According to the fundamental theorem of algebra, how many zeros does the polynomial below have?
f(x)=x^5-12x^3+7x-5

Question

amistre64 · Answer

how many?  or up to how many?

anonymous · Answer

It can have maximum five zeros...

amistre64 · Answer

x^5 gives us the options of 5 factors; each one with the possibility of producing a viable zero

amistre64 · Answer

x^5 is also an odd function; so it has to cross the x axis at least once

anonymous · Answer

at most 5, but they don't need to be all real.  must have at least one real one as per amistre

amistre64 · Answer

i always thought the fundamental thrm of algebra was: pull my finger.  :)

anonymous · Answer

that is the fundamental theorem of sensei

anonymous · Answer

can't wait not to be called this anymore

anonymous · Answer

so because the fighest exponent is 5 the answer is 5?

amistre64 · Answer

that answer doesnt fit the question asked tho; there is no way to determine HOW MANY zeros a function has.  We can determine certain conditions for it, but that is about all

anonymous · Answer

yes the maximum number of possible zeros is equal to the degree/highest index of the variable...

anonymous · Answer

every non-zero single-variable polynomial with complex coefficients has exactly as many complex roots as its degree, if each root is counted up to its multiplicity. http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra

amistre64 · Answer

complex roots, thats another story ;)

anonymous · Answer

Very important, gives closure...

And any real can be written as real + 0i

amistre64 · Answer

if there is an option for 5, id take it

anonymous · Answer

thanks guys !

anonymous · Answer

in a word, if you are living in the land of complex numbers the answer is it has 5 including multiplicity
if you are living in the world of real number the answer is "at most 5"

anonymous · Answer

if you are living in the world of real number the answer is "at most 5"

Not quite, complex root come in pairs...