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

##### 1 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#

#= (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?

Can we factor

The only possible rational zeros are

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

Then:

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