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.

Related questions
See all questions in Factorization of Quadratic Expressions
Impact of this question
23191 views around the world
You can reuse this answer
Creative Commons License