Wednesday, 17 August 2011

CBSE Class VIII Math Questions of chapter factorization


What is factorization?
We can also write any algebraic expression as the product of its factors.
Factorization: The process of expressing any polynomial as a product of its factors is called factorization.
Examples: 6x3 =   (2x) (3x2 )

A factor that cannot be factorized further is known as irreducible factor
2x2 + x – 6 =(2x – 3) (x + 2)
Here  (2x – 3) cannot be factorized further
Also  (x + 2) cannot be factorized further

Factorization by taking out the common factor
In this method, we rewrite the expression with the common factors outside brackets.
Examples:
(i) 2x+6           = 2x+(2 x 3)
                         =2 (x+3) (Note that ‘2’ is common to both terms. )
(ii) 4x2 +  20xy  = (4 x X x X)+(4 x 5 x X xY)
                         = 4x (x+5y) [Taking out the common term 4x]

Factorization by Grouping the terms
In this method, the terms in the given expression can be arranged in groups of two or three so as to get a common factor.

(i) x3 -  3x2 +  x – 3
= x2 (x- 3) +1 (x- 3) [By grouping the first two and the last two and taking out the common terms]
= (x2 + 1) (x- 3)
(ii) 2xy - 3ab + 2bx - 3ay = 2xy +2bx - 3ab - 3ay [Rearranging the terms]
= 2x (y + b) -3a (y+b) [Taking out the common terms]
= (2x-3a) (y+b)

Factorization by using Identities

(i) x2 + 6 x + 9 =
Comparing x x2 + 6 x + 9 with a2 + 2 ab+  b 2, we see that a = x, b = 3.
Now, x2 + 6 x + 9 = x2 + 2 x 3 x + 32
Using, a2 + 2 ab+  b 2 = (a + b)2  , a = x and b = 3,
We get x2 + 6 x + 9 = (x + 3)2  .
` The factors of x2 + 6 x + 9 are (x + 3)and (x + 3) .
(ii) Comparing x2 – 64 with a b 2 2 - , we see that a = x and b = 8
Using, a2  - b2        = (a + b) (a - b) ,
             x2  –  64  = x2 – 82
                          = (x + 8) (x - 8)
Factorization by using the Identity (x+ a) (x+ b) =  x2 + (a+ b)x + ab

Factorize x2 +  5x + 6
Comparing x2 +  5x + 6 with x2 + (a+ b) x + ab

For this  ab = 2 x 3 = 6  and a + b = 2 + 3 = 5
We have, ab = 6, a + b = 5 and x = x.
If ab = 6, it means a and b are factors of 6.
Let us try with a = 2 and b = 3. These values satisfy ab = 6 and a + b = 5.
Therefore the pair of values a = 2 and b = 3 is the right choice.

Using :          x2 + (a+ b) x + ab  = (x+ a) (x+ b)
 x2 +  5x + 6 =  x2 + (2 + 3) x + 2 x 3  = (x+ 2) (x+ 3)

x2 +  6x+  8 = x2 + (2 + 4) x + 2 x 4  = (x+ 2) (x+ 4)

Factors  of 8              Sum of  factors
1, 8                                  9
2, 4                                 6
Hence the correct   factors are 2, 4
Factorise the following
  1. x² + 5x + 6                                                     
  2. x² - 8x + 7
  3. x² + 6x - 7
  4. y² + 7y - 18
  5. y² - 7y - 18
  6. a² - 3a - 54
  7. 2x² - 7x + 6
  8. 6x² + 13x - 5
  9. 6x² + 11x - 10
  10. 6x² - 7x - 3
  11. x² - 2x + 1
  12. 4x² - 12x + 9
  13. 5 - 4x - 12x²
  14. x (12x + 7) - 10
  15. 2x² + 13xy - 24y²
  16. 2a²b² - 7ab - 30
  17. (a - b)² + 6 (a - b) + 8
  18. Work out the value of
    (i) (100002)² - (99998)²
    (ii)  (1001)²
    (iii  (999)²
Answers
1. (x + 2)(x + 3)             2. (x - 1)(x - 7)               3. (x + 7)(x - 1)             
 4. (y + 9)(y - 2)            5. (y - 9)(y + 2)            6. (a + 6)(a - 9)
7. (x - 2)(2x - 3)            8. (3x - 1)(2x + 5)       9. (2x + 5)(3x - 2)        
10. (3x + 1)(2x - 3)      11. (x - 1)²                   12. (2x - 3)²
13. (1 - 2x)(5 + 6x)       14. (4x + 5)(3x - 2)   15. (x + 8y)(2x - 3y)     
16. (ab -6)(2ab + 5)     17. (a - b + 4)(a - b + 2)
18.(i) 800000    (ii) 24  (iii) 998001

10 comments:

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    CBSE syllabus 

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  2. good but there should be more questions

    ReplyDelete
  3. Good...thnx but shd. be mre difficult

    ReplyDelete
  4. need to show more formulaes!!

    ReplyDelete
  5. THERE SHOULD BE MORE QUESTIONS FOR PRACTIS BY THE STUDENTS

    ReplyDelete