Originally Posted by

**HallsofIvy** By the "rational root theorem" that I quoted above, the only possible rational roots are 1, -1, 3, and -3.

$\displaystyle 1^5- 6(1)+ 3= -2$, not 0.

$\displaystyle (-1)^5- 6(-1)+ 3= 8$, not 0.

$\displaystyle (3)^5- 6(3)+ 3= 243- 18+ 3= 228$, not 0.

$\displaystyle (-3)^5- 6(-3)+ 3= -243+ 18+ 3=-222 $, not 0.

Since you say the solution is not a rational number, I presume you understand that number, which is rational, once again, is an approximation, not an exact solution. The "cubic formula" is rather complicated so I may well be wrong but I get

$\displaystyle \sqrt[3]{-\frac{1}{2}+i\sqrt{\frac{5}{108}}}+\sqrt[3]{\frac{1}{2}-i\sqrt{\frac{5}{108}}}$

Notice those "i"s? It is the fact that, even for cubics with all real roots, you may require complex numbers midway through the formula that led to the acceptance of complex numbers.

Where did you get that "5.40985"? I strongly suspect that is also an approximation. If so that is NOT a solution in radicals because it is not in terms of radicals of exact numbers.

And, in fact, for that value of x, $\displaystyle x^5- 6x+ 3$ is -8.7899913273470831550756050870503e-7, or -0.000000087899913273470831550756050870503 very, very close to 0 but NOT 0. Again, this is an approximation, not a solution.