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: 6

*x*(2^{3}=*x*) (3*x*)^{2}A factor that cannot be factorized further is known as

**irreducible factor**2

*x*^{2}+*x*– 6 =(2*x*– 3) (*x*+ 2)Here (2

*x*– 3) cannot be factorized furtherAlso (

*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) 2

*x*+6 = 2*x*+(2 x 3) =2 (

*x*+3)**(Note that ‘2’ is common to both terms. )**(ii) 4

*x*20^{2}+*xy*= (4 x*X*x*X*)+(4 x 5 x*X*x*Y*) = 4

*x*(*x*+5*y*)**[Taking out the common term 4***x*]**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)

*x*3^{3 }-*x*^{2}+ x – 3=

*x*(^{2}*x-*3) +1 (*x-*3)**[By grouping the first two and the last two and taking out the common terms]**= (

*x*1) (^{2}+*x-*3)(ii) 2

*xy*- 3*ab*+ 2*bx*- 3*ay*= 2*xy*+2*bx*- 3*ab*- 3*ay***[Rearranging the terms]**= 2

*x*(*y*+*b*) -3*a*(*y*+*b*)**[Taking out the common terms]**= (2

*x*-3*a*) (*y*+*b*)**Factorization by using Identities**

(i) x

^{2}+ 6 x + 9 =Comparing x x

^{2}+ 6 x + 9 with**a**, we see that a = x, b = 3.^{2 }+ 2^{ }ab+ b^{2}Now, x

^{2}+ 6 x + 9 = x^{2}+ 2 x 3 x + 3^{2}Using,

**a**= (a + b)^{2 }+ 2^{ }ab+ b^{2 }^{2 }, a = x and b = 3,We get x

^{2}+ 6 x + 9 = (x + 3)^{2}.` The factors of x

^{2}+ 6 x + 9 are (x + 3)and (x + 3) .(ii) Comparing

*x*^{2}– 64 with*a b*2 2 - , we see that*a*=*x*and*b*= 8Using,

**,***a*= (^{2 }- b^{2}*a*+*b*) (*a*-*b*)*x*

^{2 }– 64 =

*x*

^{2}– 8

^{2}

= (

*x*+ 8) (*x*- 8)**Factorization by using the Identity**(

*x+ a*) (

*x+ b*) =

*x*(

^{2 }+*a+ b*)

*x + ab*

Factorize

*x*5^{2 }+*x +*6Comparing

*x*5^{2 }+*x +*6 with*x*(^{2 }+*a+ b*)*x + ab*For this

*ab*= 2 x 3 = 6 and*a*+*b*= 2 + 3 = 5We 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 : x*(

^{2 }+*a+ b*)

*x + ab*= (

*x+ a*) (

*x+ b*)

*x*5

^{2 }+*x +*6 =

*x*(

^{2 }+*2 + 3*)

*x + 2 x 3*= (

*x+ 2)*(

*x+ 3*)

*x*6

^{2 }+*x+*8 =

*x*(

^{2 }+*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

- x² + 5x + 6
- x² - 8x + 7
- x² + 6x - 7
- y² + 7y - 18
- y² - 7y - 18
- a² - 3a - 54
- 2x² - 7x + 6
- 6x² + 13x - 5
- 6x² + 11x - 10
- 6x² - 7x - 3
- x² - 2x + 1
- 4x² - 12x + 9
- 5 - 4x - 12x²
- x (12x + 7) - 10
- 2x² + 13xy - 24y²
- 2a²b² - 7ab - 30
- (a - b)² + 6 (a - b) + 8
- Work out the value of

(i) (100002)² - (99998)²

(ii) (1001)²

(**iii**

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)

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)

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 (i

18.(i) 800000 (ii) 24 (i

**ii**) 998001
fantastic

ReplyDeleteVery...nyc.!!

ReplyDeleteI was looking for crucial information on this subject. The information was important as I am about to launch my own portal. Thanks for providing a missing link in my business.

ReplyDeleteCBSE syllabus

good but there should be more questions

ReplyDeleteGood...thnx

ReplyDeleteGood...thnx but shd. be mre difficult

ReplyDeleteneed more difficult

ReplyDeleteshould me more diffcult

ReplyDeleteneed to show more formulaes!!

ReplyDelete