# How do you find all the zeros of #f(x)= x^4+x^3+2x^2+4x-8# with its multiplicities?

##### 1 Answer

Aug 3, 2016

#### Explanation:

#f(x) = x^4+x^3+2x^2+4x-8#

First note that the sum of the coefficients of

#1+1+2+4-8 = 0#

Hence

#x^4+x^3+2x^2+4x-8 = (x-1)(x^3+2x^2+4x+8)#

The remaining cubic is very regular and can be factored in a number of different ways. Let's factor it by grouping:

#x^3+2x^2+4x+8#

#=(x^3+2x^2)+(4x+8)#

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

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

So

The remaining quadratic has no Real zeros since

#x = +-sqrt(-4) = +-2i#