Dwilliamsrose
Answered

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

A1, 2, 27
B3, -3, -6
C-6, 3, -6
D2, -1, 18
E3, 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)?

A1 year
B2 years
C3 years
D4 years
E6 years

3)How many negative roots can this polynomial function possibly have?

f(x) = 5x4 – 7x3 – x2 – 3x – 2

A1
B5 or 3 or 1
C3 or 1
D4 or 2 or 0
E2 or 0

Verified Answer :

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:

           \(f(x)=x^3+6x^2-9x-54\)

It could be written as:

\(f(x)=x^2(x+6)-9(x+6)\\\\\\i.e.\\\\\\f(x)=(x^2-9)(x+6)\)

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

i.e.

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

Since,

\((a-b)(a+b)=a^2-b^2\)

Hence, we get:

\(x+3=0\ or\ x-3=0\ or\ x+6=0\\\\\\i.e.\\\\\\x=-3\ or\ x=3\ or\ x=-6\)

Hence, the real zeros of the given problem is:

       B.   3, -3, -6

Question 2)

The profit function is given by:

         \(P(x)=x^3-4x^2+3x-12\)

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.

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

i.e.

\(x^2+3=0\ or\ x-4=0\\\\i.e.\\\\x^2=-3\ or\ x=4\)

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

Hence, \(x^2=-3\) is not a solution)

Hence, the solution is:

\(x=4\)

The correct answer is:

      D.  4 years    

Question 3)

The polynomial function is given by:

           \(f(x)=5x^4-7x^3-x^2-3x-2\)

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 \(f(-x)=5x^4+7x^3-x^2+3x-2\)

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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