Given x^3-4x^2+2x+1=0?

Question

A. How many possible positive roots are there?
B. How many possible negative roots are there?
C. What are the possible rational roots?
D. Using synthetic substition, which of the possible rational roots is actually a root of the equation?

Merry Christmas!

oh and

E. Find the irrational roots of the equation. (Hint: use the quadratic formula to solve the depressed equation.)

I think that a is 3, b is 0, and C is 1, but i'm just not sure! thanks!

Answer 1

f(1) = 1 - 4 + 2 + 1 = 4 - 4 = 0
Thus x - 1 is a factor of f(x).
To find other factors use synthetic division:-

1 | 1__- 4___2___1
_ | ____1__- 3__- 1
_ |1__- 3__- 1___0

(x - 1) (x² - 3x - 1) = 0

Consider (x² - 3x - 1) = 0
x = [ 3 ± √ (9 + 4) ] / 2
x = [ 3 ± √ (13) ] / 2

Roots are x = 1 , x = [ 3 ± √ (13) ] / 2

To my mind, the important part has been done and questions A to D are add ons.
I will have a guess at:-

A 2
B 1
C x = 1
D 1

Merry Christmas
Have a good day

Answer 2

A) 2 positive roots
B) 1 negative root
C,D) 1 is the only rational root
E) (1/2)*(3+√13) and (1/2)*(3-√13) ≈ [ 3.303, -.303]

Answer 3

(x-1)(x^2-3x-1)
there are 3 roots
all of them are rational roots since b^2-4ac>0
they are 1, 3+-sqrt13 all over 2.

Answer 4

= (x-1)(x^2-3x-1)=0

. (x-1)=0
x1=1
.x^2-3x-1=0
(x-x2)(x-x3)=0
x^2 -(x2+x3) +x2x3=0
x^2-3x-1=0
........................
x2+x3=3 and x2x3= -1
..........................