According to the fundamental theorem of algebra how many roots exits for the polynomial function. f(x)=8x^7-x^5+x^3+6

The Fundamental Theorem of Algebra (FTOA) tells us that any non-zero polynomial in one variable with complex (possibly real) coefficients has a complex zero.

A straightforward corollary, often stated as part of the FTOA is that a polynomial in a single variable of degree  

n

>

0

with complex (possibly real) coefficients has exactly  

n

complex (possibly real) zeros, counting multiplicity.

To see that the corollary follows, note that if  

f

(

x

)

is a polynomial of degree  

n

>

0

and  

f

(

a

)

=

0

, then  

(

x

−

a

)

is a factor of  

f

(

x

)

and  

f

(

x

)

x

−

a

is a polynomial of degree  

n

−

1

. So repeatedly applying the FTOA, we find that  

f

(

x

)

has exactly  

n

complex zeros counting multiplicity.

Discriminants

If you want to know how many real roots a polynomial with real coefficients has, then you might like to look at the discriminant - especially if the polynomial is a quadratic or cubic. Ths discriminant gives less information for polynomials of higher degree.

The discriminant of a quadratic  

a

x

2

+

b

x

+

c

is given by the formula:

Δ

=

b

2

−

4

a

c

Then:

Δ

>

0

indicates that the quadratic has two distinct real zeros.

Δ

=

0

indicates that the quadratic has one real zero of multiplicity two (i.e. a repeated zero).

Δ

<

0

indicates that the quadratic has no real zeros. It has a complex conjugate pair of non-real zeros.

The discriminant of a cubic  

a

x

3

+

b

x

2

+

c

x

+

d

is given by the formula:

Δ

=

b

2

c

2

−

4

a

c

3

−

4

b

3

d

−

27

a

2

d

2

+

18

a

b

c

d

Then:

Δ

>

0

indicates that the cubic has three distinct real zeros.

Δ

=

0

indicates that the cubic has either one real zero of multiplicity  

3

or one real zero of multiplicity  

2

and another real zero.

Δ

<

0

indicates that the cubic has one real zero and a complex conjugate pair of non-real zeros.

Step-by-step explanation:

answer for #1 is c

answer for #2 is d

can i get brainliest

answer: integers are just whole numbers: positive, negative, and zero.

i = {-∞, -2, -1, 0, 1, 2, 3, ∞}

let x = 1st integer

let y = 2nd integer

x = 3y + 17   {equation 1}

xy = -24         {equation 2}

substitute 3y+17 in place of x n equation 2

and solve for y.

(3y+17)y = -24

3y2 + 17y = -24

3y2 + 17y + 24 = 0

from quadratic equation:

y = [-17±√(172-4(3)(24))]/(2(3))

y = (-17 ±√1)/6

y = (-17+1)/6   or     y = (-17-1)/6

y = -16/6           or     y = -18/6

y = -2 2/3         or   y = -3

since we are dealing with

y = -3

x = 3y+17 = -9+17 = 8

the integers are -3, 8

step-by-step explanation: