Question

find a cubic equation whose roots are -4 and -6i

Answer 1

if -6i is a root, then so is 6i since complex roots come in complex conjugate pairs

Answer 2

so you really have these 3 roots: -4, -6i, 6i

Answer 3

ok but how do you make that into an equation...its cubic because there are three roots correct?

Answer 4

which means

x = -4, x = -6i, x = 6i

x+4 = 0, x^2 = 36i^2, or x^2 = 36i^2

x+4=0 or x^2 = -36 or x^2 = -36

x+4=0 or x^2 = -36

x+4=0 or x^2+36 = 0

(x+4)(x^2+36) = 0

keep going...

Answer 5

and yes, all cubic equations have 3 complex roots

Answer 6

any nth degree polynomial has n complex roots (fundamental theorem of algebra)

Answer 7

ok I think I understand now. thank you

Answer 8

yw

Answer 9

so is this the final answer? (x+4) (x+6)(x+6)=0

Answer 10

no

Answer 11

(x+4) (x+6)(x+6)=0 implies that x+6=0 ---> x =-6 is a root, but -6 is NOT a root

Answer 12

Too bad the question does not ask for what was provided. If we want the Complex Conjugate Pairs, the question MUST state "with real coefficients". Otherwise, (x+4)(x+6i)(x-2) = 0 would also be valid, or (x+4)(x+6i)^2 = 0 or infinitely many other options.

Please do not try to factor x^2 + 36. That makes no sense.

Answer 13

ok so just leave it as (x+4)(x^2+36)

Answer 14

no you have to expand it out

Answer 15

Make sure the question says "with Real Coefficients". If it doesn't, you might want to have a chat with the author of the question.

Answer 16

(x+4)(x^2+36)

x(x^2+36) + 4(x^2+36)

what's next?

Answer 17

x^3+36x+4x^2+144=0
x^3-4x^2+36x+144=0

Answer 18

you somehow turned +4x^2 into -4x^2

Answer 19

otherwise, it's perfect

Answer 20

sorry typo

Answer 21

yep

Answer 22

so y = x^3+4x^2+36x+144 is your cubic equation

Answer 23

oh ok it just clicked in my head.
thanks I really appreciate your help

Answer 24

ok great, yw