Question

form a third degree polynomial function with real coefficients such that 8+i and 7 are zeros

f(x) = ???

Answer 1

Since coefficients are real, the two complex roots have to be conjugates. So the simplest polynomial is
\[(x-7)(x -8 -i)(x-8+i)\]

Answer 2

The question wants the answer in a X^3 +bx^2+cX+# ?? so im confused.....

Answer 3

x = 8 +- i
x-8 = +-i
(x-8)^2 = -1
(x-8)^2 +1 = 0
x^2-16x +65 = 0

--> (x-7)(x^2-16x+65)
Now distribute

Answer 4

@dumbcow, why -i if the given is positive(+)?

Answer 5

because complex zeros always come in pairs

Answer 6

ahh, yeah. . (+) and (-)

Answer 7

x^3-23x^2-57x-455? is that right?

Answer 8

everything except the "-57x" term

Answer 9

177x?

Answer 10

yep:)

Answer 11

thanks for the direction much appreciated