Polynomials Class 9 Mathematics Exam Questions

Please refer to Polynomials Class 9 Mathematics Exam Questions provided below. These questions and answers for Class 9 Mathematics have been designed based on the past trend of questions and important topics in your class 9 Mathematics books. You should go through all Class 9 Mathematics Important Questions provided by our teachers which will help you to get more marks in upcoming exams.

Class 9 Mathematics Exam Questions Polynomials

Class 9 Mathematics students should read and understand the important questions and answers provided below for Polynomials which will help them to understand all important and difficult topics.

Very Short Answer Type Questions:

Question. Find the zeroes of the polynomial 3x + π.
Ans. 3x + π = 0 ⇒ 3x = –π ⇒ x = −π/3.
∴ x = −π/3 is the zero of the polynomial 3x + π.

Question. Find the factors of x²+ 2x – 15.
Ans. We have, x² + 2x – 15 = x² + 5x – 3x – 15
= x(x + 5) – 3(x + 5) = (x + 5) (x – 3)
∴ Factors ofx² + 2x – 15 are (x + 5) and (x – 3).

Question. Find the zero of the polynomial q(x) = 3x + 2.
Ans. We have, q(x) = 3x + 2
Put q(x) = 0 ⇒ 3x + 2 = 0 ⇒ x = −2/3.
∴ −2/3 is the zero of the polynomial q(x) = 3x + 2.

Question. Is x + 2 a zero of p(x) = 2x³ – x²+ 3?
Ans. (x + 2) will be a zero of p(x), if p (–2) = 0
∴ p(–2) = 2(–2)³ – (–2)² + 3 = –16 – 4 + 3 = –17 ≠ 0
∴ x + 2 is not a zero of p(x).

Question. Factorise : a³ – b³+ 3ba² – 3ab²
Ans. We have, a³ – b³ + 3a²b – 3ab²
= (a – b) (a² + b²+ ab) + 3ab (a – b)
= (a – b) [a²+ b²+ 4ab]

Question. How many terms are there in the polynomial 5 – 3t + 2t⁶ ?
Ans. Number of terms = 3

Question. Factorise : 14a²b – 7ab²
Ans. We have, 14a²b – 7ab² = 7ab(2a – b)

Question. Find the value of p(x) = √2x² + √2x + 6 at x = √2 .
Ans. p (√2) = √2 (√2)² + √2(√2) + 6
=2√2 + 2 + 6
=2√2 + 8

Question. If, p( − 1/4) = 0, where p(x) = 8x³ – ax² – x + 2, then find the value of a.
Ans. ∵ p ( − 1/4) = 0
⇒ 8 ( − 1/4)³ – a( − 1/4)² – ( − 1/4) + 2 = 0
⇒ – 8/64 – a/16 + 1/4 + 2 = 0
⇒ – 1/8 + 1/4 + 2 =a/16 ⇒ – 1+2+16/8 = a/16
⇒ a = 17 × 2
∴ a = 34

Question. Find the product of (lx +my) and (lx – my).
Ans. Product of (lx + my) and (lx – my) = (lx + my) (lx – my)
= (lx)² – (my)² = l²x² – m²y² 

Short Answer Type Questions

Question. If the perimeter of a rectangle is 24 units and the length exceeds the breadth by 4 units, then find the area of the rectangle.
Ans. Let l units be the length and b units be the breadth of a rectangle.
Perimeter = 2(l + b) = 24
i.e., l + b = 12 … (1)
Also, we are given that l = b + 4
i.e., l – b = 4 … (2)
From (1) and (2), we get
(l + b)² – (l – b)² = (12)² – (4)²
⇒ {(l + b) + (l – b)}{(l + b) – (l – b)} = (12 + 4)(12 – 4) [∵ x² – y² = (x + y)(x – y)]
⇒ (2l)(2b) = (16)(8)
⇒ 4(l × b) = 128 ⇒ lb = 32
Therefore, the area of the rectangle is 32 square units.

Question. If p(t) = t³ – 3t² + t – 4, then find p(1) and p(–1).
Ans. We have, p(t) = t³ – 3t² + t – 4
∴ p(1) = 1³ – 3(1)² + 1 – 4 = 1 – 3 + 1 – 4 = –5 and
p(–1) = (–1)³ – 3(–1)² + (–1) – 4 = –1 – 3 – 1 – 4 = –9

Question. If x and y be two positive real numbers such that 4x² + y² = 40 and xy = 6, then find the value of 2x + y.
Ans. We have, 4x² + y² = 40 and xy = 6
Now, consider (2x + y)² = (2x)² + 2(2x)(y) + (y)²
= 4x² + 4xy + y² = (4x² + y² ) + 4xy
= (40) + 4(6) = 40 + 24 = 64 = (8)²
Therefore, 2x + y = 8.

