What are all the possible rational zeros for #f(x)=x^3+11x^2+35x+33# and how do you find all zeros?

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Answer 1 · 2016-12-26T12:05:13

#-3#; (rational zero)
#-4+-sqrt(5)# (not rational zeros)

You would find rational zeros in the set of integer and negative numbers dividing the known term 33, which are -1; -3; -11; -33.

Using the remainder theorem, you will find that

#f(-1)!=0#

but

#f(-3)=0#

So -3 is a rational zero.

Then you will divide:

#(x^3+11x^2+35x+33):(x+3)=x^2+8x+11#

By using quadratic formula, you would find the not rational zeros:

#-4+-sqrt(5)#