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Mathematics 8 Online
OpenStudy (anonymous):

**Help!! Medals awarded! Using the given zero, find one other zero of f(x). Explain the process you used to find your solution. 2 - 3i is a zero of f(x) = x4 - 4x3 + 14x2 - 4x + 13.

OpenStudy (anonymous):

Start off by plugging in 2-3i to check that it is a zero of f(x).

OpenStudy (anonymous):

i already know it is a zero, it tells you already

OpenStudy (anonymous):

Yes, this question is a ain in the butt I'll tell you form right now @adrian111 can you synthetically divide?

OpenStudy (anonymous):

is a pain*

OpenStudy (anonymous):

does anyone want to help me?(: please

OpenStudy (anonymous):

sure @sarahjones667 thats what we're here for :)

OpenStudy (anonymous):

you people are geniuses

OpenStudy (anonymous):

i know how to, so just divide the zero by the equation ?

OpenStudy (anonymous):

Oh wait you don't need to do any of that, I thought the question said to factor it, no all you have to do is ind another zero ok.

OpenStudy (anonymous):

i know my friend, problem is i dont know how.. thats why im asking

OpenStudy (anonymous):

Ok @adrian111 there's a theorem, it's caled the fudamental theorem of Alebra, heard of it?

OpenStudy (anonymous):

yes, i have

OpenStudy (anonymous):

Algebra*, well it simply states that a polynomial with n-degree, counting multiplicity, has at least one complex root, and has EXACTLY n-roots, anywyays, this leads to the realm of imagiary numbers, the theorem, although it doesn't state, says that if 2-3i is a solution, its cojugate is also a zero

OpenStudy (anonymous):

okay, go on, i can see where you're going with this

OpenStudy (anonymous):

So if they told you 2-3i is a zero, then its cojugate is alsoo a zero which can be proved y multiplyi them, ut anyways what is the cojuagte of 2-3i?

OpenStudy (anonymous):

it would be -2+3i?

OpenStudy (anonymous):

Also you can say by the linear factorization theorem too that you get this, it says that any polnmial ca be broken up into linear factors ax^2+bx+c=(x-d)(x-e) ax^3+bx^2+cx+d=(x-e)(x-f)(x-g) or \[x^2+625=0\] \[(x-16i)(x+16i)=0\]

OpenStudy (anonymous):

not exactly, the conjugate has to do with just the imaginary number \[a-bi\] its conjugate

OpenStudy (anonymous):

would be \[a+bi\] @adrian111

OpenStudy (anonymous):

alright so would the conjugate be 2+3i?...im just flipping the signs?

OpenStudy (anonymous):

Yep, you got it!

OpenStudy (anonymous):

thanks man, i appreciate it!

OpenStudy (anonymous):

and be on the lookout, im going to be asking one more question similar to this one in a bit

OpenStudy (anonymous):

Lol and if you're wondering what the theorem of algebra has to do with algebra, it goes back to the time in math when all they were obsessed with findig zeros of a polynomial equation, THAT DIDN'T END WELL IM TELLING YOU

OpenStudy (anonymous):

@adrian111 and that what we're here for :)

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