Unit 4: Factorization - Mathematics For Class IX (Science Group) - Solved Exercise 4.1

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Unit 4: Factorization
Solved Exercise 4.1

1. Factorize the following:
(i) 4x + 16y + 4z
Solution:
⇒ 4x + 16y + 4z [taking 4 as a common from the expressions]
⇒ 4(x + 4y + z) Ans.

ii) x² + 3x²y + 4x²y²z
Solution:
⇒ x² + 3x²y + 4x²y²z [taking x² as a common from the expressions]
⇒ x²(1 + 3y + 4y²z) Ans.

iii) 3pqr + 6qpt + 3pqs
Solution:
⇒ 3pqr + 6qpt + 3pqs [taking 3pq as a common from the expressions]
⇒ 3pq(r + 2t + s) Ans.

iv) 9qr(s² + t²) + 18q²r²(s² + t²)
Solution:
⇒ 9qr(s² + t²) + 18q²r²(s² + t²) [taking 9qr(s² + t²) as a common from the expressions]
⇒ 9qr(s² + t²)(1 + 2qr Ans).

vi) a(x - y) - a²b(x - y) + a²b²(x - y)
Solution:
⇒ a(x - y) - a²b(x - y) + a²b²(x - y) [taking a(x - y) as a common from the expressions]
⇒ a(x - y)(1 - ab + ab²) Ans

2. Factorize the following:
i) 7x + xz + 7z + z²
Solution:
⇒ 7x + xz + 7z + z² [Taking x in the first pair & z in the second pair as a common from the expressions]
⇒ x(7 + z) + z(7 + z)
⇒ (x + z)(7 + z) are required factors Ans.

ii) 9a²b + 18ab² - 6ac - 12bc
Solution:
⇒ 9a²b + 18ab² - 6ac - 12bc [Taking 9ab in the first pair & -6c in the second pair as a common from the expressions]
⇒ 9ab(a + 2b) - 6c(a + 2b)
⇒ (9ab - 6c)(a + 2b) are required factors Ans.

OR

⇒ 9a²b + 18ab² - 6ac - 12bc [Taking 3 as a common from the expressions]
⇒ 3(3a²b + 6ab² - 2ac - 4bc) [Now taking 3ab in the first pair & -2c in the second pair as a common from the expressions]
⇒ 3{3ab(a + 2b) - 2c(a + 2b)}
⇒ 3(3ab - 2c)(a + 2b)
⇒ (9ab - 6c)(a + 2b) are required factors Ans.

iii) 6t - 12p + 4tq - 8pq
Solution:
⇒ 6t - 12p + 4tq - 8pq [Taking 6 in the first pair & +4q in the second pair as a common from the expressions]
⇒ 6(t - 2p) + 4q(t - 2q)
⇒ (6 + 4q)(t - 2p) are required factors Ans.

(iv) r² + 9rs - 7rs - 63s²
Solution:
⇒ r² + 9rs - 7rs - 63s² [Taking r in the first pair & -7s in the second pair as a common from the expressions]
⇒ r(r + 9s) - 7s(r + 9s)
⇒ (r - 7s)(r + 9s) are required factors Ans.

3. Find the factors of:
i) 4a² + 12ab + 9b²
Solution:
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ (2a)² + 2(2a)(3b) + (3b)²
⇒ (2a + 3b)² Ans

ii) 36x⁴ + 12x² + 1
Solution:
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ (6x²)² + 2(6x²)(1) + (1)²
⇒ (6x² + 1)² Ans

iv) 81y² + 144yz + 64z²
Solution:
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ (9y)² + 2(9y)(8z) + (8z)²
⇒ (9y + 8z)² Ans

v) 625 + 50a²b + a⁴b²
Solution:
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ (25)² + 2(25)(a²b) + (a²b)²
⇒ (25 + a²b)² Ans

vi) a² + 0.4a + 0.04
Solution:
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ (a)² + 2(a)(0.2) + (0.2)²
⇒ (a + 0.2)² Ans

4. Factorize:
i) b⁴ - 4b²c² + 4c⁴
Solution:
By using formula: [∵ a² - 2ab + b² = (a - b)²] a perfect square
⇒ (b²)² - 2(b²)(2c²) + (2c²)²
⇒ (b² - 2c²)² Ans

iii) 2a³b³ - 16a²b⁴ + 32ab⁵
Solution:
⇒ 2a³b³ - 16a²b⁴ + 32ab⁵ [By taking 2ab³ as a common from the expression]
⇒ 2ab³(a² - 8ab + 16b²)
By using formula: [∵ a² - 2ab + b² = (a - b)²] a perfect square
⇒ 2ab³{(a)² - 2(a)(4b) + (4b)²)}
⇒ 2ab³(a - 4b)² Ans

iv) 9(p+q)² - 6(p+q)r² + r⁴
Solution:
By using formula: [∵ a² - 2ab + b² = (a - b)²] a perfect square
⇒ {3(p+q)}² - 2{3(p+q)}(r²) + (r²)²
⇒ {3(p+q) - r²}² Ans

v) x²y² - 0.1xy + 0.0025
Solution:
By using formula: [∵ a² - 2ab + b² = (a - b)²] a perfect square
⇒ (xy)² - 2(xy)(0.05) + (0.05)²
⇒ (xy - 0.05)² Ans

vi) (a - b)² + 18(a - b) + 81
Solution:
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ (a - b)² + 2(a - b)(9) + (9)²
⇒ {(a - b) - 9}² Ans

