Students can Download Maths Chapter 3 Algebra Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

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

Additional Questions And Answers

Exercise 3.1

Very Short Answers [2 Marks]

Question 1.

Find the product of the following.

(i) (x, y)

(ii) (10x, 5y)

(iii) (2x^{2}, 5y^{2})

Solution:

(i) x × y = xy

(ii) 10x × 5y = (10 × 5) x × xy

= 50 xy

(iii) 2x^{2} x 5y^{2} = (2 x 5) x (x^{2} + y^{2})

= 10x^{2}y^{2}

Short Answers [3 Marks]

Question 1.

Find the product of the following.

(i) 3ab^{2} c^{3} by 5a^{3}b^{2}c

(ii) 4x^{2}yz by \(\frac{3}{2}\) x^{2}yz^{2}

Solution:

(i) (3ab^{3}c^{3}) × (5a^{3}b^{2}c)

= (3 × 5)(a × a^{3} × b^{2} × b^{2} × c^{2} × c)

= 15a^{1+3}.b^{2+2}.c^{3+1} = 15a^{4}b^{4}c^{4}

(ii) 4x^{2}yz by \(\frac{3}{2}\) x^{2}yz^{2}

= (4 × \(\frac{3}{2}\)) × (x^{2} × x^{2} × y × y × z × z^{2})

= -6x^{2+2} y^{1+1} x^{1+2} = -6x^{4}y^{2}z^{3}

Long Answers [5 Marks]

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)

= (3x^{2} – 3x – 2x + 2) × (3x + 5)

= (3x^{2} – 5x + 2) (3x + 5)

= 3x^{2} × (3x + 5) – 5x (3x + 5) + 2 (3x + 5)

= (9x^{3} + 15x^{2}) + (-15x^{2} – 25x) + (6x + 10)

= 9x^{3} + 15x^{2} – 15x^{2} – 25x + 6x + 10

= 9x^{3} – 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 + 2x^{2}) × (4 – 3x)

= (2x^{2} – 13x + 15) (4 – 3x)

= 2x^{2} × (4 – 3x) – 13x (4 – 3x) + 15 (4 – 3x)

= 8x^{3} – 6^{3} – 52x + 39x^{2} + 60 – 45x

= -6x^{3} + 47x^{2} – 97x + 60

Exercise 3.2

Very Short Answers [2 Marks]

Question 1.

Divide.

(i) 12x^{3}y^{3} by 3x^{2}y

(ii) -15a^{2} bc^{3} by 3ab

(iii) 25x^{3}y^{2} by – 15x^{2}y

Solution:

Short Answers [3 Marks]

Question 1.

Divide

(i) 15m^{2}n^{3} by 5m^{2}n^{2}

(ii) 24a^{3}b^{3} by -8ab

(iii) -21 abc^{2} by 7 abc

Solution:

Question 2.

Divide

(i) 16m^{3}y^{2} by 4m^{2}y

(ii) 32m^{2} n^{3}p^{2} by 4mnp

Solution:

Long Answers [5 Marks]

Question 1.

Divide.

(i) 9m^{5} + 12m^{4} – 6m^{2} by 3m^{2}

(ii) 24x^{3}y + 20x^{2}y^{2} – 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} = a^{2} + 2ab + b^{2}]

= 4x^{2} + 12xy + 9y^{2}

(ii) (2x – 3y)^{2}

= (2x)^{2} – 2(2x) (3y) + (3y)^{2}

[∵ using (a – b)^{2} = a^{2} – 2ab + b^{2}]

= 4x^{2} – 12xy + 9y^{2}

Short Answers [3 Marks]

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) = x^{2} + (a + b)x + ab]

= 2^{2}x^{2} + 2 × 2x + (-15)

= 4x^{2} + 4x – 15

(ii) (y – 7) (y + 3)

= y^{2} + (-7 + 3)y + (-7) (3)

[∵ (x + a)(x + b) = x^{2} + (a + b)x + ab]

= y^{2} – 4y + (-21) = y^{2} – 4y – 21

(iii) 107 × 103

= (100 + 7) × (100+ 3)

= 100^{2} + (7 + 3) × 100 +(7 × 3)

= 10000 + 10 × 100 + 21 = 10000 + 1000 + 21 = 11021

Long Answers [5 Marks]

Question 1.

If x + y = 12 and xy = 14 find x^{2} + y^{2}.

Solution:

(x + y)^{2} = x^{2} + y^{2} + 2xy

12^{2} = x^{2} + y^{2} + 2 × 14

144 = x^{2} + y^{2} + 28

x^{2} + y^{2} = 144 – 28

x^{2} + y^{2} = 116

Question 2.

If 3x + 2y = 12 and xy = 6 find the value of 9x^{2} + 4y^{2}.

Solution:

(3x + 2y)^{2} = (3x)^{2} + (2y)^{2} + 2 (3x) (2y)

= 9x^{2} + 4y^{2} + 12xy

12^{2} = 9x^{2} + 4y^{2} + 12 × 6

144 = 9x^{2} + 4y^{2} + 72

144 – 72 = 9x^{2} + 4y^{2}

∴ 9x^{2} + 4y^{2} = 72

Exercise 3.4

Question 1.

Factorize:

(i) 7(2x + 5) + 3 (2x + 5)

(ii) 12x^{3}y^{4} + 16x^{2}y^{5} – 4x^{5}y^{2}

Solution:

(i) 7(2x + 5) + 3 (2x + 5)

= (2x + 5) (7 + 3)

(ii) 12x^{3}y^{4} + 16x^{2}y^{5} – 4x^{5}y^{2}

= 4x^{2}y^{2} (3xy^{2} + 4y^{3} – x^{3})

Short Answers [3 Marks]

1. Factorize

(i) 81a^{2} – 121b^{2}

(ii) x^{2} + 8x + 16

Solution:

(i) 81a^{2} – 121b^{2}

= (9a)^{2} – (11b)^{2}

[∵ using a^{2} – b^{2} = (a + b)^{2}]

= (9a + 11b) (9a – 11b)

(ii) x^{2} + 8x + 16 = x^{2} + 2 × x × 4 + 4^{2}

[∵ using a^{2} + 2ab + b^{2} = (a + b)^{2}]

= (x + 4)^{2} = (x + 4)(x + 4)

Long Answers [5 Marks]

Question 1.

Factorize

(i) x^{2} + 2xy + y^{2} – a^{2} + 2ab – b^{2}

(ii) 9 – a^{6} + 2a^{3} – b^{6}

Solution:

(i) x^{2} + 2xy + y^{2} – a^{2} + 2ab – b^{2}.

= (x^{2} + 2xy + y^{2}) – (a^{2} – 2ab + b^{2})

= (x + y)^{2} – (a – b)^{2}

= {(x + y) + (a – b)} {(x + y) – (a – b)}

= (x + y + a – b) (x + y – a + b)

(ii) a – a^{6} + 2a^{3}b^{3} – b^{6}

= 9 – (a^{6} – 2a^{3}b^{3} + b^{6})

= 3^{2} -{(a^{3})^{2} – 2 × a^{3} × b^{3} + (b^{3})^{2}}

= 3^{2} – (a^{3} – 6^{3})^{2}

= {3 + (a^{3} – b^{3})} {3 – (a^{3} – b^{3})}

= (3 + a^{3} – b^{3}) (3 – a^{3} + b^{3})

= (a^{3} – b^{3} + 1){-a^{3} + b^{3} + 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