3.
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.
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–1
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.
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a = 5; 62
If both x – 2 and x – are factors of px2 + 5x + r, show that p = r.
Without actual division, prove that 2x4 – 5x3 + 2x2 – x + 2 is divisible by x2 – 3x + 2. [Hint: Factorise x2 – 3x + 2]
Simplify (2x – 5y)3 – (2x + 5y)3.
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–120x2y – 250y3
Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (– z + x – 2y). a 2 b2 c 2
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x3– 8y3– z3– 6xyz
If a, b, c are all non-zero and a + b + c = 0, prove that + + =3.
If a + b + c = 5 and ab + bc + ca = 10, then prove that a3 + b3 + c3 –3abc = – 25.
Prove that (a + b + c)3 – a3 – b3 – c3 = 3(a + b ) (b + c) (c + a).