Question

Form a polynomial f(x) with real coefficients having the given degrees and zeros.
a.) Degree 5; Zeros: -6,-i,-9+1
b.) Degree 4; Zeros: 5; multiplicity: 2; 2i
c.)Degree 4; zeros: 5-5i; -1; multiplicity 2

Answer 1

@satellite73 Could you help me with this problem please?

Answer 2

sure

Answer 3

which one first /

Answer 4

the first one please

Answer 5

.) Degree 5; Zeros: -6,-i,-9+1 i assume there is a typo

Answer 6

degree 5 zeros \(-6,-i,-9+i\) is my guess

Answer 7

am i right?

Answer 8

let me see sorry one sec

Answer 9

yeah you're right

Answer 10

ok since one zero is \(-6\) one factor is \(x+6\)

Answer 11

since one zero is \(-i\) the other one is its conjugate \(i\) and the factors are \[(x+i)(x-i)=x^2+1\]

Answer 12

the hardest part if finding a quadratic with zeros \(-9+i\) and \(-9-i\) but there are 3 ways to do it
one is hard, one is easy and one is really really easy

Answer 13

lets do the easy way first
work backwards starting with
\[x=-9+i\] add \(9\) to get
\[x+9=i\] square (carefully) and get
\[(x+9)^2=i^2\\
x^2+18x+81=-1\] and finally
\[x^2+18x+19\]is your quadratic

Answer 14

lots of typos here
let me fix them

Answer 15

haha ok

Answer 16

quadratic is
\[x^2+18x+82\]

Answer 17

on account of \(81+1=82\) not \(19\)
so you have to multiply out
\[(x+6)(x^2+1)(x^2+18x+82)\] i would cheat

Answer 18

http://www.wolframalpha.com/input/?i=%28x%2B6%29%28x^2%2B1%29%28x^2%2B18x%2B82%29
you can check that this is right by looking at the zeros

Answer 19

now we can find the quadratic the real real easy way if you like, when we do C

Answer 20

okayy

Answer 21

want to do B first?

Answer 22

it is the easiest of the three

Answer 23

sure!

Answer 24

Degree 4; Zeros: 5; multiplicity: 2; 2i

Answer 25

one factor is \(x-5\) but since the "multiplicity" is 2 that means the factor is \((x-5)^2\)

Answer 26

the other two are \(2i\) and \(-2i\) so the quadratic is
\[(x+2i)(x-2i)=x^2+4\] and you only have to multiply out
\[(x-5)^2(x^2+4)\]

Answer 27

c.)Degree 4; zeros: 5-5i; -1; multiplicity 2

Answer 28

\(-1\) with multiplicity 2 means one factor is \((x+1)^2\)

Answer 29

now lets find the quadratic with zeros at \(5-5i\) and \(5+5i\) the real easy way
it requires memorizing something
if the zero is \(a+bi\) then the quadratic is
\[x^2-2ax+a^2+b^2\] in your case \(a=5,b=5\) so the quadratic is
\[x^2-2\times 5x+5^2+5^2\] or
\[x^2-10x+50\]

Answer 30

final job is
\[(x+1)^2(x^2-10x+50)\]

Answer 31

it says it's wrong?

Answer 32

wait nvm sorry

Answer 33

Thank you so much! You really helped me a lot! :) @satellite73

Answer 34

yw

Answer 35

do you think you have time to help me with one more problem? Well it's one problem but it has like 3 sub problems to it

Answer 36

go ahead and post in a new thread, i will take a look

Answer 37

okay thankyou!