Question

Write a function in simplified form with the following zeros x=-4, 2+i, 2-i

Answer 1

Try doing the conjugate roots first, then multiply the resultant quadratic by (x-4)

Answer 2

Recall that a quadratic can be written as \[ax ^{2}+bx+c\], where the coefficient of b is the sum of the quadratics roots, and the coefficient of c is the product of the quadratics roots, that should help for multiplying the complex roots, then the last real root should be easy. It should look something like \[ax ^{2}\] + (sum of roots [(2+i)+(2-i)])x + (product of roots [(2+i)*(2-i)]).

Answer 3

@mightychondrion that doesn't make sense to me, the whole topic of this question is bery confusing to me

Answer 4

@hartnn could you help?

Answer 5

if 'a' is a zero of a polynomial,

then, (x-a) is a factor of that polynomial.

does this make sense?

Answer 6

Yeah, that makes sense.

Answer 7

ok, so if x =-4 is a zero,
what is the factor?

Answer 8

(x-(-x)) so (x+4) ?

Answer 9

-4*

Answer 10

(x-(-4)) = (x+4)

Answer 11

how about if 2+i
is a zero?

Answer 12

(x-2+i) ?

Answer 13

don't forget the brackets!

Answer 14

\((x-(2+i) = (x-2-i)\)
ok?

Answer 15

how about if x=2-i is a zero?

Answer 16

yeah, I got it.
(x-(2-i)) = (x-2-i)

Answer 17

(x-(2-i)) is actually (x-2+i)

Answer 18

I realized that after I posted it

Answer 19

ok, so you now have the 3 factors,
just multiply them together!

Answer 20

\(f(x) = (x+4)(x-2-i)(x-2+i) = ...\)

Answer 21

@hartnn I would just combine like terms to simplify that and it would be the final answer correct?

Answer 22

yes,
thats correct!

i suggest you start with
\((x-2-i)(x-2+i)\)
first :)

Answer 23

simplified would be x^2+4+i?

Answer 24

do you know what
\((a+b)(a-b)\)
equal?

Answer 25

No, what would be A and B in this equation?

Answer 26

a= x-2
b = i

Answer 27

\((a+b)(a-b) = a^2-b^2\)

Answer 28

x^2+4-i^2 ?

Answer 29

@hartnn