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

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Answer 1 · 2016-12-26T12:05:13

$- 3$ $-3$ ; (rational zero)
$- 4 \pm \sqrt{5}$ $-4+-sqrt(5)$ (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

$f (- 1) \neq 0$ $f(-1)!=0$

but

$f (- 3) = 0$ $f(-3)=0$

So -3 is a rational zero.

Then you will divide:

$(x^{3} + 11 x^{2} + 35 x + 33) : (x + 3) = x^{2} + 8 x + 11$ $(x^3+11x^2+35x+33):(x+3)=x^2+8x+11$

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

$- 4 \pm \sqrt{5}$ $-4+-sqrt(5)$

What are all the possible rational zeros for f(x)=x3+11x2+35x+33f(x)=x^3+11x^2+35x+33 and how do you find all zeros?