Question

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Answer 1

polynomial will have degree 5 and zeros of \(4,i,-i,1+2i\) and \(1-2i\)

Answer 2

\[f(x)=[(x-i)(x-(1+2i))]^2\]

Answer 3

you can write it in factored form as
\[(x-4)(x+i)(x-i)(x-(1+2i))(x-(1-2i))\] and see what we get
both the multiplications are actually easy

Answer 4

Degree 4*

Answer 5

(x-i)^2 * (x-(1+2i))^2

Answer 6

\[(x+i)(x-i)=x^2+1\] so now we have
\[(x-4)(x^2+1)\] as a start

Answer 7

oooh i mis read
degree 4 so ignore the \(x-4\) part, but unfortunately other two answers are wrong as well

Answer 8

start with
\[(x+i)(x-i)(x-(1+2i))(x-(1-2i))\] because you want your polynmomial to have real coefficients i am sure

Answer 9

first product is simple
\[(x+i)(x-i)=x^2+1\]
second product is not bad either and there are a couple of quick ways to do it

Answer 10

one way is to memorize that if \(a+bi\) is a zero, then so is \(a-bi\) and the quadratic it comes from is
\[x^2-2ax+(a^2+b^2)\]

Answer 11

with the factor \[\mathrm{(x^2+1)}\] you will get -i and i

Answer 12

There's a one answer.

Answer 13

therefore is \(1+2i\) is a zero, then so is \(1-2i\) and the quadratic it comes from is
\[x^2-2x+1^2+2^2\] i.e
\[x^2-2x+5\]

Answer 14

your final job is to multiply
\[(x^2+1)(x^2-2x+5)\] and that will do it

Answer 15

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Answer 16

if you do not want to memorize what i wrote above, then you can always work backards
\[x=1+2i\]
\[x-1=2i\]
\[(x-1)^2=(2i)^2=-4\]
\[x^2-2x+1=-4\]
\[x^2-2x+5=0\]

Answer 17

But that polynomial has four zeros

Answer 18

yes indeed it does

Answer 19

i am assuming the polynomial is to have rational coefficients, so if \(a+bi\) is a root, so it \(a-bi\)

Answer 20

otherwise you get polynomial with complex coefficients, which i am more or less certain is not what is being asked for

Answer 21

@some_someone this piece of algebra
\[(x^2+1)(x^2-2x+5)\] i am sure you can do

Answer 22

your answer is correct