# Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions

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## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions

Exercise 3.1

Question 1.
Find the product of the following.
(i) (x, y)
(ii) (10x, 5y)
(iii) (2x2, 5y2)
Solution:
(i) x × y = xy
(ii) 10x × 5y = (10 × 5) x × xy
= 50 xy
(iii) 2x2 x 5y2 = (2 x 5) x (x2 + y2)
= 10x2y2

Question 1.
Find the product of the following.
(i) 3ab2 c3 by 5a3b2c
(ii) 4x2yz by $$\frac{3}{2}$$ x2yz2
Solution:
(i) (3ab3c3) × (5a3b2c)
= (3 × 5)(a × a3 × b2 × b2 × c2 × c)
= 15a1+3.b2+2.c3+1 = 15a4b4c4

(ii) 4x2yz by $$\frac{3}{2}$$ x2yz2
= (4 × $$\frac{3}{2}$$) × (x2 × x2 × y × y × z × z2)
= -6x2+2 y1+1 x1+2 = -6x4y2z3

Question 1.
Simplify (3x – 2) (x – 1) (3x + 5).
Solution:
(3x – 2) (x – 1) (3x + 5)
= {(3x – 2) (x – 1)} × (3x + 5) [∴Multiplication in associative]
= {3x (x – 1) – 2 x – 1)} × (3x + 5)
= (3x2 – 3x – 2x + 2) × (3x + 5)
= (3x2 – 5x + 2) (3x + 5)
= 3x2 × (3x + 5) – 5x (3x + 5) + 2 (3x + 5)
= (9x3 + 15x2) + (-15x2 – 25x) + (6x + 10)
= 9x3 + 15x2 – 15x2 – 25x + 6x + 10
= 9x3 – 19x + 10

Question 2.
Simplify (5 – x) (3 – 2x) (4 – 3x).
Solution:
(5 – x) (3 – 2x) (4 – 3x)
= {(5 – x)(3 – 2x)} × (4 – 3x) [∴ Multiplication in association]
= {5 (3 – 2x) -x (3 – 2x)} × (4 – 3x)
= (15 – 10x – 3x + 2x2) × (4 – 3x)
= (2x2 – 13x + 15) (4 – 3x)
= 2x2 × (4 – 3x) – 13x (4 – 3x) + 15 (4 – 3x)
= 8x3 – 63 – 52x + 39x2 + 60 – 45x
= -6x3 + 47x2 – 97x + 60

Exercise 3.2

Question 1.
Divide.
(i) 12x3y3 by 3x2y
(ii) -15a2 bc3 by 3ab
(iii) 25x3y2 by – 15x2y
Solution:

Question 1.
Divide
(i) 15m2n3 by 5m2n2
(ii) 24a3b3 by -8ab
(iii) -21 abc2 by 7 abc
Solution:

Question 2.
Divide
(i) 16m3y2 by 4m2y
(ii) 32m2 n3p2 by 4mnp
Solution:

Question 1.
Divide.
(i) 9m5 + 12m4 – 6m2 by 3m2
(ii) 24x3y + 20x2y2 – 4xy by 2xy
Solution:

Exercise 3.3

Question 1.
Evaluate:
(i) (2x + 3y)2
(ii) (2x – 3y)2
Solution:
(i) (2x + 3y)2
= (2x)2 + 2 × (2x) × (3y) + (3y)2
[using (a + b)2 = a2 + 2ab + b2]
= 4x2 + 12xy + 9y2

(ii) (2x – 3y)2
= (2x)2 – 2(2x) (3y) + (3y)2
[∵ using (a – b)2 = a2 – 2ab + b2]
= 4x2 – 12xy + 9y2

Question 1.
Evaluate the following
(i) (2x – 3) (2x + 5)
(ii) (y – 7) (y + 3)
(iii) 107 × 103
Solution:
(i) (2x – 3) (2x + 5)
= (2x)2 + (-3 + 5) (2x) + (-3) (5)
[∵ (x + a) (x + b) = x2 + (a + b)x + ab]
= 22x2 + 2 × 2x + (-15)
= 4x2 + 4x – 15

