How do you find all the zeros of #f(x)=x^4+4x^3-6x^2-36x-27#?

1 Answer
George C.
Aug 13, 2016

#f(x)# has zeros:

#-1# (with multiplicity #1#)

#3# (with multiplicity #1#)

#-3# (with multiplicity #2#)

Explanation:

#f(x) = x^4+4x^3-6x^2-36x-27#

By the rational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-27# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-3, +-9, +-27#

We find:

#f(-1) = 1-4-6+36-27 = 0#

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

#x^4+4x^3-6x^2-36x-27 = (x+1)(x^3+3x^2-9x-27)#

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 cubic will factor by grouping:

#x^3+3x^2-9x-27#

#=(x^3+3x^2)-(9x+27)#

#=x^2(x+3)-9(x+3)#

#=(x^2-9)(x+3)#

#=(x^2-3^2)(x+3)#

#=(x-3)(x+3)(x+3)#

So the remaining zeros are #x=3# (with multiplicity #1#) and #x=-3# with multiplicity #2#.

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