Question

nffdhdhh

Answer 1

are you supposed to make up your own?

Answer 2

yes
can you help me

Answer 3

if so, then make it easy on yourself
if you want say a polynomial of degree 3 with two complex roots and one real root, start by multiplying out \((x^2+1)(x-2)\) for example
the real root will be \(2\) and the two complex roots will be \(\pm i\)

Answer 4

when you multiply that out , you get
\[x^3-2 x^2+x-2\]

Answer 5

then you can say something like:
" because it is a polynomial of degree 3, it has 3 roots"
descartes rule of sign: it has 3 changes in sign, from 1 to -2, from -2 to 1 and from 1 to -2

so there are either 3 or 1 positive zero

Answer 6

okay :)

Answer 7

make any sense? we are not quite done

Answer 8

yes it does

Answer 9

there are 3 changes in sign in \(f(x)=x^3-2 x^2+x-2\) so now we have to consider
\[f(-x)=-x^3-2x^2-x-1\] which has no changes in sign, so there is no negative real zero

Answer 10

this means there is either three real zeros, a positive one, or 1 real zero. there has to be once since is is a polynomial of odd degree

Answer 11

of course all the time we know what the actual zeros are, so we are just pretending

Answer 12

then i guess you can say that if there are any rational zeros, it has be be either \(2\) or \(-2\) and since it cannot be negative, it would have to be \(2\)

compute and see that \(f(2)=0\) and then to find the other zeros, factor as
\[(x-2)(x^2+1)\] as then solve \(x^2+1=0\) so get \(x=i\) or \(x=-i\)

Answer 13

now i guess you have to make up another example, but it should be more or less clear how to do it
cheat, by starting with a the polynomial in factored for so you already know the zeros, then multiply out and pretend you don't
of course if your teacher asked you how you did it, say "i wrote it in factored form and then multiplied"

Answer 14

okay thank you!

Answer 15

yw