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)#