# How do you find all the zeros of #P(x)=x^4 + 1 x^3 - 3 x^2 - 1 x + 2#?

##### 1 Answer

Jun 2, 2016

#### Explanation:

First note that the sum of the coefficients is

#1+1-3-1+2 = 0#

So

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

Looking at the remaining cubic, notice that the ratio between the first and second terms is the same as that between the third and fourth terms. So it factors by grouping:

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

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

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

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

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

Hence the remaining zeros are