How do I find all the rational zeros of #p(x)=x^3-12x-16#?

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Answer 1 · 2018-06-01T00:05:21

Check if any of the divisors of #-16# is a root of #p(x)# in our case

#x=4# is a root hence #p(4)=0#

The polynomial can be written as

#p(x)=(x-4)*(x^2+bx+c)#

It is easy to proove that #b=4# and #c=4#

Hence

#p(x)=(x-4)(x+2)^2#

So the zeros are #x=4# ,#x=-2# (double root)