Answer:
There are two rational roots for f(x)
Explanation:
We are given a function
f(x) = x^6-2x^4-5x^2+6f(x)=x6−2x4−5x2+6
To find the number of rational roots for f(x).
Let us use remainder theorem that when
f(a) =0, (x-a) is a factor of f(x) or x=a is one solution.
Substitute 1 for x
f(1) = 1-2-5+6=0
Hence x=1 is one solution.
Let us try x=-1
f(-1) = 1-2-5+6 =0
So x =-1 is also a solution and x+1 is a factor
We can write f(x) by trial and error as
f(x) = (x-1)(x+1)(x^2-3)f(x)=(x−1)(x+1)(x2−3)
We find that f(x) (x^2-3)f(x)(x2−3) factor gives two irrational solutions as
±√3.
Hence number of rational roots are 2.
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Answer:
Explanation:
Answer: the second one, aan article from www.europeanhistory.net that was updated in 2008.
i’d say an article in the Journal of European History
This source isn't very recent (but it doesnt need to be because Amaya is writing about the history of Europe), but we do know that it was updated, which implies that more relevent information was posted. The information is coming from a trustworthy source (we know this because of the url the article is from), so is likely very relevent and relaible. The last source is irrelevent becasue it's abput a carmaker who had/has nothing to do with the Columbian Exchange, and the first source is from a journal (we don't know who the journal belongs to, or what kind of information is in it).
Explanation:
labour day,madaraka day,mashujaa day... does not occur in dece
