Question

"Find a polynomial P(x) having real coefficients, with the degree and zeroes indicated.
(Assume the lead coefficient is 1.)"

No idea what it means by the "assume" part. I think what I have to do is basically turn these numbers into an equation...?

degree 4; x = -1, x = 1 + 2i

Answer 1

If x=-1 is a zero then x-(-1) is a factor of the polynomial
If x=1+2i is a zero then x-(1+2i) is a factor of the polynomial
Ok now we want real coefficients so what does that tell us another zero would be?

Answer 2

So, (x + 1) is one, but don't we have to change that 2i? On my other problems, it came out to (x^2 + 4) (of course, I'm sure I might doing it wrong...) But then, what happens to the 1?

Answer 3

Ok so if i tell you -2+2i is a zero, then -2-2i is also a zero.
Since (x-(-2+2i))(x-(-2-2i))=(x+2-2i)(x+2+2i)=([x+2]-2i)([x+2]+2i)
=(x+2)^2+2i(x+2)-2i(x+2)-4i^2=(x+2)^2-4(-1)=(x+2)^2+4 <see no imaginary parts anymore
that is what we want for this problem
So seeing this example fill in the below blank:
For this problem you say if 1+2i is a zero, then __???____ is also a zero

Answer 4

Another example:
If 2+i is a zero, then 2-i is also a zero
SINCE
(x-(2+i)(x-(2-i))=(x-2-i)(x-2+i)=([x-2]-i)([x-2]+i)=(x-2)^2+i(x-2)-i(x-2)-i^2=(x-2)^2-(-1)=(x-2)^2+1
see no imaginary parts anymore

Answer 5

Also I'm asking you to do is give the conjugate of 1+2i
Do you know what it is?

Answer 6

All* not also sorry

Answer 7

I'm sorry, but I can't follow you at all. I'm still trying to work out the string of equations you gave me.
Also, what do you mean by "conjugate"?

I'm sorry I'm being so difficult.

Answer 8

The conjugate of a+bi is a-bi
The conjugate of a-bi is a+bi
The conjugate of 2+i is 2-i
The conjugate of -2+2i is -2-2i
The conjugate of -2-2i is -2+2i
The conjugate of 2-i is 2+i
The conjugate of 41-3i is 41+3i
and so on...

Answer 9

Ah, so, if x = 1 + 2i, it also equals 1 - 2i. Is that it?
So...there's x = 1+ 2i, 1 - 2i, and -1. But since it's to the fourth degree, does this mean -1 repeats?

Answer 10

no 1+2i is not equal to 1-2i
I'm saying if one zero is x=1+2i then another zero is x=1-2i
Since
x=1+2i => x-(1+2i) is a factor
x=1-2i => x-(1-2i) is a factor
And when you multiply these you get real coefficients
like this
(x-(1+2i))(x-(1-2i))
(x-1-2i)(x-1+2i)
([x-1]-2i)([x-1]+2i)
[x-1]^2+2i[x-1]-2i[x-1]-4i^2
[x-1]^2+0-4(-1)
[x-1]^2-4(-1)
[x-1]^2+4
See we have real coefficients

Answer 11

"Equals" was bad word choice on my part, I meant to ask if it was another zero, like -1 and 1 + 2i. My bad.

I know it's taking me awhile. I'm trying to see if I can work this out better...

Answer 12

\[(x-(-1))(x-(1-2i))(x-(1+2i))(x-a)\]
I said x=a is another zero since it was a 4th degree polynomial
x=a is a zero => x-a is a factor of the polynomial
now we can clean this up a bit
we have
\[(x+1)((x-1)^2+4)(x-a)=(x+1)(x^2-2x+1+4)(x-a) \]
\[=(x+1)(x^2-2x+5)(x-a)\]
I will let you choose a and multiply the rest out

Answer 13

I think a=1 or a=0 is a good choice

Answer 14

but you can choose whatever

Answer 15

Though the question does not strictly rule out a fourth root, I would assume they want a repeated root x=-1 to exactly meet their criteria

Answer 16

Well then that would mean they should have also said x=1-2i is a zero

Answer 17

They did when they said all the coeffs are real

Answer 18

but they didn't say x=-1 multiplicity 2

Answer 19

so why does it have to be x=-1 mult. 2

Answer 20

no, but they did want a 4th degree, with the given (and implied) zeros

Answer 21

and it's not wrong to use a repeated root...

Answer 22

yeah but like i said they didn't list x=1-2i as a zero but the question implies that that should be a zero

Answer 23

but the question didn't say x=-1 mult. 2

Answer 24

it said find a polynomial
it says "a"
meaning that could be more than one answer

Answer 25

depending on what you let a equal

Answer 26

Okay, it took me awhile, but I think I got them all worked out...

\[(x+1)(x^2-2x+5)(x-1) = x^4 - 2x^3 + 4x^2 + 2x - 5 \]

\[(x+1)(x^2-2x+5)(x-0) = x^4 - x^3 + 3x^2 + 5x\]

\[(x+1)(x^2-2x+5)(x+1) = x^4 + 2x^2 + 8x + 5\]

Answer 27

lol so which one are you gonna go with?

Answer 28

phi was suggesting you go with the bottom one
it makes no difference to me based on the wording of the question

Answer 29

...I honestly have no idea. It looks like they could all be correct, but I have no way of knowing which one my professor will take.
My book actually does mention something about repeating roots, but it's pretty vague about it.

So, to play it safe: Turn in the one-question page sitting in front of me and hope for the best. Heh.

Thank you so much for helping me! I know it took forever, but I understand this a lot better now!

Answer 30

That is good :)