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
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