Students can read the important questions given below for Polynomials Class 9 Mathematics . All Polynomials Class 9 Notes and questions with solutions have been prepared based on the latest syllabus and examination guidelines issued by CBSE, NCERT and KVS. You should read all notes provided by us and Class 9 Mathematics Important Questions provided for all chapters to get better marks in examinations. Mathematics Question Bank Class 9 is available on our website for free download in PDF.
Important Questions of Polynomials Class 9
Very Short Answer Type Questions:
Question. Factorize : x2 – 3x
Ans. x2 – 3x = x(x – 3)
Question. Factorize : 8y3 – 125x3
Ans. 8y3 – 125x3 = (2y)3 – (5x)3
= (2y – 5x)(4y2+ 10xy + 25x2)
Question. Factorize: 12a2b – 6ab2
Ans. 12a2b – 6ab2 = 6ab(2a – b)
Question. If f(x) be a polynomial such that f (−1/3) = 0, then calculate one factor of f(x).
Ans. Since, f (−1/3) = 0
∴ −1/3 is a zero of polynomial f(x).
So, x + 1/3 or 3x + 1 is a factor of f(x).
Question. What is x + 1/x ?
Ans. Not a polynomial.
Question. Find the value of k, if x – 2 is a factor of p(x) = 2x2 + 3x – k.
Ans. Since, x – 2 is a factor of p(x), then p(2) = 0
i.e., p(2) = 2 × (2)2 + 3 × 2 – k = 0
or, 8 + 6 – k = 0
∴ k = 14
Question. Find the value of k, if 2x – 1 is a factor of the polynomial 6x2 + kx – 2.
Ans. Since, 2x – 1 is a factor of p(x) = 6x2 + kx – 2
Thus, p(1/2) = 0
or, 6.1/4 + k.1/2 − 2 = 0
or, k = 1
Question. Write the factors of a7 + ab6.
Ans. a7 + ab6 = a(a6 + b6)
= a[(a2)3 + (b2)3]
= a(a2 + b2)(a4 – a2b2 + b4)
Factors are a, (a2 + b2), (a4 – a2b2 + b4).
Question. Calculate the value of 833 + 173 / 832 − 83 x 17 + 172)
Ans.833 + 173 / 832 − 83 x 17 + 172) = (83 + 17) 832 − 83 x 17 + 172)
[∴ a3+ b3 = (a+ b)(a2– ab + b2 )]
= 83 + 17 = 100
Short Answer Type Questions:
Question. Find the value of k, if x – 2 is a factor of f(x) = x2 + kx + 2k.
Ans. Given, (x – 2) is a factor of f(x).
∴ f(2) = 0 1
or, (2)2 + k(2) + 2k = 0
or, 4 + 2k + 2k = 0
or, 4 + 4k = 0
or, k = – 1
Question. Expand : (1/3x − 2/3y)3
Ans. (1/3x − 2/3y)3
= (1/3X)3 − (2/3y)3 − 3 x 1/3x − 2y/3)
= x3/27 − 8y3 /27 − 2xy/3 + (x/3 − 2y/3)
= x3/27 − 8y3 /27 − 2x2y/9 +4xy2/9
Question. Factorize : 9x2 + 6xy + y2
Ans. 9x2 + 6xy + y2 = (3x)2 + 2 × (3x) × y + y2
= (3x + y)2 [∴ a2 + 2ab + b2 = (a + b)2)2]
Question. Factorize : 8a3 + 8b3
Ans. 8a3+ 8b3 = (2a)3 + (2b)3
= (2a + 2b)[(2a)2 + (2b)2 – (2a) × (2b)]
[∴ a3 + b3= (a + b)(a2 + b2 – ab)]
= 2(a + b) × 4(a2+ b2 – ab)
= 8(a + b)(a2+ b2 – ab)
Question. Factorize : 8x3 – (2x – y)3
Ans. 8x3 – (2x – y)3 = (2x)3 – (2x – y)3
= [2x – (2x – y)][(2x)2 + (2x – y)2 + 2x(2x – y)]
[Since, (a3– b3) = (a – b)(a2+ b2 + ab)]
= y[4x2 + 4x2 + y2 – 4xy + 4×2 – 2xy]
= y[12x2 + y2– 6xy]
Question. If f(x) = 3x + 5, evaluate f(7) – f(5).
Ans. Given, f(x) = 3x + 5
∴ f(7) = 3 × 7 + 5 = 26
and f(5) = 3 × 5 + 5 = 20
∴ f(7) – f(5) = 26 – 20 = 6
Question. Simplify : (2a + 3b)3 – (2a – 3b)3
Ans. Let (2a + 3b)3 – (2a – 3b)3 = x3– y3 ,
where 2a + 3b =x and 2a – 3b = y
= (x –y)(x2 + xy + y2)
= [(2a + 3b) – (2a – 3b)][(2a + 3b)2 + (2a + 3b) (2a – 3b) + (2a – 3b)2]
= 6b[(4a2 + 12ab + 9b2 ) + (4a2 – 9b2 ) + (4a2 – 12ab + 9b2 )]
= 6b(12a2 + 9b2 )
= 6b × 3 × (4a2 + 3b2 )
= 18b(4a2 + 3b2 )
Question. Classify the following as linear, quadratic and cubic polynomials :
Ans. Linear polynomial → 1 + x; degree = 1
Quadratic polynomial → x2+ x; degree = 2
Cubic polynomial → x – x3, 7x3; degree = 3
Question. Find the value of ‘a‘ for which (x – 1) is a factor of the polynomial a2x3 – 4ax + 4a – 1.
Ans. Let f(x) = a2x3 – 4ax + 4a – 1
Since, (x – 1) is a factor of f(x)
Then, f(1) = 0
or, a2– 4a + 4a – 1 = 0
or, a2– 1 = 0
or, a = ± 1
Question. Expand by using identity (2x – y + z)2.
Ans. (2x – y + z)2 = 4x2 + y2+ z2 – 4xy – 2yz + 4zx
Detailed Solution :
By using the identity, (a + b + c)2 = a2+ b2 + c2 + 2ab + 2bc + 2ca
= (2x + (–y) + z)2 = (2x)2+ (–y)2 + z2 + 2(2x)(–y) + 2(–y)(z) + 2(z)(2x)
= 4x2 + y2 + z2 – 4xy – 2yz + 4xz
Long Answer Type Questions:
Question. State Factor Theorem. Using Factor Theorem, factorize : x3 – 3x2 – x + 3.
Ans. Factor Theorem: According to Factor Theorem, if p(y), is a polynomial with degree n ≥ 1 and t is a real number, then
(i) (y – t) is a factor of p(y), if p(t) = 0, and
(ii) p(t) = 0, if (y – t) is a factor of p(y).
Let p(x) = x3 – 3x2 – x + 3
The factors of the constant term 3 are ± 1, ± 3.
p(1) = 13 – 3(1)2 – 1 + 3 = 0
∴ (x – 1) is a factor.
p(–1) = (–1)3 – 3(–1)2 – (–1) + 3 = 0
∴ (x + 1) is a factor.
p(3) = 33 – 3(3)2 – 3 + 3 = 0
∴(x – 3) is a factor.
Therefore, (x – 1)(x + 1)(x – 3) are the factors of p(x)
Question. Verify if – 2 and 3 are zeroes of the polynomial 2x3 – 3x2 – 11x + 6. If yes, factorize the polynomials.
Ans. Let p(x) = 2x3 – 3x2 – 11x + 6
For, x = – 2
p(– 2) = 2(– 2)3 – 3(– 2)2 – 11(– 2) + 6
= – 16 – 12 + 22 + 6
= – 28 + 28 = 0 1
For, x = 3
p(3) = 2(3)3 – 3(3)2 – 11(3) + 6
= 54 – 27 – 33 + 6
= 60 – 60 = 0 1
So, – 2 and 3 are zeroes of the given polynomial.
Now, p(x) = 2x3 – 3x2 – 11x + 6
(x + 2)(x – 3) = x2– x – 6 is a factor of p(x).
∴ 2x3– 3x2 – 11x + 6
= 2x3 + 4x2 – 7×2 – 14x + 3x + 6
= 2x2(x + 2) – 7x(x + 2) + 3(x + 2)
= (x + 2)(2x2 – 7x + 3)
= (x + 2)(2x2 – 6x – x + 3)
= (x + 2)[(2x(x – 3) – 1(x – 3)]
= (x + 2)(x – 3)(2x – 1)
Question. Find the value of p for which the polynomial x3 + 4x2 – px + 8 is exactly divisible by x – 2. Hence factorize the polynomial.
Ans. Let q(x) = x3 + 4x2 – px + 8
Given, p(x) is exactly divisible by x – 2.
∴ q(2) = 0
or, (2)3 + 4(2)2 – p(2) + 8 = 0
or, 8 + 16 – 2p + 8 = 0
or, 32 – 2p = 0
∴ p = 16
∴ p(x) = x3 + 4x2 – 16x + 8
Now,
x3 + 4x2 – 16x + 8 = x2(x – 2) + 6x(x – 2) – 4(x – 2)
= (x – 2) (x2 + 6x – 4)
Question. If x + 4 is a factor of polynomial x3 – x2 – 14x + 24, then find its other factors
Ans. Let p(x) = x3 – x2– 14x + 24
Since (x + 4) is a factor of polynomial p(x) Then
x3– x2– 14x + 24 = x3 + 4x2 – 5x2 – 20x + 6x +24
= x2(x + 4) – 5x(x + 4) + 6(x + 4)
= (x + 4)(x2– 5x + 6)
= (x + 4)(x2– 2x – 3x + 6)
= (x + 4)[x(x – 2) – 3(x – 2)]
= (x + 4)(x – 2)(x – 3)
Question. Factorize : x12 – y12.
Ans. x12 – y12
= (x6)2 – (y6)2
= (x6– y6)(x6 + y6)
= {(x3)2 – (y3)2}(x6 + y6)
= (x3 – y3)(x3+ y3)(x6 + y6)
= {(x)3 – (y)3}{(x)3 + (y)3}(x6+ y6)
= (x – y)(x2 + xy+ y2)(x + y)(x2 – xy+ y2)(x6+ y6)
= (x – y)(x + y)(x2 + xy + y2)(x2 – xy + y2) {(x2)3 + (y2)3}
= (x – y)(x+ y)(x2 + xy + y2)(x2 – xy+ y2)(x2 + y2) (x4 – x2y2 + y4)
Question. Factorize : x3+ 2x2 – 5x – 6
Ans. Let p(x) = x3+ 2x2 – 5x – 6
Factor of 6 = (± 1, ± 2, ± 3, ± 6)
p(– 1) = (– 1)3 + 2(– 1)2 – 5(– 1) – 6
= – 1 + 2 + 5 – 6
= 7 – 7 = 0
∴ x = – 1 is zero of p(x) or (x + 1) is a factor of p(x).
∴ x3 + 2x2 – 5x – 6
= x2(x + 1) + x(x + 1) – 6(x + 1)
= (x + 1)(x2 + x – 6)
= (x + 1)(x2+ 3x – 2x – 6)
= (x + 1)[x(x + 3) – 2(x + 3)]
= (x + 1)(x + 3)(x – 2)
