Emma H.

asked • 12/11/15# Find the nth degree polynomial with real coefficients satisfying the given conditions

Find the nth degree polynomial with real coefficients satisfying the given conditions.

n=3

f(-1)=80

zeros are 3 and 3i

The expanded and simplified polynomial is f(x)=______________

n=3

f(-1)=80

zeros are 3 and 3i

The expanded and simplified polynomial is f(x)=______________

*****Got the answer 2x

^{3}- 6x^{2 }+ 18x - 54 multiple times and it is not correct!
More

## 2 Answers By Expert Tutors

Mark M. answered • 12/11/15

Tutor

4.9
(883)
Retired math prof. Very extensive Precalculus tutoring experience.

f(x) = a(x-3)(x-3i)(x-(-3i)

= a(x-3)(x-3i)(x+3i)

= a(x-3)(x

^{2}+9) = a(x

^{3}+9x-3x^{2}-27)Since f(-1) = 80, we have a(-1-9-3-27) = 80

-40a = 80

So, a = -2

f(x) = -2(x

^{3}+9x-3x^{2}-27) = -2x^{3}+ 6x^{2}-18x^{ }+54Derek F. answered • 12/11/15

Tutor

4.9
(8)
An excellent tutor must be clear, concise and patient.

Updated.

Two roots are given: 3 and 3i. The third root must be -3i.

The form of the factorized polynomial is

f(x) = a (x-3) (x-3i) (x+3i)

Applying f(-1) = 80

80 = a ((-1) - 3) ((-1) - 3i) ((-1) + 3i)

a = -2

Not sure why your answer is listed as incorrect.

Mark M.

The third root is the conjugate of 3i, -3i. Complex numbers always appear in conjugate pairs!!!!!

The function you present has imaginary coefficients. It does not have real coefficients as required in the problem.

Report

12/11/15

## Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.

Most questions answered within 4 hours.

#### OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.

Mark M.

12/11/15