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Tuesday, 23 July 2024

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



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