# How do you find the zeros of #f(x) = x^4 - x^3 - 6x^2 + 4x + 8#?

##### 1 Answer

Jun 10, 2016

#x = -1#

#x = 2# with multiplicity#2#

#x = -2#

#### Explanation:

First note that

So

#x^4-x^3-6x^2+4x+8 = (x+1)(x^3-2x^2-4x+8)#

Note that in the remaining cubic, the ratio of the first and second terms is the same as that between the third and fourth terms. So this will factor 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)#

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

Hence the other zeros are:

#x = 2# with multiplicity#2#

#x = -2#