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.
The colonists could export goods only to Britain, according to the Navigation Acts.
According to these laws, no goods originating from the English colonies in Asia, Africa or America could be imported or exported by foreign ships.
This meant that foreign ships could not conduct any trade operations in the ports of English overseas possessions, nor transport goods to England, Ireland or Wales.
This increased the price of imported goods, thus causing a great harm to the economy of the Thirteen Colonies.
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