Question. Verify whether − 1/2 and 1/5 are the zeroes of the polynomial p(y) = 5y – 1.
Ans. We have, p(y) = 5y – 1
p ( − 1/2) = 5( − 1/2) – 1 = − 5/2 – 1 = – 7/2 ≠ 0
∴ 1/2 is a zero of p(y) = 5y – 1
p(1/5) = 5(1/5) – 1 = 1– 1 = 0 
∴ 1/5 is a zero of p(y) = 5y – 1

Question. Factorize x¹⁶ – y¹⁶.
Ans. We have, x¹⁶ – y¹⁶ = {(x⁸ )² – (y⁸)²}
= (x⁸– y⁸) (x⁸+ y⁸) = {(x⁴)² – (y⁴)²} (x⁸+ y⁸)
= (x⁴– y⁴)(x⁴+ y⁴) (x⁸+ y⁸) = {(x²)² – (y² )²}(x⁴+ y⁴) (x⁸+ y⁸)
= (x²– y² )(x²+ y² )(x⁴+ y⁴) (x⁸+ y⁸)
= (x – y)(x + y)(x²+ y² ) (x⁴+ y⁴) (x⁸+ y⁸)

Question. For what integral values of ‘a’, p(a) = 0, where p(x) = xⁿ – aⁿ ?
Ans.We have, p(x) = xⁿ – aⁿ
P(a) = aⁿ– aⁿ = 0, for any value of a.
∴ p (a) = 0, for any value of a.

Question. Factorise 36/25x⁴ – y⁴/16 
Ans. We have,

Question. If f(x) = 7x² – 3x + 7, find f(2) + f(–1) + f(0).
Ans. f(x) = 7x² – 3x + 7
Now, f(2) = 7(2)² – 3(2) + 7 = 28 – 6 + 7 = 29
Also, f(–1) = 7(–1)² – 3(–1) + 7 = 7 + 3 + 7 = 17
Also, f(0) = 7
∴ f(2) + f(–1) + f(0) = 29 + 17 + 7 = 53

Question. Identify whether the following are polynomials or not. Justify your answer. 
(i) x²/2 – 2/x²
(ii) √2x −1
(iii) x² + 3x^3/2/√x
(iv) x – 1/x + 1
Ans. (i) The given polynomial can be written as, x²/2 – 2x^–2
It is not a polynomial, because one of the exponents of x is – 2, which is not a whole number.
(ii) The given polynomial can be written as √2x^1/2 − 1
It is not a polynomial, because exponent of x is 1/2 , which is not a whole number.
(iii) The given polynomial can be written as x² + 3x^{3/2 – 1/2} i.e., x² + 3x
It is a polynomial, because each exponent of x is a whole number.
(iv) We have, x –1/x +1
It is not a polynomial because it is not defined for x = –1. 

Question. Write the coefficients of x and x³ in polynomial 2x⁴ + 3x² – 7x + 5.
Ans. The given polynomial can be written as
2x⁴ + 0 ⋅ x³ + 3x² – 7x + 5
∴ Coefficient of x = –7
And coefficient of x³ = 0

Question. Factorise the following using appropriate identities:
(i) 27y³ – 27y² + 9y – 1
(ii) 40x³ – 135
Ans. (i) We have, 27y³ – 27y² + 9y – 1
= (3y)³ – 3 ⋅ (3y)² ⋅ 1 + 3 ⋅ 3y ⋅ 1² – 1³ = (3y – 1)³
= (3y – 1) (3y – 1) (3y – 1)
(ii) 40x³ – 135 = 5(8x³ – 27) = 5[(2x)³ – 3³]
= 5(2x – 3) [(2x)² + 2x ⋅ 3 + 3²]
= 5(2x – 3)(4x² + 6x+ 9)

Question. Factorise the following polynomials by splitting the middle term:
(i) 2x² + 7x + 6 (ii) 6x² + 7x – 10.
Ans. (i) Let p(x) = 2x² + 7x + 6
By splitting the middle term, we get
p(x) = 2x² + 3x + 4x + 6
= x (2x + 3) + 2 (2x + 3) = (x + 2) (2x + 3)
(ii) Let p(x) = 6x² + 7x – 10
By splitting the middle term, we get
p(x) = 6x² + 12x – 5x – 10
= 6x (x + 2) – 5(x + 2) = (6x – 5) (x + 2)

