Practice Set 3.6 Polynomials Class 9th Mathematics Part I MHB Solution
Practice Set 3.6
Question 1.
Find the factors of the polynomials given below.
2x2 + x – 1
Answer:
2x2 + x – 1
⟹ 2x2 + 2x - x – 1
⟹ 2x (x + 1) – 1 (x + 1)
⟹ (x + 1) (2x – 1)
Therefore, the factors of the given polynomial = (x + 1) (2x – 1)
Question 2.
Find the factors of the polynomials given below.
2m2 + 5m – 3
Answer:
2m2 + 5m – 3
⟹ 2m2 + 6m - m – 3
⟹ 2m (x + 3) – 1 (m + 3)
⟹ (m+ 3) (2m – 1)
Therefore, the factors of the given polynomial = (m+ 3) (2m – 1)
Question 3.
Find the factors of the polynomials given below.
12x2 + 61x + 77
Answer:
12x2 + 61x + 77
⟹ 12x2 + 28x + 33x + 77
⟹ 4x (3x + 7) + 11 (3x + 7)
⟹ (4x + 11) (3x + 7)
Therefore, the factors of the given polynomial = (4x + 11) (3x + 7)
Question 4.
Find the factors of the polynomials given below.
3y2 – 2y – 1
Answer:
3y2 – 2y – 1
⟹ 3y2 – 3y + y – 1
⟹ 3y (y - 1) + 1 (y - 1)
⟹ (3y + 1) (y – 1)
Therefore, the factors of the given polynomial = (3y + 1) (y – 1)
Question 5.
Find the factors of the polynomials given below.
√3x2 + 4x + √3
Answer:
√3x2 + 4x + √3
⟹ √3x2 + 3x + x + √3
⟹ √3x (x + √3) + 1 (x + √3)
⟹ (x + √3) (√3x + 1)
Therefore, the factors of the given polynomial = (x + √3) (√3x + 1)
Question 6.
Find the factors of the polynomials given below.
1/2 x2 - 3x + 1
Answer:
1/2 x2 - 3x + 1
⟹ 1/2x2 - 2x - x + 4
⟹ 1/2x (x - 4) – 1 (x - 4)
⟹ (x - 4) (1/2x – 1)
Therefore, the factors of the given polynomial = (x - 4) (1/2x – 1)
Question 7.
Factorize the following polynomials.
(x2 – x)2 – 8(x2 – x) + 12
Answer:
Put (x2 – x) = a
⟹ a2 – 8a + 12
⟹ a2 – 2a – 6a + 12
⟹ a (a-2) – 6(a-2)
⟹ (a-6) × (a-2)
⟹ but a = (x2 – x)
⟹ ((x2 – x)-6) × ((x2 – x) – 2
⟹ (x2 – x -6) × (x2 – x -2)
⟹ (x2 –3x + 2x – 6) × (x2 – 2x + x -2)
⟹ (x (x-3) + 2(x – 3)) × (x(x-2) + 1(x-2))
⟹ (x + 2)(x-3)(x-2)(x+1)
Therefore, the factorized form = (x + 2)(x-3)(x-2)(x+1)
Question 8.
Factorize the following polynomials.
(x-5)2–(5x-25)-24
Answer:
(x-5)2 – 5(x-5) -24
Put (x – 5) = a
⟹ a2 – 5a - 24
⟹ a2 – 8a + 3a -24
⟹ a (a-8) + 3(a-8)
⟹ (a-8) × (a+3)
⟹ But a = (x – 5)
⟹ (x – 5 - 8) × (x-5 +3)
⟹ (x – 13) × (x-2)
Therefore, the factorized form of the polynomial = (x – 13) × (x-2)
Question 9.
Factorize the following polynomials.
(x2 – 6x)2 – 8(x2 – 6x + 8) – 64
Answer:
(x2 – 6x) 2 – 8(x2 – 6x + 8) – 64
⟹ (x2 – 6x) 2 – 8(x2 -6x) - 64 – 64
⟹ (x2 – 6x) 2 – 8(x2 -6x) – 128
Put (x2 -6x) = a
⟹ (a) 2 – 8(a) – 128
⟹ a2 – 8a – 128
⟹ a2 – 16a + 8a - 128
⟹ a (a-16) + 8(a – 16)
⟹ (a + 8) × (a-16)
⟹ But a = (x2 -6x)
⟹ ((x2 -6x) + 8) × ((x2 -6x) – 16)
⟹ (x2 -6x + 8) × (x2 -6x – 16)
⟹ (x2 -4x – 2x + 8) × (x2 -8x + 2x – 16)
⟹ (x(x-4) – 2(x-4)) × (x(x-8) + 2(x-8)
⟹ (x-2)(x-4)(x-8)(x+2)
Therefore, the factorized form = (x-2) (x-4) (x-8) (x+2)
Question 10.
