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

##### 1 Answer

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

#= (x-1)(x+1-sqrt(3)i)(x+1+sqrt(3)i)#

#### Explanation:

First note that the sum of the coefficients is zero.

That is

So

#x^3+x^2+2x-4 = (x-1)(x^2+2x+4)#

The remaining quadratic factor is of the form

This has discriminant

#Delta = b^2-4ac = 2^2-(4*1*4) = -12#

Since this is negative, the quadratic has no linear factors with Real coefficients.

If we want to proceed further, we can use the quadratic formula to find the Complex roots and derive factors:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#= (-b+-sqrt(Delta))/(2a)#

#= (-2+-sqrt(-12))/2#

#= (-2+-sqrt(12)i)/2#

#= (-2+-2sqrt(3)i)/2#

#=-1+-sqrt(3)i#

Hence:

#x^2+2x+4 = (x+1-sqrt(3)i)(x+1+sqrt(3)i)#