RBSE Solutions for Class 8 Maths Chapter 10 Factorization In Text Exercise is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 10 Factorization In Text Exercise.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 10 |

Chapter Name |
Factorization |

Exercise |
In Text Exercise |

Number of Questions |
4 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 8 Maths Chapter 10 Factorization In Text Exercise

**Page No: 119**

Question 1.

Find the two integers a and b such that

a+b | ab | a | b |

8 | 15 | 5 | 3 |

13 | 12 | ||

-1 | -20 | -5 | 4 |

-5 | 4 | ||

10 | 21 | ||

-1 | -12 | ||

-11 | 10 | ||

-7 | 10 |

Solution:

(a + b)² = a² + 2ab + b²,

(a – b)² = a² – 2ab + b²

(a + b)² – (a – b)² – 4ab

⇒ (a – b)² = (a + b)² = 4ab

⇒ a – b = [latex]\sqrt { { (a+b) }^{ 2 }-4ab } [/latex]

Now (i) a + b = 13, and ab = 12

then a – b = [latex]\sqrt { { (13) }^{ 2 }-4\times 12 } [/latex]

= [latex]=\sqrt { 169-48 } =\sqrt { 121 } [/latex] = 11

then a + b + a – b = 13 + 11 = 24

⇒ 2a = 24

⇒ a = 12

and b = 13 – a = 13 – 12 = 1

⇒ b = 1

(ii) When a + b = – 5 and ab = 4

then a – b = [latex]\sqrt { { (13) }^{ 2 }-4\times 12 } [/latex]

Filling the(RBSESolutions.com)above values in table given below

a+b | ab | a | b |

8 | 15 | 5 | 3 |

13 | 12 | 12 | 1 |

-1 | -20 | -5 | 4 |

-5 | 4 | -4 | -1 |

10 | 21 | 7 | 3 |

-1 | -12 | -4 | 3 |

-11 | 10 | -10 | -1 |

-7 | 10 | -5 | -2 |

**Page No: 123**

Question 2.

3x + x + 4x = 56

7x = 56

x = [latex]\frac { 56 }{ 7 }[/latex]

= 8

Find the error.

Solution:

3x + x + 4x = 56

=> 3x + 1x + 4x = 56

=> (3 + 1 + 4) x = 56

=> 8x = 56

=> x = [latex]\frac { 56 }{ 8 }[/latex] = 7 (correct value)

Question 3.

Find the value of 5x at x = – 2 = 5 – 2 = 3

Find the(RBSESolutions.com)error and also find the correct value.

Solution:

Value of 5x at x = – 2

= 5 x (- 2) | 5x = 5 × x

= – 10 (correct value)

Question 4.

Solutions of the expression is given in the column A and B. Find which of the solution is correct.

Expression | A | B |

3(x-4) | 3x-4 | 3x-12 |

(2x)² | 2x² | 4x² |

(x+4)² | x²+16 | x²+8x+16 |

(x-3)² | x²-9 | x²-6x+9 |

[latex]\frac { y+1 }{ 5 }[/latex] | y+1 | [latex]\frac { y }{ 5 }+1[/latex] |

Solution:

(i) 3(x – 4)

= 3 × x – 3 × 4 =

3x – 12

Hence B is correct.

(ii) (2x)²

= (2x) × (2x)

= (2 × x) × (2 × x)

= 2 × 2 × x × x

= 4 × x²

= 4x²

Hence B is correct.

(iii) (x + 4)² = x² + 2(x) (4) + (4)²

= x² + 8x + 16

Hence B is correct.

(iv) (x – 3)² = x² – 2 (x) (3) + (3)²

= x² – 6x + 9

Hence B is correct.

(v)

Hence B is correct.

We hope the given RBSE Solutions for Class 8 Maths Chapter 10 Factorization In Text Exercise will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 10 Factorization In Text Exercise, drop a comment below and we will get back to you at the earliest.

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