Question

Please help me figure this problem out someone please....
Form a polynomial f(x) with real coefficients having the given degree and zeros.
Degree 5; Zeros: -3;-i; 9+i

Answer 1

complex zeroes come in conjugate pairs

Answer 2

that simply means,

if 9+i is a zero, 9-i will also be a zero

Answer 3

if -i is a zero, +i will also be a zero

Answer 4

Zeros: -3;-i; 9+i

meaning
x = -3 => (x+3) =0
x = -i => (x+i) = 0
x = 9+i => (x-9-i) = 0

one thing to keep in mind with conjugates, they do not come all by their lonesome, they always have a CONJUGATE companion
so
x = -3 => (x+3) =0
x = -i , x = i => (x-i)(x+i) = 0
x = 9+i , x = 9-i => (x-9+i)(x-9-i) = 0

Answer 5

one thing to keep in mind with complex roots I meant

Answer 6

okay so I should enter f(x)=a(x-9+i)(x-9-i)=0

Answer 7

well, the roots are all those, so you multiply all those 5 roots
and you'd get your polynomial
since it has 5 roots, it will be a 5th degree equation

Answer 8

okay so whats my answer, if you don't mind me asking... horrible at math.... Thanks for your patient....

Answer 9

well, you'd need to multiply the roots