How do you find all the zeros of #f(x)=x^4+5x^3+5x^2-5x-6#?

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Answer 1 · 2016-07-09T21:46:15

Find zeros: #x=1#, #x=-1#, #x=-2# and #x=-3#

#f(x)=x^4+5x^3+5x^2-5x-6#

First note that the sum of the coefficients is #0#. That is:

#1+5+5-5-6 = 0#

So #f(1) = 0#, #x=1# is a zero and #(x-1)# a factor:

#x^4+5x^3+5x^2-5x-6 = (x-1)(x^3+6x^2+11x+6)#

Next note that if you reverse the signs of the coefficients of odd degree in the remaining cubic, then the sum is #0#. That is:

#-1+6-11+6 = 0#

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

#x^3+6x^2+11x+6 = (x+1)(x^2+5x+6)#

To factor the remaining quadratic, note that #2+3 = 5# and #2xx3=6#.

Hence:

#x^2+5x+6 = (x+2)(x+3)#

giving zeros #x=-2# and #x=-3#