aX2 + bx + c
to choose 2 numbers such as m. in such a way that
m+ n = - b/a
m. n = c/a
If this equation has solution with integer values, then equation can be factored to
a X2 + bX + c = a( x -m )( x-n)
Example: a =1:
X2 - 23X + 38 =
2 numbers ( m.n) whose sum is 23 and product 38 are ( 19 , 4), therefore:
X2 - 23X + 76 = (X -19) ( X-4)
In order for a quadratic to be factorable to binomial with integer values.
It is necessary and sufficient that -b to be Sum of the products of 2 factors of C:
In here we write the number 38 as product of its prime factors as:
76= 2 . 2 . 19 factors are (2, 4, 19)
we see that 76 = 4 *19 4+ 19 =23 , therefore the quadratic is factorable as above.
Now consider the following:
3 X2 - 14 X + 15 =
Now we have to have 2 numbers is 14, and product is 45.
The numbers by calculating in our mind are (5,9)
We break up the 14 into 5+9
3X2 - 9X -5X +15 =
We factor by part:
3X( X -3 ) - 5 ( X - 3) =
Now Common factor is ( X-3) , therefor
3 X2 - 14X + 15 = ( X- 3) ( 3X - 5 )
X2 + 5X - 7
Here there is no 2 integers where their Sum is +5, and product is -7, in this case we
factor by what is called : Factoring by Completing square which gives us factors of
irrational and complex numbers.