How do you factor #x^3-3x^2+x-55#?

1 Answer
Oct 7, 2015

Answer:

#(x-5)(x^2+2x+11)#

Explanation:

If #x^3-3x^2+x-55# has an integer root #r# then
#r# must be a factor of #55#
i.e. #r in {+-5,+-11}#

A quick check reveals #r=5# satisfies the requirement
so
#(x-r) = (x-5)# is a factor

Dividing #x^3-3x^2+x-55# by #(x-5)#
#color(white)("XXX")#(for example by using synthetic division:)

#{:(," | ", 1, -3,color(white)("X")1, -55), (+5," | ",,color(white)("X")5,10,color(white)("X")55),(,"---","---","---","---","----"),(,,color(red)(1),color(red)(2),color(red)(11),color(white)("X")color(green)(0)):}#

gives
#(x-5)(color(red)(1)x^2+color(red)(2)x+color(red)(11))#

Checking the discriminant of #(x^2+2x+11)# reveals that no further Real factoring is possible.