relation of delta and factorization?

Quote:

Originally Posted by kenny1999

for a quadratic "expression"

i.e. ax^2 + bx + c

with any real and integral value of a, b and c

how to show that if delta (b square minus 4ac) is a perfect square

then such a quadratic expression can be factorized by cross method?

into something like (mx-n)(px-q) where m,n,p,q are integers.

I had been thinking about it for a whole night.

Hope someone can help

If $b^2-4ac = d^2$ then $c = \frac{b^2-d^2}{4a}$ (note if a = 0, then we don't have a quadratic). So

$ax^2+bx+c = ax^2 + bx + \frac{b^2-d^2}{4a} = \frac{4a^2x^2 + 4abx + b^2 -d^2}{4a}$

From the first three terms in the numerator and then the difference of squares

$<br /> \frac{(2ax +b)^2 -d^2}{4a} = \frac{(2ax + b - d)(2ax + b + d)}{4a}<br />$