Question

Find a polynomail equation with real coefficients that has the given zeros.
1-5i and 1+5i

Answer 1

1-5i + n = 0

1+5i +m = 0

n= -1+5i
m= -1-5i

(1-5i -1+5i) (1+5i -1-5i)

Answer 2

1 +5i -1 -5i
-5i -25i^2 +5i +25i^2
-1 -5i +1 +5i
5i +25i^2 -5i -25i^2

Answer 3

guess it aint gonna work out like i thought :) gotta head to marketing class now; Ciao :)

Answer 4

? Ok.

Answer 5

\[ f(x) = (x-(1+5i))(x-(1-5i)) \]
\[ = x^2 - x(1-5i) -x(1+5i) + (1+5i)(1-5i)\]
\[ = x^2 -x +5xi - x - 5xi + 1 -5i + 5i - 25i^2\]
\[ = x^2 - 2x + 1 + 25 \]
\[ = x^2 - 2x + 26 \]
\[ \rightarrow f(1-5i) = 0, f(1+5i) = 0\]

Answer 6

When you are doing this problem: I understand the first equation is just subtracting from x
then the secong equation you are distributing... why do you add the additonal (1+5i)(1-5i)??

Answer 7

It's all the same equation. I didn't add the (1+5i)(1-5i). That term is the result of multiplying the last term in each factor..
\[ (a - b)(a - c) = a^2 -ac -ab +bc\]
Where b in this case is (1+5i) and c is (1-5i)

Answer 8

Ok..I dont understand. But thank you so much for you help. I have the answer just not sure how I got it... thanks again.

Answer 9

Well wait. Which part don't you understand?

Answer 10

It doesn't do any good to have the answer and not understand how you got it. Otherwise you won't be able to use this tool on more complicated tasks.

Answer 11

Do you agree that
\[ (x-a)(x-b) = x^2 - ax -bx +ab \]

Answer 12

Yes I agree. That would be just simply distrubtng....

Answer 13

Right. Do you also agree that if x=a or x=b then the product (x-a)(x-b) would be 0 because one of the factors would be 0?

Answer 14

Did I lose you?

Answer 15

Kinda.. because I dont know where the zero comes from??

Answer 16

\[ If\ x = a, then\ (x-a) = (a-a) = 0 \rightarrow (x-a)(x-b) = 0*(a-b) = 0\]

Answer 17

Thanks for your help.. I dont want to waste anymore of your time...

Answer 18

... You're not wasting my time. Did you understand my last sentence?

Answer 19

\[ \rightarrow \] just means 'therefore'

Answer 20

I wish I did...

Answer 21

Let x = a. Then (x-a) = (a-a). (a-a) = ?

Answer 22

0?

Answer 23

Yes. a - a = 0. No matter what 'a' is.

Answer 24

Ok.. so now that I know that.. what signifigance does that have in my problem?

Answer 25

So what we're saying is that when x = a, (x-a) = 0. If (x-a)=0 then (x-a)(x-b) would be 0*(x-b) which is?

Answer 26

The point of the problem is to find a polynomial that has 2 roots. That means that there are two different values for x at which the polynomial will be 0. We construct such a polynomial by writing an expression which has two factors

(x-a)(x-b)

Where a and b are the two roots (0's). When x is equal to either a or b, the whole expression will be equal to 0.

Answer 27

Does that make sense?

Answer 28

Thanks for your help. But I really need to go. Im not getting it. But appreciate your help so much...

Answer 29

Give it some thought. And try to follow through what I've written. If you have questions, feel free to come back and ask.