Question

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Please Help!!!!!!
Using complete sentences, describe how you would find all possible rational zeros of the polynomial function f(x) = 9x^4 – 17x^3 + 2x^2 – 3x + 33.

Be sure to provide the answer in your explanation.

Answer 1

I neeed help please

Answer 2

Take a look at the Rational Zero Theorem. It states, basically, that if some rational number x=p/q is a root of your polynomial, f(x), then p must be a multiple of your leading coefficient (in this case, that's 9), and q must be a multiple of your constant term (which, in this case, is 33).

Answer 3

I'm pretty sure that this polynomial has no rational zeros (or any zeros whatsoever, for that matter), so it's unclear to me how one would show conclusively with only Algebra 2 knowledge that there are no rational zeros.

Answer 4

So what is the answer.

Answer 5

According to Wolfram|Alpha, there are no rational zeros.

Answer 6

The possible rational zeros have the form p/q where p = a factor of 33, and q = a factor of 9.

Answer 7

Oh, sorry - thanks mathstudent55, I seem to have remembered the theorem incorrectly. That makes it much easier to tackle. You can just try out all variations of the fraction p/q to see if any of them yield zeros.

Answer 8

Yes I need the @whpalmer4 seal of confirmation

Answer 9

"I'm pretty sure that this polynomial has no rational zeros (or any zeros whatsoever, for that matter)," Of course it has zeros, 4 of them, two conjugate pairs.

Answer 10

Can you just sum it for me in one post please as the question asks it. Also do not leave I haave 2 more question if you can.

Answer 11

Whoops, I meant to include "real" there. The roots certainly aren't rational.

Answer 12

Using complete sentences, describe how you would find all possible rational zeros of the polynomial function f(x) = 9x^4 – 17x^3 + 2x^2 – 3x + 33.
the possible rational roots are numbers \(\frac{p}{q}\) where \(p\) divides \(33\) and \(q\) divides \(9\)

Answer 13

None of them are rational, though!

\[ 9x^4 – 17x^3 + 2x^2 – 3x + 33\]Applying Descartes' Rule of Signs:

Signs are +-+-+ -> 4 sign changes. That means either 4 positive real roots, 2 positive real roots, or 0 positive real roots.

Rewrite as f(-x):
\[9x^4+17x^3+2x^2+3x+33\]
Signs are +++++ -> 0 sign changes. That means there are 0 negative real roots.

So, Descartes tells us that we either have 4 positive real roots, or 2 positive real roots and 2 complex roots, or 0 positive real roots and 4 complex roots. How to decide?

Here's where the rational root theorem comes in. Any rational roots (and remember, we aren't guaranteed to have any!) will be factors of the constant term (33) divided by factors of the leading term (9), with either sign possible. You would have to go through and try the various possibilities. If you don't find any that make the polynomial evaluate to 0, then you know that it is all complex roots. If you do find some that make the polynomial evaluate to 0, then you can divide by (x-root) and simplify the polynomial before restarting the process to find the other roots.

Answer 14

such numbers look like
\[\pm1,\pm3,\pm,11,\pm33\] and
\[\pm\frac{1}{3}, \pm\frac{11}{3}\]
and a few more

Answer 15

I am lost can I just have the answer please.

Answer 16

repost

Answer 17

Me

Answer 18

yes, you

Answer 19

I recommend trying to understand what we're trying to say instead of just asking for someone to tell you the answer that you can copy down... This is a site for homework help, not homework answers.

Answer 20

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