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

1. amistre64

1-5i + n = 0 1+5i +m = 0 n= -1+5i m= -1-5i (1-5i -1+5i) (1+5i -1-5i)

2. amistre64

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

3. amistre64

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

4. anonymous

? Ok.

5. anonymous

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

6. anonymous

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)??

7. anonymous

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)

8. anonymous

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.

9. anonymous

Well wait. Which part don't you understand?

10. anonymous

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.

11. anonymous

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

12. anonymous

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

13. anonymous

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?

14. anonymous

Did I lose you?

15. anonymous

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

16. anonymous

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

17. anonymous

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

18. anonymous

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

19. anonymous

$\rightarrow$ just means 'therefore'

20. anonymous

I wish I did...

21. anonymous

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

22. anonymous

0?

23. anonymous

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

24. anonymous

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

25. anonymous

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?

26. anonymous

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.

27. anonymous

Does that make sense?

28. anonymous

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

29. anonymous

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