How do you factor #x^3-x^2-4x+5#?

1 Answer
May 20, 2016

Answer:

Factoring by grouping does not work with this polynomial, but read on...

Explanation:

Factoring by grouping does not immediately help with this cubic polynomial. If it were #x^3-x^2-4x+4# then it would be much easier:

#x^3-x^2-4x+4#

#= (x^3-x^2)-(4x-4)#

#= x^2(x-1)-4(x-1)#

#= (x^2-4)(x-1)#

#= (x-2)(x+2)(x-1)#

Was there a typo in the question?

#color(white)()#
Can we factor #f(x) = x^3-x^2-4x+5# ?

The only possible rational zeros are #+-1#, #+-5#, neither of which work.

This cubic turns out to have one Real zero and two Complex zeros.

It is possible, if a little lengthy, to find them using Cardano's method, which will give you:

Real zero:

#x_1 = 1/3(1-root(3)((97-3sqrt(69))/2)-root(3)((97+3sqrt(69))/2))#

Complex zeros:

#x_2 = 1/3(1-omega root(3)((97-3sqrt(69))/2)-omega^2 root(3)((97+3sqrt(69))/2))#

#x_3 = 1/3(1-omega^2 root(3)((97-3sqrt(69))/2)-omega root(3)((97+3sqrt(69))/2))#

where #omega = -1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#.

Then:

#x^3-x^2-4x+5=(x-x_1)(x-x_2)(x-x_3)#