Question. Factorise: (2y + x)² (y – 2x) + (2x + y)² (2x – y)
Ans. We have, (2y + x)² (y – 2x) + (2x + y)² (2x – y)
= (2y + x)² (y – 2x) – (2x + y)² (y – 2x)
= (y – 2x) [(2y + x)² – (2x + y)²]
= (y – 2x) [(2y + x) + (2x + y)] [(2y + x) – (2x + y)]
= (y – 2x) (3x + 3y) (y – x)
= (y – 2x) 3(x + y) (y – x) = 3(y – 2x) (y + x) (y – x)

Question. If (2a + b) = 12 and ab = 15, then find the value of 8a³ + b³.
Ans. We have, (2a + b) = 12 … (i)
Cubing both sides of (i), we get
(2a + b)³ = (12)³
⇒ (2a)³ + b³+ 3 × 2a × b(2a + b) = 1728
⇒ 8a³ + b³+ 3 × 2 × 15(12) = 1728 [Substituting ab = 15 and (2a + b) = 12]
⇒ 8a³ + b³+ 1080 = 1728
⇒ 8a³ + b³ = 1728 – 1080 = 648.
⇒ 8a³ + b³ = 648.

Question. If (3x – 1)³ = 6a₃x³ + a₂x² + a₁x + a₀, then find a₃+ a₂ + a₁ + a₀ .
Ans. We have, (3x – 1)³ = 6a₃x³ + a₂x² + a₁x + a₀
⇒ (3x)³ – (1)³ – 3 × 3x × 1(3x – 1) = 6a₃x³ + a₂x² + a₁x + a₀ [∵ (a – b)³ = a³ – b³ – 3ab(a – b)]
⇒ 27x³ – 1 – 9x(3x – 1) = 6a₃ x³ + a₂x² + a₁ x + a₀
⇒ 27x³ – 27x² + 9x – 1 = 6a₃ x³ + a₂x² + a₁ x + a₀
Comparing the coefficient of x³, x², x and x⁰, we get
6a₃  = 27 ⇒ a₃ = 9/2, a₂ = –27, a₁ = 9 and a₀ = –1
Now, a₃ +a₂ + a₁ + a₀ = 9/2 – 27 + 9 – 1
= − 9/2 – 19 9 – 38/2 = – 29/2

Question. Give one example of a
(i) monomial of degree 100
(ii) binomial of degree 50
(iii) trinomial of degree 205
Ans. (i) –8y¹⁰⁰
(ii) 3z⁵⁰ + z²
(iii) 9x²⁰⁵ – 7x³ + 4

Question. Find the coefficient of x in the expansion of (x + 7)³.
Ans. We have, (x + 7)³ = x³ + (7)³ + 3(x)(7)(x + 7)
= x³ + 343 + 21x(x + 7) [∴ (a + b)³ = a³ + b³+ 3ab(a + b)]
= x³ + 21x² + 147x + 343
∴ Coefficient of x in the expansion of (x + 7)³ is 147

Question. Factorise: x³– 12x² + 47x – 60.
Ans. Let p(x) = x³ – 12x² + 47x – 60
Factors of 60 are ± 1, ± 2, ± 3, ± 4, ± 5, ± 6, ± 10, ± 12, ± 15, ± 20, ± 30, ± 60
By trial method, we see that
p(3) = (3)³ – 12(3)² + 47(3) – 60 = 27– 108 + 141 – 60
= 168 – 168 = 0
p(4) = 4³ – 12(4)² + 47(4) – 60 = 64 –192 + 188 – 60 = 0
p(5) = 5³ – 12(5)² + 47(5) – 60 = 125 – 300 + 235 – 60 = 0
∴ By factor theorem, (x – 3), (x – 4) and (x – 5) are the factors of p(x).
Thus, x³ – 12x² + 47x – 60 = (x – 3)(x – 4)(x – 5)

