Question

Precalc question please help
question posted below

Answer 1

Find a cubic function with the given zeros.
\(\sqrt{6}\sqrt{-6}~-3\)

\(f(x) = x^3 - 3x^2 - 6x - 18\)
\(f(x) = x^3 + 3x^2 - 6x - 18\)
\(f(x) = x^3 + 3x^2 + 6x - 18\)
\(f(x) = x^3 + 3x^2 - 6x + 18\)

Answer 2

test them out ....

Answer 3

i got the 2nd one @amistre64

Answer 4

are your zeros posted correctly?

Answer 5

yes

Answer 6

sqrt(6) -x
sqrt(-6)-x
-----------
sqrt(-36) -xsqrt(6)
-xsqrt(6) + x^2
-----------------------
sqrt(-36)-2x sqrt(6) + x^2
-3 - x
------------------------
your roots dont match up as is, unless ive made some error

Answer 7

-x^3+(1+i) sqrt(6) x^2-3 x^2+(3+3 i) sqrt(6) x-6 i x-18 i

or its negative

Answer 8

again, are your roots posted correctly?

-sqrt(6) or is it sqrt(-6) ??

Answer 9

sqrt(-6)

Answer 10

then none of these options work, you have ONE complex root

Answer 11

whoops just looked again it is supposed to be -sqrt

Answer 12

yeah .... thats what i thought, if so then the second one is fine

Answer 13

Thank you!

Answer 14

http://www.wolframalpha.com/input/?i=x%5E3+%2B+3x%5E2+-+6x+-+18

youre welcome