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### Practice Set 3.6 Polynomials Class 9th Mathematics Part I MHB Solution

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

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

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

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

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

√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

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

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

(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

(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

(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

(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

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

(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)