Factorize the following polynomials.
(x2 – 2x + 3) (x2 – 2x + 5) – 35
Answer:
(x2 – 2x + 3) (x2 – 2x + 5) – 35
Put (x2 – 2x) = a
⟹ (a + 3) (a + 5) – 35
⟹ (a2 + 5a + 3a + 15) – 35
⟹ a2 + 8a + 15 – 35
⟹ a2 + 8a – 20
⟹ a2 + 10a – 2a – 20
⟹ a (a + 10) – 2 (a + 10)
⟹ (a – 2) (a + 10)
⟹ But a = (x2 – 2x)
⟹ (x2 – 2x) + 10) ((x2 – 2x) – 2)
⟹ (x2 – 2x + 10) (x2 – 2x – 2)
Therefore, the factorized form = (x2 – 2x + 10) (x2 – 2x – 2)
Question 11.
Factorize the following polynomials.
(y+2) (y+3) (y-3) (y + 8) + 56
Answer:
(y+2) (y+3) (y-3) (y + 8) + 56
⟹ (y2 + 3y + 2y + 6) (y2 + 8y - 3y - 24) + 5
⟹ (y2 + 5y + 6) (y2 + 5y - 24) + 56
Put (y2 + 5y) = a
⟹ (a + 6) (a – 24) + 56
⟹ a2 -24a + 6a – 144 + 56
⟹ a2 – 18a – 88
⟹ a2 -22a + 4a – 88
⟹ a (a-22) + 4 (a-22)
⟹ (a+4) (a-22)
⟹ But a = (y2 + 5y)
⟹ ((y2 + 5y) + 4) ((y2 + 5y)-22)
⟹ (y2 + 5y + 4) (y2 + 5y -22)
⟹ (y2 + 4y + y + 4) (y2 + 5y -22)
⟹ (y (y+4) + 1(y+4)) (y2 + 5y -22)
⟹ (y+1) (y+ 4) (y2 + 5y -22)
Therefore, the factorized form = (y+1) (y+ 4) (y2 + 5y -22)
Question 12.
Factorize the following polynomials.
(y2 + 5y)(y2 + 5y – 2) - 24
Answer:
Put (y2 + 5y) = a
⟹ a (a -2) – 24
⟹ a2 -2a – 24
⟹ a2 -6a + 4a – 24
⟹ a (a – 6) + 4 (a – 6)
⟹ (a + 4) (a – 6)
⟹ But a = (y2 + 5y)
⟹ ((y2 + 5y) + 4) ((y2 + 5y) – 6)
⟹ (y2 + 5y + 4) (y2 + 5y – 6)
⟹ (y2 + 4y + y + 4) (y2 + 6y - y – 6)
⟹ (y (y + 4) + 1 (y + 4)) (y (y + 6) – 1 (y + 6))
⟹ (y + 4) (y + 1) (y + 6) (y – 1)
Therefore, the factorized form = (y + 4) (y + 1) (y + 6) (y – 1)
Question 13.
Factorize the following polynomials.
(x – 3) (x- 5) (x – 4)2 – 6
Answer:
(x – 3) (x- 5) (x – 4)2 – 6
⟹ (x2 – 8x + 15) (x2 – 8x + 16) – 6
⟹ Put (x2 – 8x) = a
⟹ (a + 15) (a + 16) – 6
⟹ a2 + 15a + 16a + 240 – 6
⟹ a2 + 31a + 234
⟹ a2 + 13a + 18a + 234
⟹ a (a + 13) + 18 (a + 13)
⟹ (a + 18) (a + 13)
⟹ But a = (x2 – 8x)
⟹ ((x2 – 8x) + 18) ((x2 – 8x) + 13)
⟹ (x2 – 8x + 18) (x2 – 8x + 13)
Therefore, the factorized form =(x2 – 8x + 18) (x2 – 8x + 13)
