Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.4 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.4

Question 1.

Factorise the following by taking out the common factor

(i) 18xy – 12yz

Answer:

18xy – 12yz = (2 × 3 × 3 × y × x) – (2 × 2 × 3 × y × z)

Taking out the common factors 2, 3, y, we get

= 2 × 3 × y(3x – 2z) = 6y(3x – 2z)

(ii) 9x^{5}y^{3} + 6x^{3}y^{2} – 18x^{2}y

Answer:

9x^{5} + 6x^{3}y^{2} – 18x^{2}y = (3 × 3 × x^{2} × x^{3} × y × y) + (2 × 3 × x^{2} × x × y × y) – (2 × 3 × 3 × x^{2} × y)

Taking out the common factors 3, x^{2}, y, we get

= 3 × x^{2} × y (3x^{3}y^{2} + 2xy – 6)

= 3x^{2}y (3x^{3}y^{2} + 2xy – 6)

(iii) x(b – 2c) + y(b – 2c)

Answer:

Taking out the binomial factor (b – 2c) from each term, we have

= (b – 2c)(x + y)

(iv)(ax + ay) + (bx + by)

Answer:

Taking at ‘a’ from the first term and ‘b’ from the second term we have

(ax + ay )+ (bx + by) = a(x + y) + b(x + y)

Now taking out the binomial factor (x + y) from each term

= (x + y) (a + b)

(v) 2x^{2}(4x – 1) – 4x + 1

Answer:

Taking out -1 from last two terms

2x^{2} (4x – 1) – 4x + 1 = 2x^{2} (4x – 1) – 1 (4x – 1)

Taking out the binomial factor 4x – 1, we get

= (4x – 1) (2x^{2} – 1)

(vi) 3y(x – 2)^{2} – 2(2 – x)

Answer:

3y(x – 2)^{2} – 2(2 – x) = 3y(x – 2)(x – 2) – 2( -1)(x – 2)

[∵ Taking out – 1 from 2 – x]

= 3y(x – 2)(x – 2) + 2(x – 2)

Taking out the binomial factor x – 2 from each term, we get

= (x – 2) [3y(x – 2) + 2]

(vii) 6xy – 4y^{2} + 12xy – 2yzx

Answer:

= 6xy + 12xy – 4y^{2} – 2yzx [∵ Addition is commutative]

= (6 × x × y) + (2 × 6 × x × y) + (-1) (2) (2) y + y) + ((-1) (2) (y) (z) (x))

Taking out 6 x x x y from first two terms and (-1) × 2 × y from last two terms we get

= 6 × x × y(1 + 2) + (-1) (2) y [2y + zx]

= 6 × y(3) – 2y(2y + zx)

= (2 × 3 × 3 × x × y) – 2xy(2y + zx)

Taking out 2y from two terms

= 2y(9x – (2y + zx))

= 2y (9x – 2y – xz)

(viii) a^{3} – 3a^{2} + a – 3

Answer:

a^{2} – 3a^{2} + a – 3 = a^{2}(a – 3) + 1(a – 3) [:Groupingthetermssuitably]

= (a – 3) (a^{2} + 1)

(ix) 3y^{3} – 48y

Answer:

3y^{2} – 48y = 3 × y × y^{2} – 3 × l6 × y

Taking out 3 × y

= 3y(y^{2} – 16) = 3y(y^{2} – 4^{2})

Comparing y^{2} – 4^{2} with a^{2} – b^{2}

a = y, b = 4

a^{2} – b^{2} = (a + b) (a – b)

y^{2} – 4^{2} = (y + 4) (y – 4)

∴ 3y(y^{2} – 16) = 3y(y + 4)(y – 4)

(x) ab^{2} – bc^{2} – ab + c^{2}

Answer:

ab^{2} – bc^{2} – ab + c^{2}

Grouping suitably

ab^{2} – bc^{2} – ab + c^{2} = b (ab – c^{2}) – 1 (ab – c^{2})

Taking out the binomial factor ab – c^{2} = (ab – c^{2}) (b – 1)

Question 2.

