How do you factor #x^2-4x+24#?

1 Answer
George C.
May 16, 2015

#x^2-4x+24# is of the form #ax^2+bx+c# with #a=1#, #b=-4# and #c=24#.

This has discriminant

#Delta = b^2-4ac#
# = (-4)^2-(4xx1xx24) = 16-96 = -80#

Since #Delta < 0#, #x^2-4x+24=0# has 2 complex, non-real roots and #x^2-4x+24# has no linear factors with real coefficients.

In other words #x^2-4x+24# is already in simplest terms for real coefficients.

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