Question

PLEASE HELP!! SHOW ALL WORK
Write a polynomial function with rational coefficients so that P(x)=0 has the given roots.
1. 3i2
2. 1,2,-i
3.2+squareroot ,3
4. 2,4+i
THANKS A LOT!!!!

Answer 1

what is the first one?

Answer 2

just \(3i\) ?

Answer 3

ok that is the easiest of the bunch
if \(3i\) is a zero, then so is \(-3i\) and in factored form the polynomial will be
\[P(x)=(x+3i)(x-3i)\]

Answer 4

when you multiply that out you get
\[P(x)=x^2+9\] which you might have guessed at to begin with since if you go to solve
\[x^2+9=0\] you get
\[x^2+9=0\\
x^2=-9\\
x=\pm 3i\]

Answer 5

\(1,2,-i\) is similar
if \(-i\) is a zero then so is \(i\) so you have in factored form
\[P(x)=(x-1)(x-2)(x-i)(x+i)\] and you multiply that mess out

Answer 6

not as bad as it looks, since \((x-i)(x+1)=x^2+1\) so your real job is to try and multiply
\[(x-1)(x-2)(x^2+1)\] with out making an algebra mistake
i would cheat

Answer 7

how do you do it? @satellite73

Answer 8

how do i multiply it out?

Answer 9

can you tell me the answers to all the questions too? plzzz @satellite73

Answer 10

not sure what the question is
is it how do i multiply out \((x-1)(x-2)(x^2+1)\) ? besides cheating, i would start with
\[(x-1)(x-2)=x^2-3x+2\] and then multiply
\[(x^2-3x+2)(x^2+1)\] using six multiplications

Answer 11

can you do that or not?

Answer 12

what about the others?

Answer 13

no im confused and i need to bring the grade up tomoorow is the last day to turn it in

Answer 14

is the third one \(2+\sqrt3\) ?

Answer 15

i really need a good grade i have an F :(

Answer 16

you don't want an F that is for sure

Answer 17

[2+sqrt{5} 3

Answer 18

yah i dont thats why i need to get this done so atleast it goes to a D or a C

Answer 19

ok lets take is slow and finish the first one
if you want to multiply \((x-1)(x-2)(x^2+1)\) and make sure you don't get a mistake (which is also making sure you don't actually learn how to do it, but whatever) use this
http://www.wolframalpha.com/input/?i=%28x-1%29%28x-2%29%28x^2%2B1%29

Answer 20

the second one i take it is \(2+\sqrt5,3\) is that right?

Answer 21

yes

Answer 22

that site is so confusing

Answer 23

then if \(2+\sqrt5\) is a zero, so is \(2-\sqrt5\)
oh no the answer is right there
right where it says "alternate form"
do you see it?

Answer 24

you get two graphs, and right underneath it says "alternate form" which means they multiplied it out for you and you see
\[x^4-3 x^3+3 x^2-3 x+2\]which is your answer

Answer 25

ready for the one with zeros \(2+\sqrt5,2-\sqrt5,3\) ?

Answer 26

what?

Answer 27

number 3
it says the zeros are \(2+\sqrt5\) and \(3\) right?

Answer 28

yes

Answer 29

ok so you know one factor is \((x-3)\) since \(3\) is a zero
your hard job is to come up with the quadratic whose zero is \(2+\sqrt5\) but there is an easy way to do this
do you know it?

Answer 30

no

Answer 31

set \[x=2+\sqrt5\] then subtract \(2\) from both sides and get
\[x-2=\sqrt5\]
then square both sides
you get
\[(x-2)^2=5\]or
\[x^2-4x+4=5\] and finally subtract \(5\) from both sides to get
\[x^2-4x-1=0\] so your quadratic is
\[x^2-4x-1\] and you final job is to multiply
\[(x^2-4x-1)(x-3)\]

Answer 32

how to multiply?

Answer 33

take \(x^2-4x-1\) and multiply each term by \(x\)
then multiply each term by \(-3\)
then combine like terms

Answer 34

you know how to do that?

Answer 35

so than what is the answers?

Answer 36

if you just want the answer, do this
http://www.wolframalpha.com/input/?i=%28x^2-4x-1%29%28x-3%29
and read where it says "alternate form"

Answer 37

there is no alternate form for number 4

Answer 38

lol we didn't do number 4 yet
that is \(2,4+i\) right?

Answer 39

yes

Answer 40

ok for this one set
\[x=4+i\] subtract 4 get
\[x-4=i\] square and get
\[x^2-8x+16=-1\] add \(1\) to get
\[x^2-8x+17\] then multiply
\[(x^2-8x+17)(x-2)\]

Answer 41

u know i just want to get this done and its getting late and i have to sleep and i really appreciate u helping me with this! ur the only one THANKS

Answer 42

alternate form says \(x^3-10 x^2+33 x-34\) for that one
http://www.wolframalpha.com/input/?i=%28x^2-8x%2B17%29%28x-2%29

you got them all now?

Answer 43

yw

Answer 44

thanks alot ur great and im saying it in a good way not a bad way

Answer 45

thanks
my pleasure
hope you have all the right answers and don't get an F

Answer 46

there is something more if u can help?

Answer 47

sure go ahead and ask

Answer 48

Find the real and/or imaginary solution of each polynomial equation
1. \[4x^3+4=0\]
2. x^4-x^2-72=0
3.8x^3=-1
4. x^3=64
5.8x^3+27=0
6.2x^4+16x^2=40
7.2x^4-x^2=3
8.x^4-7x^2-8=0

Answer 49

first one
divide by 4 and start with
\[x^3+1=0\] then factor as
\[(x+1)(x^2-2x+1)=0\]so \(x=-1\) or \(x^2-2x+1=0\) which does not factor, so you need to use the quadratic formula

Answer 50

unless it was \(4x^3+4x=0\) but i assume what you wrote was correct

Answer 51

i made a typo there, it should be \((x+1)(x^2-x+1)=0\)

Answer 52

there is no x after the second 4

Answer 53

the zeros of \(x^2-x+1\) are \(\frac{1\pm\sqrt{3} i}{2}\)

Answer 54

so you have 3 zeros:
\(-1,\frac{1+\sqrt{3}i}{2},\frac{1-\sqrt{3}i}{2}\)

Answer 55

second one factors
\[x^4-x^2-72=(x^2-9)(x^2+8)\]

Answer 56

set \(x^2-9=0\) get \((x-3)(x+3)=0\) so \(x=3\) or \(x=-3\)

Answer 57

set \(x^2+8=0\) get \(x^2=-8\) so \(x=\pm\sqrt{8}i\)

Answer 58

or if you prefer \(x=\pm2\sqrt{2}i\)

Answer 59

\(8x^3=-1\) gives \[x^3=-\frac{1}{8}\]so \(x=-\frac{1}{2}\)

Answer 60

oh you need the complex zeros too?

Answer 61

idk it doesnt say