Say you have the equation x^2 + 6x + 9 = 0. Couldn't (x

>>8523190
What is this even supposed to accomplish?

You still have the exact same quadratic to solve (in the numerator) you just tacked on some useless terms that cancel to 1 for no reason

You would create a discontinuity at x = -1, so it wouldn't be the same function.

>>8523190
x^2 + 6x + 9 = (x+1)((x^2 + 6x + 9)/(x + 1)), for all x != -1.

Thus if 0 = x^2 + 6x + 9 shares solutions with
0 = (x+1)((x^2 + 6x + 9)/(x + 1)) when x != -1.

>>8523190
no, for the same reason that 7 is not considered a factor of 10, even though 10=7*(10/7)

>>8523454
this is essentially the answer.

When we talk about factoring polynomials, we mean factoring in the ring of polynomials, not the field of rational functions. Just like we factor integers and not rational numbers.