CBSE Class 9 Mathematics Polynomials Worksheet Set H

Very Short Answer Type Questions:

Question. Factorize : x² – 3x
Answer: x² – 3x = x(x – 3)

Question. Factorize : 8y³ – 125x³
Answer: 8y³ – 125x³ = (2y)³ – (5x)³
= (2y – 5x)(4y²+ 10xy + 25x²)

Question. Factorize: 12a²b – 6ab² 
Answer: 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).
Answer: 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 ?
Answer: Not a polynomial.

Question. Find the value of k, if x – 2 is a factor of p(x) = 2x² + 3x – k.
Answer: 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.
Answer: 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.
Answer: 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)
Answer: 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.
Answer: 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
Answer: (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
Answer: 9x² + 6xy + y² = (3x)² + 2 × (3x) × y + y²
= (3x + y)² [∴ a² + 2ab + b² = (a + b)2)²]

Question. Factorize : 8a³ + 8b³
Answer: 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
Answer: 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).
Answer: 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
Answer: 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 :
Answer: 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.
Answer: 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.
Answer: (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.
Answer: 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.
Answer: 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.
Answer: 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
Answer: 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.
Answer: 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
Answer: 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)