Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.5 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.5

Question 1.

Subtract: -2(xy)^{2} (y^{3} + 7x^{2}y + 5) from 5y^{2} (x^{2}y^{3} – 2x^{4}y + 10x^{2})

Answer:

Question 2.

Multiply (4x^{2} + 9) and (3x – 2).

Answer:

(4x^{2} + 9)(3x – 2) = 4x^{2}(3x – 2) + 9(3x – 2)

= (4x^{2})(3x) – (4x^{2}) (2) + 9(3x) – 9(2)

= (4 × 3 × x × x^{2}) – (4 × 2 × x^{2}) + (9 × 3 × x) – 18

= 12x^{3} – 8x^{2} + 27x – 18(4x^{3} + 9)(3x – 2)

= 12x^{3} – 8x^{2} + 27x – 18

Question 3.

Find the simple interest on Rs. 5a^{2}b^{2} for 4ab years at 7b% per annum.

Answer:

Question 4.

The cost of a note book is Rs. 10ab. If Babu has Rs. (5a^{2}b + 20ab^{2} + 40ab). Then how many note books can he buy?

Answer:

For ₹ 10 ab the number of note books can buy = 1.

Number of note book he can buy = \(\frac { 1 }{ 2 }\)a + 2b + 4

Question 5.

Factorise: (7y^{2} – 19y – 6)

Answer:

7y^{2} – 19y – 6 is of the form ax^{2} + bx + c where a = 7; b = – 19; c = – 6

Product = – 42 | Sum = -19 |

1 × – 42 = -42 | 1 + (-42) = – 41 |

2 × – 21 = – 42 | 2 + (-21) = – 19 |

The product a × c = 7 × – 6 = – 42

sum b = – 19

The middle term – 19 y can be written as – 21y + 2y

7y^{2} – 19y – 6 = 7y^{2} – 21y + 2y – 6

= 7y(y – 3) + 2(y – 3)

= (y – 3)(7y + 2)

7y^{2} – 19y – 6 = (y – 3)(7y + 2)

Question 6.

A contractor uses the expression 4x^{2} + 11x + 6 to determine the amount of wire to order when wiring a house. If the expression comes from multiplying the number of rooms times the number of outlets and he knows the number of rooms to be (x + 2), find the number of outlets in terms of ’x’. [Hint : factorise 4x^{2} + 11x + 6]

Answer:

Given Number of rooms = x + 2

Number of rooms × Number of outlets = amount of wire.

(x + 2) × Number of outlets = 4x^{2} + 11x + 6

Number of outlets = \(\frac{4 x^{2}+11 x+6}{x+2}\) … (1)

Now factorising 4x^{2} + 11x + 6 which is of the form ax^{2} + bx + c with a = 4 b = 11 c = 6.

The product a × c = 4 × 6 = 24

sum b = 11

Product = 24 | Sum = 11 |

1 × 24 = 24 | 1 + 24 = 25 |

2 × 12 = 24 | 2 + 12 = 14 |

3 × 8 = 24 | 3 + 18 = 11 |

The middle term 11x can be written as 8x + 3x

∴ 4x^{2} + 11 x + 6 = 4x^{2} + 8x + 3x + 6

= 4x(x + 2) + 3 (x + 2)

4x^{2} + 11x + 6 = (x + 2)(4x + 3)

Now from (1) the number of outlets

∴ Number of outlets = 4x + 3

Question 7.

A mason uses the expression x^{2} + 6x + 8 to represent the area of the floor of a room. If the decides that the length of the room will be represented by (x + 4), what will the width of the room be in terms of x?

Answer:

Given length of the room = x + 4 .

Area of the room = x^{2} + 6x + 8

Length × breadth = x^{2} + 6x + 8

breadth = \(\frac{x^{2}+6 x+8}{x+4}\) ….. (1)

Factorizing x^{2} + 6x + 8, it is in the form of ax^{2} + bx + c

Where a =1 b = 6 c = 8.

The product a × c = 1 × 8 = 8

sum = b = 6

Product = 8 | Sum = 6 |

1 × 8 = 8 | 1 + 8 = 9 |

2 × 4 = 8 | 2 + 4 = 6 |

The middle term 6x can be written as 2x + 4x

∴ x^{2} + 6x + 8 = x^{2} + 2x + 4x + 8

= x(x + 2) + 4(x + 2)

x^{2} + 6x + 8 = (x + 2)(x + 4)

Now from (1)

∴ Width of the room = x + 2

Question 8.

Find the missing term: y^{2} + (-)x + 56 = (y + 7)(y + -)

Answer:

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

56 = 7 × 8

∴ y^{2} + (7 + 8)x + 56 = (y + 7) (y + 8)

Question 9.

Factorise : 16p^{4} – 1

Answer:

16p^{4} – 1 = 2^{4}p^{4} – 1 =(2^{2})^{2} (p^{2})^{2} – 1^{2}

= (2^{2}p^{2})^{2} – 1^{2}

Comparing with a^{2} – b^{2} (a + b)(a – b) where a = 2^{2}p^{2} and b= 1

∴ (2^{2}p^{2})^{2} – 1^{2} = (2^{2}p^{2} + 1)(2^{2}p^{2} – 1)

= (4p^{2} + 1)(4p^{2} – 1)

∴ 16p^{4} – 1 = (4p^{2} + 1)(4p^{2} – 1) = (4p^{2} + 1)(2^{2}p^{2} – 1^{2})

= (4p^{2} + 1) [(2p)^{2} – 1^{2}] = (4p^{2} + 1) (2p + 1)(2p – 1) [∵ using a^{2} – b^{2} = (a + b)(a – b)]

∴ 16p^{4} – 1 = (4p^{2} + 1)(2p + 1)(2p – 1)

Question 10.

Factorise : 3x^{3} – 45x^{2}y + 225xy^{2} – 375y^{3}

Answer:

= 3x^{3} – 45x^{2}y + 225xy^{2} – 375y^{3}

= 3(x^{3} – 15x^{2}y + 75xy^{2} – 125y^{3})

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

= 3(x – 5y)^{3}