( X + a) ( X + b) = X^{2} + ( a +b) X + ab (1) ( X + a) 2 = X^{2} + 2aX +a^{2} (2)
( X +a ) ( X a )= X^{2 }a^{2} (3)
The 3 above identities are key to factoring quadratic, and finding the roots
In general every quadratic, trinomial aX^{2} + bX + c is generated by multiplication of 2 binomial (linear) expression in the form of 3 identity given above, and subsequently any given Ttrinomial ( quadratic), can
be factored to binomial factors i.e.( X + a) ( X + b) = X^{2} + ( a +b) X + ab, that is used in evaluation of roots of the quadratic.
, .
In other words identities (1), (2) , ( 3) can work both ways.
given a quadratic like : X^{2} + 7X + 10
here we see that ( a + b) = 7 ab =10
a = 2 b =5 is the answer, therefore X^{2} + 7X + 10 = ( X +2) ( X + 5 )
equation (2) is a special case of (1) where a=b , a+b = 2a , ab=a2
equation (3) is a special case of ( 1) where b = a, a + ( a) = 0 , ab = a2
These 3 identities are used in factoring a quadratic.
Equation (1) is factorable if there exists 2 whole number whose sum is ( a+ b), and product =ab.
If the answer of the system of equation is not a whole number, then have to do factoring by competing the square, and come up with a factors of irrational and complex numbers, yielding to Irrational and complex roots .
9/12/2013

Parviz F.