What is the factors of x²+2xy-y²?
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Answer: x2+ax+b=(x−−a+D√2)(x−−a−D√2), where D=a2−4b.
What happens if a and b are not constants, but some functions of a different (not even necessarily) variable or variables? This formula still works!
x2−2yx+2y2=(x−(y+−y2−−−−√))(x−(y−−y2−−−−√))=(x−(1+−1−−−√)y)(x−(1−−1−−−√)y)
Uh-oh, what's that −1in the discriminant?
If we are working over the reals or some other ring without a root of −1, this means that our polynomial is irreducible (why?).
If we do have a root of −1 (say in C or Z/5Z), namely i, we have already written our factorization:
x2−2yx+2y2=(x−(1+i)y)(x−(1−i)y
Explanation: