only one zero= -3.1045
f(x) = 5x⁴ + 3x³ + 3x² + 6x +95.
Assuming that the first term is 5x⁴.
f(x) = (x²+1) (3x² + 6x) + x² (2x²- 3x) is always positive for x < -2.5.
Between 0 and -2.5 there's no zero.
This function doesn't have any zeroes.
+++++++++++
If f(x) = 5x + 3x³ + 3x² + 6x +95.
then we have some zeroes.
f(x) = 3x³ + 3x² + 11 x + 95.
= 3 [(x+1/3)³ + (8/3)(x+1/3)+(830/27)]
Let t = x+1/3.
f(t) = 3 (t³ + 8/3 t + 830/27) .
g(t) = t³ + 8/3 t + 830/27.
This is in the form of a depressed cubic equation of the form t³+px+q. p=8/3. q=830/27.
As the determinant ∆ = 4p³+27q² > 0, there's only one real zero of g(t). Other two are imaginary. We use the hyperbolic functions to find the zeroes.
t₀ = -2√(p/3) × Sinh [ 1/3× ArcSinh (3q√3 / (2 p √p) )]
= (-4√2 /3) Sinh [ 1/3× ArcSinh (415/(16√2))]
= -3.1045
where Sinh x = (eˣ + e⁻ˣ)/2 ,
and ArcSinh x = Log [ x + √(1+x²)].