Let us take some examples for this:

1. x² - 7x + 10

Here factor of 10 is 2, 5

Now we use 2 and 5 in such a way that sum of the coefficient of x become -7 and multiplication of constant term is 10.

So x² - (2+5)x + 2*5

= x² - 2x - 5x + 2*5

= x(x-2) - 5(x-2)

= (x-2)*(x-5)

2.  2x² + 5x + 3

Here coifficenet of x² is 2 and constant term is 3.

Now using these two values in such a way that sum of the coefficient of x becomes 5 and the product becomes 6 (since the product of the coefficient of x² and the constant term is 6).

So   2x² + (2+3)x + 3

=>  2x² + 2x + 3x + 3

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

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

3.  2x² + 5x + 3

First make coefficent of x² is 1. So in this example, divide by 2, we get 

 x² + 5x/2 + 3/2

Now take coefficent of x and find half of it

i.e. (5/2)/2 = 5/4

Now square it 

i.e. (5/4)² = 25/16

Now add and subtract these values in the polynomial i.e.

    x² + 5x/2 + 3/2 + 25/16 - 25/16

= x² + 5x/2 + 25/16 + 3/2 - 25/16

= x² + 5x/2 + (5/4)² + (24 - 25)/16

= (x + 5/4)² - 1/16                                { since (x+a)² = x² + 2ax + a² }

= (x + 5/4)² - (1/4)²

= (x + 5/4 - 1/4)*(x + 5/4 + 1/4)                   { since x² - a² = (x-a)*(x+a) }

= (x + 4/4)*(x+ 6/4)

= (x + 1)*(x+ 3/2)

So, now apply these three methods and practice the questions, then you will defenately find the factors within few seconds.