Question

(1 point) List all possible rational roots for the function

f(x) = 5x + 3x³ + 3x² + 6x +95.
Give your list in increasing order. Beside each possible rational root, type "yes" if it is a
root and "no" if it is not a root. Leave any unnecessary answer blanks empty.
Possible rational root: 1
Possible rational root: -1
Possible rational root: 1/5
Possible rational root: -1/5
Possible rational root: 5
Possible rational root: -5
Possible rational root: 19
Possible rational root: -19
19/5
Possible rational root:
Possible rational root: -19/5
Possible rational root: 95
Possible rational root: -95
radit o
Is it a root? no
Is it a root?
no
Is it a root?
no
Is it a root?
Is it a root?
Is it a root?
Is it a root?
Is it a root?
Is it a root?
Is it a root?
Is it a root?
Is it a root?
this pmblom
no
no
no
no
no
no
no
no
no

Answer 1

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)].