Question

Find the polynomial function with roots 11 and 2i.

Answer 1

(x-11)(x-2i)

Answer 2

If complex allowed...
Else assume -2i as well.

Answer 3

ok but how do you do it

Answer 4

Romero gave a complex polynomial, if you want real then you need 3 roots, 2 of which have to be complex. Usually complex roots are in pairs so assume -2i as well as 2i.

Answer 5

ok i need u to show me not explain please i cant understand it if u just tell me

Answer 6

(x-11)(x-2i)(x+2i)

Answer 7

ok so i multiply everything together?

Answer 8

Use the foil method to multiple the three factors. Multiply the first two factors first and then multiply that result by the third one.

Just remember that i^2=-1

Answer 9

thats the main thing i dont understand

Answer 10

-2xi=(-2)i(x) this will subtract out when you multiply (x)(2i)

Notice that (x-2i)(x+2i) is a difference of squares.

So the product of these factors will be x^2-4i^2

Answer 11

x^2- (11+2i)x + 22i is what i got

Answer 12

There wouldn't be any i term. Note that (x-2i)(x+2i) =x^2+[(-4)(i^2)

Which becomes x^2-4i^2

Since i^2=-1

It becomes x^2+4

Now multiply (x^2+4)(x-11)

Answer 13

so the i just goes away?

Answer 14

Yes

Answer 15

x^3-11x^2+4x-44

Answer 16

Looks good to me.

Answer 17

ok thank you so much

Answer 18

One more thing. We are assuming the coefficients of the polynomial are real numbers.

If you can have coefficients that are no real, then I believe you just multiply (x-2i)(x-11)

x^2+(-2i-11)x+22i