How do you find all the zeros of #P(x) = x^3 + 5x^2 - 2x - 24# where one zero is 2?

1 Answer
George C.
Aug 7, 2016

#P(x)# has zeros #2#, #-3# and #-4#.

Explanation:

#P(x) = x^3+5x^2-2x-24#

SInce we are told that #x=2# is one of the zeros, #(x-2)# must be a factor:

#x^3+5x^2-2x-24#

#=(x-2)(x^2+7x+12)#

The remaining quadratic can be factored by noticing that #3+4=7# and #3xx4=12#. So:

#x^2+7x+12 = (x+3)(x+4)#

Hence the other zeros are #x=-3# and #x=-4#

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