How do you find all zeros of the function #f(x)=3x^3+x^2-48x-16#?

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Answer 1 · 2016-05-01T03:11:25

Factor #f(x)# to find that #f(x) = 0# if and only if #x in {-4, -1/3, 4}#

#f(x) = 3x^3+x^2-48x-16#

#= 3x(x^2-16)+x^2-16#

#=(3x+1)(x^2-16)#

#=(3x+1)(x+4)(x-4)#

Thus #f(x) = 0# if and only if #3x+1 = 0# or #x+4 = 0# or #x-4 = 0#.

Solving, this gives us

#3x+1 = 0 => x = -1/3#

#x+4 = 0 => x = -4#

#x-4 = 0 => x = 4#

Thus the solution set to #f(x) = 0# is #x in {-4, -1/3, 4}#