x^3 - 4x^2 + 2x + 1 = 0 solving for roots

Question

using x^3 - 4x^2 + 2x + 1 = 0

 

1.  how many possible positive roots are there

 

2.  how many possible negative roots are there

 

3.  what are the possilbe rational roots

 

4.  using synthetic subtitution, which of the possible rational roots is actually a root of the equation

 

5.  find the irrational roots of the equation using the quadratic formula to solve the depressed equation.

Answer 1

The product of three roots equals to the reciprocal of the free coefficient, that is coefficient without x, in this case, 1. 

 

x₁x₂x₃=-1. Since product of three numbers is negative when either all of them are negative or two of them are positive, we can have either 2 positive roots or no positive roots. Hence the answer to part 1 is 0 or 2.

 

Answer to 2 is 1 or 3.

 

Possible rational roots are -1 or 1, since the product of roots is -1.

 

4. x=1 is the root, as seen by the substitution:

 

1³-4*1²+2*1+1=1-4+2+1=0--true identity. x=-1 is not a root at all.

 

5. Divide x³-4x²+2x+1 by (x-1) to obtain:

 

x²-3x-1;

 

Solve x²-3x-1=0 using quadratic formula:

 

x₁=[3-√(9+4)]/2

x₂=[3+√(9+4)]/2

 

Upon simplification, we obtain:

 

x₁=(3-√13)/2

x₂=(3+√13)/2