Well, according to the Rational Roots Theorem, which applies to all polynomials with integer coefficients, we take the integer factors of both the lead coefficient and the constant term
The lead term is 4 and has factors 1, 2, 4
The constant term is 6 and has factors 1, 2, 3, 6
Divide each factor of the constant term by each factor of the lead term:
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Find all possible rational zeros.
Well, according to the Rational Roots Theorem, which applies to all polynomials with integer coefficients, we take the integer factors of both the lead coefficient and the constant term
The lead term is 4 and has factors 1, 2, 4
The constant term is 6 and has factors 1, 2, 3, 6
Divide each factor of the constant term by each factor of the lead term:
1/1, 1/2, 1/4, 2/1, 2/2, 2/4, 3/1, 3/2, 3/4, 6/1, 6/2, 6/4
Reduce and consolidate the list
1, 1/2, 1/4, 2, 3, 3/2, 3/4, 6
Take the positive and negatives of each
−6, −3, −2, −3/2, −1, −3/4, −1/2, −1/4, 1/4, 1/2, 3/4, 1, 3/2, 2, 3, 6
These are the ONLY possible rational roots of the polynomial.
And as the previous respondent pointed out, none of them work as roots. This means that the polynomial has NO rational real roots at all.
There are none. The zeros of that polynomial are all irrational.
For example, the real zero is at x = (1/12) ( 5 - 83 / cuberoot{ 611 + 18sqrt(2917)} + cuberoott\{ 611 + 18sqrt(2917)} )
The other two roots are complex, but they are irrational also.