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

1 Answer
George C.
Jun 10, 2016

Zeros:

#x = -1#

#x = 2# with multiplicity #2#

#x = -2#

Explanation:

First note that #f(-1) = 1+1-6-4+8 = 0#.

So #x=-1# is a zero and #(x+1)# a factor:

#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#

Related questions
See all questions in Zeros
Impact of this question
3201 views around the world
You can reuse this answer
Creative Commons License