Question

Find the polynomial function of lowest degree with rational coefficients thaat has the given numbers as some of its zeros.
4+i,2

Answer 1

other zero is
\[4-i\] so your job is to multiply
\[(x-2)(x-(4+i))(x-(4-i))\]

Answer 2

which is not as hard as it might look. do you know how to do it? the snap way i mean?

Answer 3

No i get stuck after the first part of solving it

Answer 4

i can get x\[x ^{3^{?}}\]

Answer 5

ok first off we have to multiply
\[(x-(4+i))(x-(4-i))\] and it is pretty straight forward. the "last" term, that is the constant, will be
\[(4+i)(4-i)\] clear?

Answer 6

yes

Answer 7

and it is always true that
\[(a+bi)(a-bi)=a^2+b^2\] so you get a constant of
\[4^2+1^2=17\] notice we do not compute with any i's

Answer 8

the "outer , inner" will be
\[-(4+i)x -(4-i)x=8x\]

Answer 9

so we have
\[(x-(4+i))(x-(4-i))=x^2-8x+17\] done

Answer 10

now you have to multiply this by x -2 but that is straghtforward

Answer 11

you should convince yourself that
\[(x-(a+bi))(x-(a-bi))=x^2-2ax+a^2+b^2\] so it is not hard

Answer 12

your "final answer" will be
\[(x-2)(x^2-8x+17)\] whatever that is

Answer 13

after solving that equation I will have my answer?

Answer 14

yes. here it is, and you can even see if you scroll down that we have the correct zeros
http://www.wolframalpha.com/input/?i=%28x-2%29%28x^2-8x%2B17%29

Answer 15

real root is 2, complex root is 4 + i and its conjugate 4 - i

Answer 16

okay I beileve I got it

Answer 17

Thank you.