Answer :

**For quadratic polynomial. We factorize the middle term such that the product of the terms is equal to the product of the last term and coefficient of x ^{2} and sum or difference is equal to the middle term.(i)**

12x

^{2}- 7x + 1

For the equation given above:

7 is to be factorized such that, the product of two terms is 12 x 1 = 12 and the sum is equal to 7

This can be done by 3 and 4

as 3 x 4 = 12

and 3 + 4 = 7

Therefore,

12x^{2} - 7x + 1 = 12x^{2} – 4x – 3x + 1

= 4x (3x – 1) – 1 (3x – 1)

= (3x – 1) (4x – 1)

**(ii) **

2x^{2 }+ 7x + 3

For the equation given above:

7 is to be factorized such that, the product of two terms is 3 x 2 = 6 and the sum is equal to 7

This can be done by 6 and 1

as 6 x 1 = 6

and 6 + 1 = 7

Therefore,

2x^{2 }+ 7x + 3 = 2x^{2} + 6x + x + 3

= 2x (x + 3) + 1 (x + 3)

= (x + 3) (2x + 1)

**(iii)**

6x^{2} + 5x – 6

7 is to be factorized such that, the product of two terms is 6 x 6 = 36 and the difference is equal to 5

This can be done by 9 and 4

as 9 x 4 = 36

and 9 - 4 = 5

Therefore,

6x^{2} + 5x – 6 = 6x^{2} + 9x – 4x – 6

= 3x (2x + 3) – 2 (2x + 3)

= (2x + 3) (3x – 2)

**(iv)**

3x^{2} - x - 4

For the equation given above:

7 is to be factorized such that, the product of two terms is 4 x 3 = 12 and the difference = 1

This can be done by 4 and 3

as 4 x 3 = 12

and 4 - 3 = 1

Therefore,

3x^{2} - x - 4=3x^{2} – 4x + 3x – 4

= x (3x – 4) + 1 (3x – 4)

= (3x – 4) (x + 1)

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