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:

$-4+-sqrt(5)$

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