SOLUTION: Okay this is a really long problem that my math teacher gave me :-( and i'm stuck. The problem is: 2x^6+3x^5+8x^4+15-2+12x-8=0 1) How many solutions are there and why? *The

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: Okay this is a really long problem that my math teacher gave me :-( and i'm stuck. The problem is: 2x^6+3x^5+8x^4+15-2+12x-8=0 1) How many solutions are there and why? *The      Log On


   



Question 27213: Okay this is a really long problem that my math teacher gave me :-( and i'm stuck.
The problem is: 2x^6+3x^5+8x^4+15-2+12x-8=0
1) How many solutions are there and why?
*There are 6 because thats the highest degree.
2) How many real and imaginary answers are there?
* I think there are 3, 1, or 0 real positive and 3,1, or 0 real negative numbers... and i have no idea how many imaginary...
3) Solve for all 6 answers (and show wokr):
* I punched it into my calculator and figured out the first two answers (x-(1/2)) and (x+2)then i used each with synthetic division with the problem and broke it down to:
2x^4+10+8
Then I took out a 2
2(x^4+5+4)
Then i broke it down:
2(x^2+4)(x^2+1)
So thats what I have so far, (x-(1/2)) and (x+2) i only need 4 more but i have no idea where to go from here. Any help would be great! Thanks!

Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
I CAN SYMPATHISE WITH YOU.IT IS A BIG PROBLEM AND YOU REALLY TRIED WELL..I WOULD REALLY LIKE TO HELP YOU OUT BUT THERE ARE SOME OBVIOUS TYPING MISTAKES IN YOUR PROBLEM..I CANOT IMAGINE WHAT IS THE CORRECT PROBLEM AQND TRY TO HELP YOU ..I AM GIVING BELOW THE MISTAKES..CORRECT THEM AND I SHALL HELP YOU OUT..
SEE BELOW
Okay this is a really long problem that my math teacher gave me :-( and i'm stuck.
The problem is: 2x^6+3x^5+8x^4+15-2+12x-8=0 ...WHAT IS THIS 15-2?IS IT 15X^3 AND -2X^2?
1) How many solutions are there and why?
*There are 6 because thats the highest degree....OK
2) How many real and imaginary answers are there?
* I think there are 3, 1, or 0 real positive and 3,1, or 0 real negative ..HOW 0 ?AS LONG THERE IS -8 IN THE EQN.X=0 CANNOT BE A SOLUTION?
numbers... and i have no idea how many imaginary...
3) Solve for all 6 answers (and show wokr):
* I punched it into my calculator and figured out the first two answers (x-(1/2)) and (x+2)then i used each with synthetic division with the problem and broke it down to:
2x^4+10+8
Then I took out a 2
2(x^4+5+4)
Then i broke it down:
2(x^2+4)(x^2+1)
So thats what I have so far, (x-(1/2)) and (x+2) i only need 4 more but i have no idea where to go from here. Any help would be great! Thanks!