Dwilliamsrose
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1) What are the real zeroes of x3 + 6 x2 – 9x – 54?

A. 1, 2, 27
B. 3, -3, -6
C. -6, 3, -6
D. 2, -1, 18
E. 3, 3, -6


2) An entrepreneur estimates that the total profit (profit = total revenue minus total cost) from his proposed company will be given by the function P(x) = x3 – 4x2 + 3x – 12,
where P is in hundreds of dollars and x is the number of years elapsed after the company starts operations. In how many years (x) will the company break even
(no profit, no loss)?

A. 1 year
B. 2 years
C. 3 years
D. 4 years
E. 6 years

3)How many negative roots can this polynomial function possibly have?
f(x) = 5x4 – 7x3 – x2 – 3x – 2

A. 1
B. 5 or 3 or 1
C. 3 or 1
D. 4 or 2 or 0
E. 2 or 0

Respuesta :

meerkat18
A polynomial with a degree of 3 is understood to have three zeroes or roots. In the given problem in 1, x3 + 6 x2 – 9x – 54, the roots of the polynomial using the calculator are B. B. 3, -3, -6. The same applies for number 2 given the polynomial x3 – 4x2 + 3x – 12. The answer in 2 is D. 4 years. The roots of the polynomial in 3 are: 

x1=−0.45117

x2=1.77474

x3=0.03821+0.70576∗i

x4=0.03821−0.70576∗i. Hence there is A. 1 negative root.

Answer with explanation:

Question 1)

The polynomial function is:

           [tex]f(x)=x^3+6x^2-9x-54[/tex]

It could be written as:

[tex]f(x)=x^2(x+6)-9(x+6)\\\\\\i.e.\\\\\\f(x)=(x^2-9)(x+6)[/tex]

The roots of a function are the possible values of x where f(x)=0

i.e.

[tex](x^2-9)(x+6)=0\\\\i.e.\\\\\\(x-3)(x+3)(x+6)=0[/tex]

Since,

[tex](a-b)(a+b)=a^2-b^2[/tex]

Hence, we get:

[tex]x+3=0\ or\ x-3=0\ or\ x+6=0\\\\\\i.e.\\\\\\x=-3\ or\ x=3\ or\ x=-6[/tex]

Hence, the real zeros of the given problem is:

       B.   3, -3, -6

Question 2)

The profit function is given by:

         [tex]P(x)=x^3-4x^2+3x-12[/tex]

We are asked to find the point where the company break even i.e. there is no profit or loss i.e. the possible value of x such that P(x)=0

i.e.

[tex]x^3-4x^2+3x-12=0\\\\\\i.e.\\\\\\x^2(x-4)+3(x-4)\\\\\\i.e.\\\\\\(x^2+3)(x-4)=0[/tex]

i.e.

[tex]x^2+3=0\ or\ x-4=0\\\\i.e.\\\\x^2=-3\ or\ x=4[/tex]

( As the square of a number can't be negative.

Hence, [tex]x^2=-3[/tex] is not a solution)

Hence, the solution is:

[tex]x=4[/tex]

The correct answer is:

      D.  4 years    

Question 3)

The polynomial function is given by:

           [tex]f(x)=5x^4-7x^3-x^2-3x-2[/tex]

We know that the number of sign changes of function f(x) gives the possible number of positive real roots

and  the number of sign changes of function f(-x) gives the possible number of negative real roots.

The number of sign changes in f(x) is 1

(Hence, f(x) has at most 1 positive real root)

and [tex]f(-x)=5x^4+7x^3-x^2+3x-2[/tex]

The number of sign changes in f(x) is 3

(Hence, f(x) has at most 3 negative real root)

              The correct answer is:

                 C.      3 or 1

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