anonymous one year ago Write a polynomial with rational coefficients having roots 3, 3+i, and 3-i. Part 1: Write the factors (in the form x-a) that are associated with the roots (a) given in the problem. Part 2: Multiply the 2 factors with complex terms to produce a quadratic expression. Part 3: Multiply the quadratic expression you just found by the 1 remaining factor to find the resulting cubic polynomial.

1. anonymous

nah don' t do that

2. anonymous

$(x-(3+i))(x-(3-i))$ is a pain to multiply there is a much easier wy

3. anonymous

you want a quadratic with zeros $$3+i$$ and $$3-i$$ there is an easy way to find it also a real real easy way which would you like to use?

4. anonymous

Well I am pretty sure I can find the easier way, but the packet I'm doing requires me to do it the hard way, unfortunately. Can you help me with the harder way?

5. anonymous

what difference does it make? you have to end up with a quadratic, might as well just write it

6. anonymous

Well each of the steps is graded separately as a different question. So if I skip the first two steps (or do them differently) then I get points taken off.

7. anonymous

if the zeros are $$a\pm bi$$ the quardratic is $x^2-2ax+(a^2+b^2)$

8. anonymous

ok then lets multiply

9. anonymous

So how do I find my factors for this problem?

10. anonymous

if a root is $$r$$ then a factor is $$(x-r)$$

11. anonymous

your roots are 3, 3 + i, 3 - i so your factors are $(x-3)(x-(3+i))(x-(3-i))$

12. anonymous

Okay yes that's what I got

13. anonymous

that is the answer to A

14. anonymous

for B they want you to multiply $(x-(3+i))(x-(3-i))$

15. anonymous

So the complex roots only? I see.

16. anonymous

that is what is says right?

17. anonymous

Yep, thanks.

18. anonymous

Do I multiply this by FOIL?

19. anonymous

i wouldn't

20. anonymous

well ok i guess, but keep $$3+i$$ and $$3-i$$ together lets do the first outer inner last business, but to it the easy way

21. anonymous

so don't distribute the x to the roots?

22. anonymous

first is a no brainer, it is $$x^2$$

23. anonymous

lets skip to the last when you multiply a complex number $$a+bi$$ by its complex conjugate $$a-bi$$ you get the real number $$a^2+b^2$$ to the "last" is $3^2+1^2=10$

24. anonymous

Oh okay I've never seen that before.

25. anonymous

then "outer and inner"$-(3+i)x-(3-i)x$combine like terms, the $$i$$ part goes bye bye and you are left with $$-6x$$

26. anonymous

So did you distribute the x first, then subtract?

27. anonymous

if you haven't seen that then you should learn it now $(a+bi)(a-bi)=a^2+abi-abi-b^2i^2=a^2+b^2$

28. anonymous

yes i would call it "distribute then combine like terms" but it is the same thing

29. anonymous

Okay thanks. Okay and so the last part is to multiply the quadratic we found by x-3?

30. anonymous

final answer $x^2-6x+10$ as promised

31. anonymous

yes i will let you do that yourself, it is easy enough right?

32. anonymous

Yeah it'll take me a while because I'm rusty but I can do it. Thank you so much!

33. anonymous

yw