Question. Simplify : (x/3 + y/5)³ – (x/3 – y/5)³
Ans. We have,

Question. Factorise : (x – y)² – 9(x² – y² ) + 20(x + y)²
Ans. We have, (x – y)² – 9(x² – y² ) + 20(x + y)²
= (x – y)² – 9(x + y) (x – y) + 20(x + y)²
= (x – y)² – 4(x + y) (x – y) – 5(x + y) (x – y) + 20(x + y)²
= (x – y) [x – y – 4x – 4y] – 5(x + y)[x – y – 4x – 4y]
= (x – y) [–5y – 3x] – 5(x + y) [ – 5y – 3x]
= (–5y – 3x) [x – y – 5 (x + y)] = –(5y + 3x) (–4x – 6y)
= (5y + 3x) (4x + 6y) = 2(2x + 3y) (5y + 3x)

Question. Simplify : (y – 1/y)(y + 1/y)(y² + 1/y²)(y⁴ + 1/y⁴)
Ans. We have, (y – 1/y)(y + 1/y)(y² + 1/y²)(y⁴ + 1/y⁴)
= (y² – 1/y² )(y² + 1/y²)(y⁴ + 1/y⁴) [∵ (a + b) (a – b) = a² – b²]
= (y⁴ – 1/y⁴)(y⁴ + 1/y⁴) = (y⁸ – 1/y⁸ )

Long Answer Type Questions

Question. If x + 1/x = 6 , find
(i) x² + 1/x² (ii) x⁴ + 1/x⁴
Ans. (i) We have, x + 1/x = 6
⇒ (x + 1/x)² = 6² [On squaring both sides]
⇒ x² + 1/x² + 2 X x X 1/x = 36
⇒ x² + 1/x² + 2 = 36 ⇒ x² + 1/x² = 36 – 2 = 34
(ii) We have, x² + 1/x² = 34
⇒ (x² + 1/x²)² = (34)²  [On squaring both sides]
⇒ (x²)² + (1/x²)² + 2 X x² + 1/x² = 1156
⇒ x⁴ + 1/x⁴ + 2 = 1156 ⇒ x⁴ + 1/x⁴ = 1156 −2 =1154

Question. If x = 2 and x = 0 are zeroes of the polynomial 2x³ – 5x² + ax + b, then find the values of a and b.
Ans. Let p(x) = 2x³ – 5x² + ax + b
∵ x = 2 and x = 0 are zeroes of p(x).
∴ p(2) = 0 and p(0) = 0
p(2) = 0 ⇒ 2(2)³ – 5(2)² + 2a + b = 0
⇒ 16 – 20 + 2a + b = 0 ⇒ 2a + b = 4 …(i)
p(0) = 0 ⇒ 2(0)³ – 5(0)² + a ⋅ 0 + b = 0
⇒ 0 – 0 + 0 + b = 0 ⇒ b = 0
Put b = 0 in (i)
∴ 2a + 0 = 4 ⇒ a = 2
∴ a = 2 and b = 0.

Question. Verify that x³ + y³+ z³ – 3xyz = 1/2 (x + y + z) [(x − y)² + (y − z)² + (z − x)² ]
Ans. R.H.S.
= 1/2 (x + y + z)[(x – y)² (y – z)² (z – x)² ]
= 1/2 (x + y + z)[(x² + y² -2xy) (y² + z² +2yz) +(z²+ x² − 2zx)]
= 1/2 (x + y + z)(x² + y² + y² + z² + z²+ x² – 2xy – 2yz – 2zx)
= 1/2 (x + y + z)[2(x² + y² + z² – xy – yz – zx)]
= 2 × 1/2 x (x + y + z)(x² + y² + z² – xy – yz – zx)
= (x + y + z)(x²+ y² + z²– xy – yz – zx)
= x³+ y³+ z³ – 3xyz = L.H.S.
Hence, verified.

Question. Prove that : (a + b + c)³ – a³ – b³ – c³ = 3(a + b)(b + c)(c + a).
Ans. L.H.S. = [(a + b + c)³ – a³] – (b³ + c³)
= (a + b + c – a)[(a + b + c)² + a² + (a + b + c) a] – [(b + c) (b²+ c² – bc)]
[ ∵ x³ − y³ = (x – y)(x² − y² + xy) and x³ + y³ = (x + y)(x² + y^2 – xy) ]
= (b + c)[a²+ b²+ c² + 2ab + 2bc + 2ca + a² + a² + ab + ac] – (b + c)(b²+ c² – bc)
= (b + c)(3a² + 3ab + 3ac + 3bc) = (b + c)[3(a² + ab + ac + bc)] 
= 3(b + c)[a(a + b) + c(a + b)] = 3(a + b)(b + c)(c + a) = R.H.S.