Question

can anyone please explain to me how to find a polynomial of degree three with the given numbers as zeros? -6,-√ 3,√ 3

Answer 1

(you do not have to multiply out)

Answer 2

'the given numbers as zeros? -6,-√ 3,√ 3 ' what do u mean by this?

Answer 3

are these the roots of the polynomial?
-6,-√ 3,√ 3

Answer 4

yes I believe that's what it means, that's just how it was worded

Answer 5

(x+6)(x+√ 3)(x-√ 3)

Answer 6

Just put them in like that. You can multiply them out if you so wish.

Answer 7

Well, that would be the polynomial if you multiplied it out.

Answer 8

(x+√ 3)(x-√ 3)
x^2-x√ 3+x√ 3-3
(X+6)(X^2-3)
X^3-3x+6x^2-18
\[x^3+6x^2-3x-18\]

Answer 9

@Jacquelinenorth
did that kid 'explain' it so you understand...?

Answer 10

I think so...I'm trying the steps right now by myself

Answer 11

k

Answer 12

okay I understand this one but my next one has i and -i in it

Answer 13

Basically, if a zeros are -5 and 2, it would be (x+5)(x-2)
Therefore, it is (x+√ 3)(x-√ 3)(x+6)
The first is for the -√3 zero, the second is for the √3 zero, and the third is for the -6 zero.
If you multiply these three out, they give you the full equation of the polynomial.

Answer 14

Oh. Well. Grr. Those would theoretically be graphed in complex coordinate planes, because they involve non-real numbers.

Answer 15

oh....yeah....all that stuff you said

Answer 16

Well, anyways, hit me with your best shot. I'll try.

Answer 17

okay it says: find a polynomial of degree four with the given numbers as zeros:
i,-i,-2+√ 5,-2-√ 5

Answer 18

so I would have (x+i) (x-i) (x-2+√ 5)(x+2-√ 5)?

Answer 19

I'm not entirely certain that it would be x. Because once we switch over to the other coordinate plane it is now in the (a,b) format, not (x,y) which complicates things.

Answer 20

oh geez louise

Answer 21

Actually.

Answer 22

If \[i=\sqrt-1\]
then
\[i^2=\sqrt-1 \times \sqrt -1\]
\[=-1\]

Answer 23

Hold on, let me try.

Answer 24

I think I murdered you with math on accident

Answer 25

(x+i) (x-i)
x^2-xi+xi-sqrt(-1^2)
\[x^2-x*i+x*i-\sqrt(-1)^2\]
So I was correct.
X^2--1
x^2+1=(x+i)(x-i)
Now, all we have to do is multiply out the other two factors, which are (x-2+√ 5)(x+2-√ 5)
X^2+2x-x√ 5-2x-4+2√ 5+x√ 5+2√ 5-5
\[x^2+2x-2x-x√ 5+x√ 5+2√ 5+2√ 5-5-4\]
Now this is really nasty looking.
\[x^2-x√ 5+x√ 5+4√ 5-5-4\]
Sweet Jesus
Combine them nasty like terms now.
\[x^2+4√ 5-9\]
Oh my effing god I'm not even done yet, where is my gun because I'm dying over here.

Answer 26

Yes you did
I'm dying over here thank you ;)

Answer 27

I'm sorry!!!! do not die without finishing this equation haha

Answer 28

\[(x^2+1)(x^2+4√ 5-9)\]
x^4+4√ 5x^2-9x^2+x^2+4√ 5-9
\[x^4+(4√ 5)x^2 -8x^2+4√ 5-9\]
I don't really even know how to get rid of that √ 5 without having the redo every single bit of this, and quite frankly I don't think I need to do that all over again. That was seriously one of the most mind-intensive things I've ever done. Even though it isn't really anything that new, it took all over my attention and all of my effort just to do that. O.o and it's probably wrong too but at this point it is what it is, I'm dyin' over here.

Answer 29

@halorazer haha well thanks for trying, I'll double check it with my instructor tomorrow. thanks for being so patient, I won't force you to do any more math. and sorry for the math murder

Answer 30

I don't have a problem with the math as long as it isn't that. ;o

Answer 31

deal!