## 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
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

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CBSE syllabus

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