# x^3-4x^2+2x+1=0 how many postive roots are there? how many negative? What are the possible rational roots

*print*Print*list*Cite

f(x)= x^3 - 4x^2 + 2x + 1 = 0

Let us try an substitute x=1

==> 1-4 + 2 + 1 = 4-4=0

Then 1 is a root , then (x-1) is a factor for f(x)

==> f(x) = (x-1)*Q(x)

==> Q(x) = f(X)/(X-1)= (x^3 -4x^2 + 2x + 1)/(x-1)

= x^2 -3x -1

==> F(X) = (x-1)(x^2 -3x -1) = 0

==> x1= 1 (positive rational)

==> x2= (3+sqrt(13))/2 (positive irrational)

==> x3= (3-sqrt13)/2 (negative irrational)

Observe that the sum of the coefficients are zero.f(1) = 0. So x-1 is a factor.

Therefore, x^3-4x^2+2x+1 = (x-1){x^2+kx-1}. Equating the coefficients of the like powers on both sides,

X^3 : 1 on both sides.

x^2: -4 = k-1. So k = -4+1 = -3

x: 2 = -1-k, For k =-3

constant : 1 on both sides.

Therefore x^3-4x^2+2x+1 = (x-1) (x^2-3x-1)

x^2-3x+1 = 0 has roots: x = {-3+sqrt((-3)^2 - 4*1*(-1))}/2

= {-3 +sqrt(13)}/2 Or {-3-sqrt13}/2.

So there are two positive roots: 1 and (-3+sqrt13)}/2 and -{3+sqrt13}/2 is the single negative root.

Only 1 is the rational root. The other two roots are irrational