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Question

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anonymous · Answer

try 9

aripotta · Answer

how do i try 9? 

i need to know how to do it. like, how'd you come up with 9 ._.

e.mccormick · Answer

Set to 0 and fator.  You may need to see if there is something you can divide out, or do factor by grouping to get it from a trinomeal to a binomeal to start.

e.mccormick · Answer

Well, 3rd power to 2 power... so polynomeal to trinomeal, actually....

aripotta · Answer

can you show me?

e.mccormick · Answer

Have you done polynomeal long division r synthetic division?  Have you covered those yet?

aripotta · Answer

yea i have

e.mccormick · Answer

Basically, it is trying a fw likely divison choices, like $(x\pm 1)$, $(x\pm 3)$, or what he suggested, $(x\pm 9)$.

aripotta · Answer

but how did you come up with those? :/

e.mccormick · Answer

I tried $(x-3)$ and it had a remainder, so that is not a factor.  Once you get the first factor out of these,, they are usually not too hard.  He suggeste 9 because the last number is 9.  I tried 3 because 3 is a divisor of 9.  1 divides anything.

e.mccormick · Answer

So if the last number is even, $(x\pm 2)$ is a good one to try, but because it is odd, 2 is usually out for the first one to try.

aripotta · Answer

ok so... x - 9 is a factor, so that means 9 is one of the zeros? now how do i find the others?

e.mccormick · Answer

Once you have divided it out, you will have a new formula to work with.  It may be easy to factor, or it may need the quadratic formula.

aripotta · Answer

x^2 - 3x + 1?

e.mccormick · Answer

Yah, I got that as the other factor, which does not look nice.  So you coud plug in the part of it to the determinant of the quadratic formula and see if you get a negative or not.  (The determinant of the quadratic fomula is the part under the radical.  If it is negative, no need to go any furthur.)

aripotta · Answer

it's positive

e.mccormick · Answer

Then finish the quadratic and you will have the other 2 factors and it will be in fully factorded form.  Then, each factor tells you a zero.

aripotta · Answer

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e.mccormick · Answer

Yeo, so those are two of the roots, or zeros, and if you solve the first one you divided out you will have the third.

aripotta · Answer

9 is one of the zeros, right?

e.mccormick · Answer

Well that was no fun! Yes. $$y=0 \mathrm{\;when\;} x=\{9, \frac{3\pm \sqrt{5}}{2}\}$$ And here is a graph so you can see what they mean by zeros. https://www.desmos.com/calculator/e6xhpshadu

aripotta · Answer

can you check this one @e.mccormick ?

e.mccormick · Answer

@AriPotta Sorry I could not answer yesterday.  All I got all afternoon was a loading thing spinning around.  Real pain in the anatomy!

You copied one thing wrong.  At the bottom you list $3, -1, 3i$, but in the work you find $2, -1, 3i$.  The $2, -1, 3i$ is mostly correct, so this is just a copy error.

I say mostly because typically zeros only deal with the real numbers.  3i is a complex number and typically not seen as a zero.  However, if dealing with complex numbers, then the conjugate is also a zero!  This goes back to $x^2=a\Rightarrow x=\pm\sqrt{a}$.  Therefore, if using complex, it would be $2, -1, \pm 3i$, or for people that really get specific: $2, -1, 0\pm 3i$, where the 0 is a critical part of the complex conjugates.

Finally, a word on notation.  Some teachers will want this sort of answer in set notation.  That would be something along the lines of:

$f(x)=0 \mathrm{\;for\;} x=\{2, -1, 0\pm 3i\}$  

This just makes it clearer what you mean when you say a zero.

aripotta · Answer

i already turned it in :/ hopefully it'll be marked as correct :s