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.
Thermometer and refrigerant Identifier
