{ x  }^{ 4  }  +16 { x  }^{ 3  }  - { x  }^{ 2  }  +4x+29 =0

To solve the equation, we can use the Rational Root Theorem to find possible rational roots. The Rational Root Theorem states that if a polynomial equation has a rational root, it must be of the form p/q, where p is a factor of the constant term (in this case, 29) and q is a factor of the leading coefficient (in this case, 1).

The factors of 29 are 1 and 29, and the factors of 1 are 1 and -1. So, the possible rational roots are ±1, ±29.

We can try these values one by one to see if they are roots of the equation.

For x = 1:
(1)^4 + 16(1)^3 - (1)^2 + 4(1) + 29 = 1 + 16 - 1 + 4 + 29 = 49 ≠ 0

For x = -1:
(-1)^4 + 16(-1)^3 - (-1)^2 + 4(-1) + 29 = 1 - 16 - 1 - 4 + 29 = 10 ≠ 0

For x = 29:
(29)^4 + 16(29)^3 - (29)^2 + 4(29) + 29 = 707281 + 134456 + 841 + 116 + 29 = 842723 ≠ 0

For x = -29:
(-29)^4 + 16(-29)^3 - (-29)^2 + 4(-29) + 29 = 707281 - 134456 + 841 - 116 + 29 = 574579 ≠ 0

None of the possible rational roots are actual roots of the equation. Therefore, the equation does not have any rational roots.