Answered

According to the Rational Root Theorem, the following are potential roots of f(x) = 6x^4 + 5x^3 – 33x^2 – 12x + 20.


-5/2, -2, 1, 10/3


Which is an actual root of f(x)?

A-5/2
B–2
C1
D10/3

Verified Answer :

A graphical analysis of this function indicates that out of those possible roots, -5/2 is the only true root

Answer:

\(\frac{-5}{2}\)                              Option A

Step-by-step explanation:

Given that a function,

f(x) = 6x⁴ + 5x³ - 33x² - 12x + 20

put the numbers in place of x, if function become 0 it means that number is root of the given function.

For \(\frac{-5}{2}\)

\(f(\frac{-5}{2})=6(\frac{-5}{2})^{4}+5(\frac{-5}{2})^{3}-33(\frac{-5}{2})^{2}-12(\frac{-5}{2})+20\)

\(f(\frac{-5}{2})=0\)

So, \(\frac{-5}{2}\) is the root of given function.

That's the final answer.

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