Factorise the following expressions

(i) x^{2} + 14x + 49

Answer:

x^{2} + 14x + 49 = x^{2} + 14x + 72

Comparing with a^{2} + 2ab + b^{2} = (a + b)^{2} we have a = x and b = 7

⇒ x^{2} + 2(x)(7) + 7^{2} = (x + 7)^{2}

∴ x^{2} + 14x + 49 = (x + 7)^{2}

(ii) y^{2} – 10y + 25

Answer:

y^{2} – 10y + 25 = y^{2} – 10y + 5^{2}

Comparing with a^{2} – 2ab + b^{2} = (a – b)^{2} we get a = y; b = 5

⇒ y^{2} – 2(y) (5) + 5^{2} = (y – 5)^{2}

∴ y^{2} – 10y + 25 = (y – 5)^{2}

(iii) c^{2} – 4c – 12

Answer:

This is of the form ax^{2} + bx + c

Where a = 1, b = -4 c = -12, x = c

Now the product ac = 1 × – 12 = -12 and the sum b = -4

Product = – 72 | Sum = 1 |

1 × (-12) = -12 | 1 + (-12) = -11 |

2 × (-6) = – 12 | 2 + (-6) = – 4 |

∴ The middle term – 4c can be written as 2c – 6c

∴ c^{2} – 4c – 12 = c^{2} + 2c – 6c – 12

= c(c + 2) -6 (c + 2)

Taking out (c + 2)

⇒ (c + 2)(c – 6)

∴ c^{2} – 4c – 12 = (c + 2)(c – 6)

(iv) m^{2} + m – 72

Answer:

m^{2} + m – 72

This is of the form ax + bx + c

where a = 1, b = 1, c = -72

Product = – 72 | Sum = 1 |

1 × -72 = – 72 | 1 + (-72) = -71 |

2 × – 36 = – 72 | 2 + (-36) = – 34 |

3 × (-24) = – 72 | 3 + (-24) = – 21 |

4 × (-18) = -72 | 4 + (-18) = – 14 |

6 × (-12) = -72 | 6 + (-12) = – 6 |

8 × (-9) = -72 | 8 + (-9) = – 1 |

9 × (-8) = – 72 | 9 + (-8) = 1 |

Product a × c = 1 × -72 = -72

Sum b = 1

The middle term m can be written as 9m – 8m

m^{2} + m – 72 = m^{2} + 9m – 8m – 72

= m(m + 9) – 8(m + 9)

Taking out (m + 9)

= (m + 9)(m – 8)

∴ m^{2} + m – 72 = (m + 9)(m – 8)

(v) 4x^{2} – 8x + 3

Answer:

4x^{2} – 8x + 3

This is of the form ax^{2} + bx + c with a = 4 b = -8 c = 3

Product ac = 4 × 3 = 12

sum b = -8

Product = 12 | Sum = -8 |

(-1) × (-12) = 12 | (-1) + (-12) = – 13 |

(-2) × (-6) = 12 | (-2) + (-6) = – 8 |

The middle term can be written as – 8x = – 2x – 6x

4x^{2} – 8x + 3 = 4x^{2} – 2x – 6x + 3

= 2x (2x – 1) – 3 (2x – 1)

= (2x – 1)(2x – 3)

4x^{2} – 8x + 3 = (2x – 1) (2x – 3)

Question 3.

Factorise the following expressions using (a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3} identity

(i) 64x^{3} + 144x^{2} + 108x + 27

(ii) 27p^{3} + 54p^{3}q + 36pq^{2} + 8q^{3}

Answer:

(i) 64x^{3} + 144x^{2} + 108x + 27

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

= (4x + 3)^{3}

(ii) 27p^{3} + 54p^{3}q + 36pq^{2} + 8q^{3}

= (3p)^{3} + 3(3p)^{2} (2q) + 3(3p) (2q)^{2} + (2q)^{3}

= (3p + 2q)^{3}

Question 4.

Factorise the following expressions using (a – b)^{3} = a^{3} – 3a^{2}b + 3ab^{2} – b^{3} identity

(i) y^{3} – 18y^{2} + 108y – 216

(ii) 8m^{3} – 60m^{2}n + 150mn^{2} – 125n^{3}

Answer:

(i) y^{3} – 18y^{2} + 108y – 216

= y^{3} – 3y^{2}(6) + 3(6)^{2}y – 6^{3}

= (y – 6)^{3}

(ii) 8m^{3} – 60m^{2}n + 150mn^{2} – 125n^{3}

= (2m)^{3} – 3(2m)^{2} (5) + 3(2m)(5n)^{2} – (5n)^{3}

= (2m – 5n)^{3}