Question

Find an nth degree polynomial function with real coefficients satisfying the conditions:
N=3;1and -2+3i are zeros;leading coefficients is 1

Answer 1

@satellite73

Answer 2

@iambatman

Answer 3

@Loser66

Answer 4

@FibonacciChick666

Answer 5

there are a couple methods for finding a polynomial of degree 2 with zero at \(-2+3i\)
a hard way,
an easy way
and a real real easy way
take your pick

Answer 6

go with sat here

Answer 7

the real real easy way requires just memorizing something
if the zeros are \(a+bi\) and \(a-bi\) then the quadratic is
\[x^2-2ax+(a^2+b^2)\]

Answer 8

the fairly easy way is to take
\[x=-2+3i\] and work bacwards
add \(2\)
\[x+2=3i\] square (carefully) to get rid of the \(i\)
\[(x+2)^2=(3i)^2\\
x^2+4x+4=-9\]

Answer 9

that makes the quadratic
\[x^2+4x+13\] by adding 9
would have gotten the same answer with
\[x^2-2\times (-2)x+(2^2+3^2)\]

Answer 10

Method

Answer 11

we have n =3 , right?

Answer 12

Yes

Answer 13

that is the polynomial has degree 3, right?

Answer 14

Yes

Answer 15

so that it is in the form ax^3 + bx^2 + cx + d =0
get it?

Answer 16

Yes

Answer 17

all that writing didn't help? dang

Answer 18

leading coefficient =1 means a =1,
oh, I am sorry @satellite73

Answer 19

satellite could u help me again and accepted me as your friends

Answer 20

@kristinak Let read satelite's guide

Answer 21

@satellite sorry I need a step by step interaction process

Answer 22

Okay @Loser66

Answer 23

What will be the conjugate of -2+3i

Answer 24

@satellite73

Answer 25

the conjugate of \(-2+3i\) is \(-2-3i\)

Answer 26

not sure what is going on here with this question
you want a polynomial with zero at \(-2+3i\) and \(1\)
the quadratic that has \(-2+3i\) as a zero is
\[x^2+4x+13\] so your final job is to multiply
\[(x-1)(x^2+4x+13)\] to get your polynomial of degree 3

Answer 27

When I set it upside it'll be (x-1)(x-2-3i)

Answer 28

oh no

Answer 29

you want a polynomial with real coefficients, of degree 3, not complex coefficients of degree 2

Answer 30

one factor will be \((x-1)\) for sure
but you also want to find the quadratic polynomial (degree 2) that has zero at \(-2+3i\) and \((-2-3i\)

Answer 31

No this is just to set it up

Answer 32

if you want ,(i wouldn't recommend it) you can write it as
\[(x-1)(x-(-3+2i))(x-(-3-2i))\]

Answer 33

if your goal is to really find the answer, a polynomial with real coefficients of degree 3, then your actual job is to multiply
\[(x-1)(x^2+4x+13)\]

Answer 34

That's how my teacher taught me

Answer 35

Buts what's a easier way to get to the answer

Answer 36

yeah i see the question
i tried to explain how to get the answer before

Answer 37

would you like me to try again?

Answer 38

Yes

Answer 39

ok here we go
you want a polynomial of degree 2 with a zero at \(-2+3i\) right?

Answer 40

No degree 3

Answer 41

ok lets go slow

Answer 42

one of the zeros is \(1\)
so one of the factors of your polynomial is \((x-1)\)
is that ok?

Answer 43

Yes

Answer 44

damn typo there
one of the zeros is \(-2+3i\) the other is its conjugate \(-2+3i\)
is that clear?

Answer 45

ack , conjugate \(-2-3i\) that is what i meant

Answer 46

so your next job is to find a polynomial of degree 2 with zeros at \(-2+3i\) and\(-2-3i\)

Answer 47

Okay I got that

Answer 48

lets to that

Answer 49

start by setting
\[x=-2+3i\] and work backwards
add \(2\) to both sides, get
\[x+2=3i\]

Answer 50

then to get rid of the \(i\) square both sides (carefully) and get
\[(x+2)^2=(3i)^2\\
x^2+4x+4=-9\]

Answer 51

add \(9\) and get
\[x^2+4x+13=0\] and that is your quadratic \(x^2+4x+13\)

Answer 52

then to get your polynomial of degree 3, multiply
\[(x-1)(x^2+4x+13)\]

Answer 53

you should get \[x^3+3 x^2+9 x-13\] which i hope is one of your answers

Answer 54

Let me try to do it on paper so I can show you what I get

Answer 55

@satellite73 I get x^3+3x^2+9x-13

Answer 56

yes

Answer 57

help me please