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

1 Answer
Oct 7, 2015

(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.