12 X^{2} - 72X -108 = 0
first factor 12:
12( X^{2} - 6X + 9) =0
Now to factor quadratic, think of 2 numbers whose sum is 6 and their product 9. Have to do this in your head or scratch paper, or .... the numbers are 3, 3 , 3*3 = 9 , 3+3 =6, then the quadratic factors to:
12 ( X -3 ) ^{2 }= 0 x=3 , is repeated roots of the quadratic
63 X^{2} - 112 =0
7 ( 9X^{2 }- 16 ) =0
9 X^{2} - 16 =0
X^{2} = 16/9 X =±4/3
( 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
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=a^{2}
equation (3) is a special case of ( 1) where b = -a, a + ( -a) = 0 , ab = a^{2}
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 .