(ii) (y – 7) (y + 3)
= y2 + (-7 + 3)y + (-7) (3)
[∵ (x + a)(x + b) = x2 + (a + b)x + ab]
= y2 – 4y + (-21) = y2 – 4y – 21

(iii) 107 × 103
= (100 + 7) × (100+ 3)
= 1002 + (7 + 3) × 100 +(7 × 3)
= 10000 + 10 × 100 + 21 = 10000 + 1000 + 21 = 11021

Question 1.
If x + y = 12 and xy = 14 find x2 + y2.
Solution:
(x + y)2 = x2 + y2 + 2xy
122 = x2 + y2 + 2 × 14
144 = x2 + y2 + 28
x2 + y2 = 144 – 28
x2 + y2 = 116

Question 2.
If 3x + 2y = 12 and xy = 6 find the value of 9x2 + 4y2.
Solution:
(3x + 2y)2 = (3x)2 + (2y)2 + 2 (3x) (2y)
= 9x2 + 4y2 + 12xy
122 = 9x2 + 4y2 + 12 × 6
144 = 9x2 + 4y2 + 72
144 – 72 = 9x2 + 4y2
∴ 9x2 + 4y2 = 72

Exercise 3.4

Question 1.
Factorize:
(i) 7(2x + 5) + 3 (2x + 5)
(ii) 12x3y4 + 16x2y5 – 4x5y2
Solution:
(i) 7(2x + 5) + 3 (2x + 5)
= (2x + 5) (7 + 3)
(ii) 12x3y4 + 16x2y5 – 4x5y2
= 4x2y2 (3xy2 + 4y3 – x3)

1. Factorize
(i) 81a2 – 121b2
(ii) x2 + 8x + 16
Solution:
(i) 81a2 – 121b2
= (9a)2 – (11b)2
[∵ using a2 – b2 = (a + b)2]
= (9a + 11b) (9a – 11b)

(ii) x2 + 8x + 16 = x2 + 2 × x × 4 + 42
[∵ using a2 + 2ab + b2 = (a + b)2]
= (x + 4)2 = (x + 4)(x + 4)

Question 1.
Factorize
(i) x2 + 2xy + y2 – a2 + 2ab – b2
(ii) 9 – a6 + 2a3 – b6
Solution:
(i) x2 + 2xy + y2 – a2 + 2ab – b2.
= (x2 + 2xy + y2) – (a2 – 2ab + b2)
= (x + y)2 – (a – b)2
= {(x + y) + (a – b)} {(x + y) – (a – b)}
= (x + y + a – b) (x + y – a + b)

(ii) a – a6 + 2a3b3 – b6
= 9 – (a6 – 2a3b3 + b6)
= 32 -{(a3)2 – 2 × a3 × b3 + (b3)2}
= 32 – (a3 – 63)2
= {3 + (a3 – b3)} {3 – (a3 – b3)}
= (3 + a3 – b3) (3 – a3 + b3)
= (a3 – b3 + 1){-a3 + b3 + 3)

Question 2.
Factorize
(i) 100 (x + y )2 – 81 (a + b)2
(ii)(x + 1)2 – (x – 2)2
Solution:
(i) 100 (x + y)2 – 81 (a + b)2
= {10 (x + y)}2 – {(a (a + b)}2
= {10 (x + y) + 9 (a + b)}
{10 (x + y) – 9(a + b)}
= (10x + 10y + 9a – 9b)}
(10x + 10y – 9a – 9b)

(ii) (x – 1)2 – (x – 2)2
= {(x – 1 +(x – 2)}
{(x – 1) – (x – 2)}
= (2x – 3) – (x – 1 – x + 2)
= (2x – 3) × 1 = 2x – 3
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