5. Factorize:
i) 4a² - 9b²
Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (2a)² - (3b)²
⇒ (2a - 3b)(2a + 3b) Ans

ii) 16x² - 25y²
Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (4x)² - (5y)²
⇒ (4x - 5y)(4x + 5y) Ans

iii) 100x²z² - y⁴
Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (10xz)² - (y²)²
⇒ (10xz - y²)(10xz + y²) Ans

6. Find the factors of:
i) (2x + z)² - (2x - z)²
Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (2x + z)² - (2x - z)²
⇒ {(2x + z) - (2x - z)} {(2x + z) + (2x - z)} Ans

ii) (4a - 9b)² - (2a + 5b)²
Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (4a - 9b)² - (2a + 5b)²
⇒ {(4a - 9b) - (2a + 5b)} {(4a - 9b) + (2a + 5b)} Ans

iii) 169x⁴ - (3t + 4)²
Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (13x²)² - (3t + 4)²
⇒ {13x² - (3t + 4)} {13x² + (3t + 4)]}Ans

iv) (9x² - 4y²)² - (4x² - y²)²
Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (9x² - 4y²)² - (4x² - y²)²
⇒ {(3x)² - (2y)²)}² - {(2x)² - (y²)}²
⇒ {(3x - 2y)(3x + 2y)}² - {(2x - y)(2x + y)}²
[again by using formula: a² - b² = (a - b)(a + b)]
⇒ {(3x - 2y)(3x + 2y) - (2x - y)(2x + y)} {(3x - 2y)(3x + 2y) + (2x - y)(2x + y)} Ans

OR

Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (9x² - 4y²)² - (4x² - y²)²
⇒ {(9x² - 4y²) - (4x² - y²)} {(9x² - 4y²) + (4x² - y²)}
[again by using formula: a² - b² = (a - b)(a + b)]
⇒ {(3x)² - (2y)²)} - {(2x)² - (y)²)} {(3x)² - (2y)²)} + {(2x)² - (y)²)}
⇒ {(3x - 2y)(3x + 2y) - (2x - y)(2x + y)} {(3x - 2y)(3x + 2y) + (2x - y)(2x + y)} Ans

7. Find the factors of:
i) (x² + 2xy + y²) - 9z⁴
Solution:
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ (x)² + 2(x)(y) + (y)²) - (3z²)²
⇒  (x + y)² - (3z²)²
[again by using formula: a² - b² = (a - b)(a + b)] Difference of two squares
⇒ {(x + y) - 3z²} {(x + y) + 3z²}
⇒ (x + y - 3z²)(x + y + 3z²) Ans

ii) (4a² + 8ab² + 4b⁴) - 9c²
Solution:
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ {(2a)² + 2(2a)(2b²) + (2b²)²)} - (3c)²
⇒ (2a + 2b²)² - (3c)²
[Again by using formula: a² - b² = (a - b)(a + b)] Difference of two squares
⇒ {(2a + 2b²) - 3c} {(2a + 2b²) + 3c}
⇒ (2a + 2b² - 3c)(2a + 2b² + 3c) Ans

iii) 16d⁴ - (c⁴ - 2c²d + d²)
Solution:
By using formula: [∵ a² - 2ab + b² = (a - b)²] a perfect square
⇒ (4d²)² - {(c²)² - 2(c²)(d) + (d)²)}
⇒ ∴ (4d²)² - (c² - d)²
[Again by using formula: a² - b² = (a - b)(a + b)] Difference of two squares
⇒ {4d² - (c² - d)} {4d² + (c² - d)
⇒ (4d² - c² + d)(4d² + c² - d) Ans

iv) 4(x² + 2xy² + y⁴) - 9y⁶
Solution:
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ 4{(x)² + 2(x)(y²) + (y²)²} - (3y³)²
⇒ ∴ 4{(x + y²)²} - (3y³)²
[Again by using a formula: a² - b² = (a - b)(a + b)] Difference of two squares
⇒ {2(x + y²)}² - (3y³)²
⇒ {2(x + y²) + 3y³} {2(x + y²) - 3y³}Ans

v) x² - y² - 4x - 2y + 3
Solution:
[we rearrange the terms]
⇒ (x² - 4x + 3) - y² - 2y [Taking -1 as common from second expression]
⇒ (x² - 4x + 3) - (y² + 2y)
[Add 1 in both the terms expression]
⇒ (x² - 4x + 3 + 1) - (y² + 2y +1)
⇒ (x² - 4x + 4) - (y² + 2y +1)
By using formula: [∵ a² 土 2ab + b² = (a + b)²] a perfect square
⇒ {(x)² - 2(x)(2) + (2)²} - {(y)² + 2(y)(1) +(1)²}
⇒ (x - 2)² - (y + 1)²
[Again by using formula: a² - b² = (a - b)(a + b)] Difference of two squares
⇒ {(x - 2) + (y + 1)} {(x - 2) - (y + 1)} Ans

vi) 4x² - y² - 2y - 1
Solution:
⇒ 4x² - (y² + 2y + 1) [Taking -1 as common from second expression]
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ 4x² - {(y)² + 2(y)(1) + (1)²}
⇒ 4x² - (y + 1)²
[Again by using formula: a² - b² = (a - b)(a + b)] Difference of two squares
⇒ (2x)² - (y + 1)²
⇒ {(2x - (y + 1)} {(2x + (y + 1)}
⇒ (2x - y - 1)(2x + y + 1) Ans

8. Factorize:
i) (√ ab)²z² - (√ c)²
Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (√ abz)² - (√ c)²
⇒ (√ abz - √ c)(√ abz + √ c) Ans

i) (√ 4x)² - (√ 9y)²
Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (√ 4x)² - (√ 9y)²
⇒ (√ 4x - √ 9y)(√ 4x + √ 9y) Ans