SOLUTION: I actually have three questions. Two of the same type. Different numbers of roots, and I'm not even sure this is the right heading, but I couldn't find one that fit better, if you

Algebra ->  Trigonometry-basics -> SOLUTION: I actually have three questions. Two of the same type. Different numbers of roots, and I'm not even sure this is the right heading, but I couldn't find one that fit better, if you       Log On


   



Question 254301: I actually have three questions. Two of the same type. Different numbers of roots, and I'm not even sure this is the right heading, but I couldn't find one that fit better, if you could help that would be great. If they're the same thing in the end, one being solved is fine, I just need a better explanation than the notes I was given for my online class. :(. The third question has imaginary numbers in it, and after a failed, and very irritating attempt I thought I would seek assistance.
1. Determine all roots of P(x) = 3x^4+5x^3-9x^2-10x+8.
2. Determine all the roots of P(x) = 3x^5+5x^4+10x^3+14x^2+3x-3.
3. Determine a polynomial P(x) with real coefficients and of degree 5 for which three of its roots are x= -1, x = 1+2i and x= -3i.
Thank you!

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
1. Determine all roots of P(x) = 3x^4+5x^3-9x^2-10x+8.
I had to graph it to find zeroes:
x = 1.6371972...
x = -2.409123
The other two roots seem to be complex.
------------------------------------------
2. Determine all the roots of P(x) = 3x^5+5x^4+10x^3+14x^2+3x-3.
x = 1/3 is the only Real root
The other 4 roots seem to be complex.
---------------------------------------------
3. Determine a polynomial P(x) with real coefficients and of degree 5 for which three of its roots are x= -1, x = 1+2i and x= -3i.
----
Since the coefficients are Real x = 1-2i and x = +3i
are also roots
---------------------
P(x) = (x+1)((x-1)-2i)((x-1)+2i)(x-3i)(x+3i)
---
P(x) = (x+1)((x-1)^2+4)(x^2+9)
---
P(x) = (x+1)(x^2-2x+5)(x^2+9)
=====================================
Cheers,
Stan H.