The rational roots of f(x) = 3x^3 + 2x^2 + 3x + 6 are ±(1, 2, 3, 6, 1/3, 2/3)
<h3>How to determine the rational root of the function f(x)?</h3>
The function is given as:
f(x) = 3x^3 + 2x^2 + 3x + 6
For a function P(x) such that
P(x) = ax^n +...... + b
The rational roots of the function p(x) are
Rational roots = ± Possible factors of b/Possible factors of a
In the function f(x), we have:
a = 3
b = 6
The factors of 3 and 6 are
a = 1 and 3
b = 1, 2, 3 and 6
So, we have:
Rational roots = ±(1, 2, 3, 6)/(1, 3)
Split the expression
Rational roots = ±(1, 2, 3, 6)/1 and ±(1, 2, 3, 6)/3
Evaluate the quotient
Rational roots = ±(1, 2, 3, 6, 1/3, 2/3, 1, 2)
Remove the repetition
Rational roots = ±(1, 2, 3, 6, 1/3, 2/3)
Hence, the rational roots of f(x) = 3x^3 + 2x^2 + 3x + 6 are ±(1, 2, 3, 6, 1/3, 2/3)
The complete parameters are:
The function is given as:
f(x) = 3x^3 + 2x^2 + 3x + 6
The rational roots of f(x) = 3x^3 + 2x^2 + 3x + 6 are ±(1, 2, 3, 6, 1/3, 2/3)
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