SOLUTION: If one of the zeros of y=x^3+bx+1 is 1, determine the value of b. then solve x^3+bx+1=0 express your answer as exect values.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: If one of the zeros of y=x^3+bx+1 is 1, determine the value of b. then solve x^3+bx+1=0 express your answer as exect values.      Log On


   

Question 308790: If one of the zeros of y=x^3+bx+1 is 1, determine the value of b. then solve x^3+bx+1=0 express your answer as exect values.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
0=1%5E3%2Bb%281%29%2B1
b%2B2=0
b=-2
.
.
.
Then solve for
%28x%5E3-2x%2B1%29%2F%28x-1%29
The first multiplier would be x%5E2 and yield x%5E2%28x-1%29=x%5E3-x%5E2.
Subtracting from the polynomial,
%28x%5E3-2x%2B1%29-%28x%5E3-x%5E2%29=-x%5E2-2x%2B1
The second multiplier would be -x and yield -x%28x-1%29=-x%5E2%2Bx
%28-x%5E2-2x%2B1%29-%28-x%5E2%2Bx%29=-x%2B1
The final multiplier would be -1 and yield -1%28x-1%29=-x%2B1
%28-x%2B1%29-%28-x%2B1%29=0
So then,
%28x%5E3-2x%2B1%29%2F%28x-1%29=x%5E2-x-1
Use the quadratic formula to solve for these roots.
x+=+%28-%28-1%29+%2B-+sqrt%28+1%5E2-4%2A1%2A%28-1%29+%29%29%2F%282%2A1%29+
x+=+%281+%2B-+sqrt%28+5+%29%29%2F%282%29+
The three roots of x%5E3-2x%2B1 are then
(1, %281%2Bsqrt%285%29%29%2F2,%281-sqrt%285%29%29%2F2)