SOLUTION: Please help me! I have to find the remaing roots (zeros) but I'm confused. The problem is f(x)=x^3+2x^2-3x+20. One root is there and it is -4. I have to find two more roots (ze

Algebra ->  Rational-functions -> SOLUTION: Please help me! I have to find the remaing roots (zeros) but I'm confused. The problem is f(x)=x^3+2x^2-3x+20. One root is there and it is -4. I have to find two more roots (ze      Log On


   



Question 30958: Please help me! I have to find the remaing roots (zeros) but I'm confused. The problem is f(x)=x^3+2x^2-3x+20. One root is there and it is -4. I have to find two more roots (zeros). I tried to work it out using synthetic division:
4/ 1 , 2 , -3 , 20
____4 , 24 , 84
= 1 , 6 , 21 , 104
The problem is really confusing when I type it out but I hope you can help me...

Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
Please help me! I have to find the remaing roots (zeros) but I'm confused. The problem is f(x)=x^3+2x^2-3x+20. One root is there and it is -4.
OK ..GO AHEAD..YOU SHOULD DIVIDE WITH -4 ..NOT +4....
I have to find two more roots (zeros). I tried to work it out using synthetic division:
-4|......... 1.......... , 2......... , -3.......... , 20
............0.............-4.............8............-20
--------------------------------------------------------------------------____
.............1............-2.............5.............0
ANSWER IS
X^2-2X+5=THIS HAS IMAGINARY ROOTS.AS DISCRIMINANT IS NEGATIVE...USE QUADRATIC FORMULA...
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%282+%2B-+sqrt%28+%28-2%29%5E2-4%2A1%2A5+%29%29%2F%282%2A1%29+
x+=+%282+%2B-+sqrt%28+-16+%29%29%2F%282%29+
x+=+%281+%2B-+%282%2Ai%29%29+