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## Main Question or Discussion Point

I thought this was rather odd, and wanted to just show it to see what you all thought of it. Well, also, if anyone knows what I should read to exactly understand what I did.

1: Lets define each answer of a polynomial such as (ax + b)(cx + d) as x1 = (-b/a), x2 = (-d/c).

2: The polynomial written out is qx^2 + rx + s

3: Change the polynomial from qx^2 + rx + s to where x = s, and rewrite the problem as qS^2 + rS = S.

4: Devide S out, which leaves qS + r = 1

5: subtract r, qS + r - r = 1 - r. qS = 1 - r

6: Devide q, qS/q = (1 - r)/q. S = (1-r)/q

7: Now if the equation (q x1 + 1)/q is subtracted from S will solve for x2. Or if x1 is replaced with x2 the equation then solves for x1.

(1-r)/q - (q x1 + 1)/q = x2

(1-r)/q - (q x2 + 1)/q = x1

I did simplify this, but I still think it is rather interesting as to 'how' I did this, rather than the 'simplicity' of it.

(Darn it, I forgot what these kinds of polynomials are called. Is it 2nd order polynomials, or 2nd order quadratics?)

Thanks

1: Lets define each answer of a polynomial such as (ax + b)(cx + d) as x1 = (-b/a), x2 = (-d/c).

2: The polynomial written out is qx^2 + rx + s

3: Change the polynomial from qx^2 + rx + s to where x = s, and rewrite the problem as qS^2 + rS = S.

4: Devide S out, which leaves qS + r = 1

5: subtract r, qS + r - r = 1 - r. qS = 1 - r

6: Devide q, qS/q = (1 - r)/q. S = (1-r)/q

7: Now if the equation (q x1 + 1)/q is subtracted from S will solve for x2. Or if x1 is replaced with x2 the equation then solves for x1.

(1-r)/q - (q x1 + 1)/q = x2

(1-r)/q - (q x2 + 1)/q = x1

I did simplify this, but I still think it is rather interesting as to 'how' I did this, rather than the 'simplicity' of it.

(Darn it, I forgot what these kinds of polynomials are called. Is it 2nd order polynomials, or 2nd order quadratics?)

Thanks