Is it possible to factor $y=x^2 - 9x+36$ ? If so, what are the factors?

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Answer 1 · 2016-01-07T21:39:15

Yes, but only with Complex coefficients:

$x^2-9x+36 = (x-9/2-(3sqrt(7))/2 i)(x-9/2+(3sqrt(7))/2 i)$

$y=x^2-9x+36$ is in the form $ax^2+bx+c$ with $a=1$ , $b=-9$ and $c=36$ .

This has discriminant $Delta$ given by the formula:

$Delta = b^2-4ac = (-9)^2-(4*1*36) = 81-144 = -63$

Since this is negative, this quadratic has no factors with Real coefficients.

If you still want to factor it, use the quadratic formula to find the Complex zeros:

$x = (-b+-sqrt(b^2-4ac))/(2a)$

$= (-b+-sqrt(Delta))/(2a)$

$=(9+-sqrt(-63))/2$

$=(9+-sqrt(63) i)/2$

$=(9+-sqrt(3^2*7)i) / 2$

$=(9+-3sqrt(7)i)/2$

So:

$x^2-9x+36 = (x-9/2-(3sqrt(7))/2 i)(x-9/2+(3sqrt(7))/2 i)$

Is it possible to factor y=x^2 - 9x+36y=x^2 - 9x+36? If so, what are the factors?