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

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Answer 1 · 2016-03-15T22:38:21

Look at coefficient sums and divide by the factors found to simplify the problem and find zeros:

#x=1#, #x=-1#, #x=-2# and #x=-3#

First note that the sum of the coefficients is zero.

That is: #1+5+5-5-6 = 0#

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

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

Next note the if you reverse the signs of the terms of the remaining cubic factor with odd degree then the sum of the coefficients is zero.

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

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

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

Then note that #2+3 = 5# and #2 xx 3 = 6#, so the remaining quadratic factor factorises as follows:

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

Putting this all together, we find:

#f(x) = (x-1)(x+1)(x+2)(x+3)#

with zeros #x=1#, #x=-1#, #x=-2# and #x=-3#