Question

Can someone please explain how to do this to me?

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form.

8, -14, and 3 + 9i

Answer 1

there isn't one... if the polynomial has real coefficients then complex roots will come in conjugate pairs

Answer 2

These are the choices

f(x) = x4 - 11x3 + 72x2 - 606x + 10,080

f(x) = x4 - 303x2 + 1212x - 10,080

f(x) = x4 - 11x3 - 72x2 + 606x - 10,080

f(x) = x4 - 58x2 + 1212x - 10,080

Answer 3

yeah, sorry. there isn't a 3rd degree... you'd have to include the conjugate which would give a 4th degree.

so the zeros would have to be 8, -14, 3+9i and 3-9i

=> (x-8)(x+14)(x^2+90)

multiply that out and you'll have your polynomial

Answer 4

Where did you get the 90?

Answer 5

3^2 + 9^2 because (x + 3 + 9i)(x + 3 - 9i) = x^2 +3x -9ix +3x + 9 -27i + 9ix +27i +81= x^2 +6x +90

oops, forgot the 6x. sorry.

so it should be (x-8)(x+14)(x^2 + 6x +90)

Answer 6

it should be -6x

Answer 7

no

Answer 8

it is (x-(3+i9))(x-(3-i9))=(x-3+i9)(x-3-i9)=x^2-6x+90

Answer 9

oh sorry, i must not be awake... i didn't change the signs


[x-(3+9i)][x-(3-9i)]

Answer 10

so yes @le0n ... -6x

Answer 11

you did all the job i gust corrected one step

Answer 12

Oh okay now I get it! Thank you so much!!