Question

Suppose a polynomial function of degree 4 with rational coefficients has the given number as zeros. Find the other zeroes. 2+3i,2+(sqrt)3

Answer 1

well you actually have 4 zeros
\[\{2+3i,2-3i, 2+\sqrt{3}, 2-\sqrt{3}\}\]

Answer 2

so your job is to multiply out
\[(x-(2+3i))(x-(2+3i))(x-(2+\sqrt{3}))(x-(2-\sqrt{3})\] a seemingly annoying task, but not that hard actually

Answer 3

we can start with
\[(x-(2+3i))(x-(2-3i))\] using "first outer inner last"

Answer 4

first is easy.

\[x^2\]
last is also easy because
\[(a+bi)(a-bi)= a^2+b^2\] so
\[(2+3i)(2-3i)=4+9=13\]

Answer 5

and all that is left is the outer and inner which is
\[-(2+3i)x-(2-3i)x=-4x\]

Answer 6

so first product is
\[(x-(2+3i))(x-(2-3i))=x^2-4x+13\]

Answer 7

similarly with
\[(x-(2+\sqrt{3})(x-(2-\sqrt{3}))\] we get
first
\[x^2\]
last
\[(2+\sqrt{3})(2-\sqrt{3})=4-3=1\]

Answer 8

im not understanding.. :(

Answer 9

and "outer inner " is
\[(-2+\sqrt{3})x-(2-\sqrt{3})x=-4x\]

Answer 10

giving
\[x^2-4x+1\]

Answer 11

ok we go slow.

Answer 12

from the top slowly. you are given 2 zeros yes?

Answer 13

correct

Answer 14

one is
\[2+3i\] but you have rational coefficients. so what must the other zero be?

Answer 15

or is it not clear what the other zero must be?

Answer 16

and 2+(sqrt)3?

Answer 17

well you are given that one. i am asking if
\[2+3i\] is one zero, what must the other one be?

Answer 18

and i see that it is not clear. if
\[a+bi\] is a zero of a polynomial with real coefficients, then so is
\[a-bi\]

Answer 19

im sorry satellite73, this all just doesn't seem familiar at all :(

Answer 20

how do we know this? because of the quadratic formula. it say the solutions to a quadratic are
\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] so is
\[b^2-4ac<0\] you get two complex zeros, the plus and the minus

Answer 21

for example if you solve
\[x^2-4x+13=0\] what are the solutions?

Answer 22

you can use the quadratic formula to get them or you can complete the square. your choice

Answer 23

im still not following, i shoulve paid attention in class during spring.. ;*

Answer 24

do you know the quadratic formula?

Answer 25

yes from intermediat algrebra but my test is due in a couple of hours online.. :S im just really stressed

Answer 26

well in that case to hell with the explanation and allow me to give you the answer. in other words we will just cheat

Answer 27

:D

Answer 28

dude, i am desperate i had to drop it in spring lol now summers kickin my butt

Answer 29

didn't you give the answer in your first post?

Answer 30

ok first product is
\[x^2-4x+13\]
second one is
\[x^2-4x+1\] so multiply these two together to get
\[x^4-8 x^3+30 x^2-56 x+13\]

Answer 31

and that is your answer. no charge

Answer 32

also no implied warranty of fitness but i am sure it is right

Answer 33

lol ok thanks you!! im hoping i will get it done

Answer 34

that is the correct polynomial...though I don't believe it was required

Answer 35

does that mean its just the first one then the second one?

Answer 36

oh crap. if i could read i would be dangerous
it says "find the other zeros" not "find a possible polynomial"
answer to
"find the other zeros" is
\[2-3i\] and
\[2-\sqrt{3}\]

Answer 37

oh lol thanks!

Answer 38

maybe if you read my posts ;)