Question 702084
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In general, find p, q, r, and s such that pq = a, rs = c, and ps + rq = b.  Then your factors are (px - r)(qx - s)


Again, in general, this is a trial and error process.  A couple of hints:  If the sign on the constant term is positive, then the signs in the binomials must be the same.  Then, if the sign on the first degree term is negative, they are both negative, or if the sign on the first degree term is positive, then they are both positive.  If the sign on the constant term is negative, the the signs in the binomials are opposite and the value of b is the difference between ps and rq.


There is a method that always works, even if the quadratic does not factor over the integers.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{r}{p}\ = \frac{-b\ +\ \sqrt{b^2\ -\ 4ac}}{2a}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{s}{q}\ = \frac{-b\ -\ \sqrt{b^2\ -\ 4ac}}{2a}]


Then since *[tex \LARGE x\ = \frac{r}{p}], a factor of the quadratic is *[tex \LARGE x\ -\ \frac{r}{p}] which is equivalent to *[tex \LARGE px\ -\ r]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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