The zeros of the polynomial 3x^4 - 5x^3 - 62x^2 - 92x - 24 are x = {-2, -1/3, 6}. Determine the intervals where the value of f(x) is a negative value. Check all that apply.

a. -∞ < x < -2
b. -2 < x < -1/3
c. -1/3 < x < 6
d. 6 < x < ∞

Step-by-step explanation:

There are 3 zero's but we see the polynomial is of degree 4.

It means it has 2 same zero's. We can verify it is -2. Since -2 is doubled, it reflects the local minimum and it is on the x-axis.

In reality we need to consider the other two zero's.

It is obvious the negative interval is between -1/3 and 6 since the polynomial is of even degree and has positive leading coefficient.

Correct choice is c.

The graph is attached to confirm the theory.

the entire expression should be subtracted from 30. to subtract the quantity, distribute the negative sign, which changes the signs on the terms. so, the –2 changes to positive and gets added to 30 instead of subtracted from it.

hope this ! ^w^

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: