SOLUTION: Given x^3-4x^2+2x+1=0 A. How many possible negative roots are there? B. Using synthetic substitution, which of the possible rational roots is actually a root of the equation?

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Question 176615: Given x^3-4x^2+2x+1=0
A. How many possible negative roots are there?
B. Using synthetic substitution, which of the possible rational roots is actually a root of the equation?
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
A) you must use f(-x) to find out the number of negative root possibles....f(-x)=-x^3-4x^2-2x+1=0
number of negative real roots is less than or equal to the number of variations in the function f (- x).
:
there is 1 variation in f(-x) therefore there is one negative real root
:
1 or -1 could be roots because they are factors of the constant which is 1
:x%5E3-4x%5E2%2B2x%2B1=0
1%5E3-4%281%29%5E2%2B2%281%29%2B1=0 therefore 1 is a root
%28-1%29%5E3-4%28-1%29%5E2%2B2%28-1%29%2B1=-8 therefore -1 is NOT a root
:
This relationship is always true: If a polynomial has rational roots, then those roots will be fractions of the form (plus-or-minus) (factor of the constant term) / (factor of the leading coefficient). However, not all fractions of this form are necessarily zeroes of the polynomial. Indeed, it may happen that none of the fractions so formed is actually a zero of the polynomial.

since 1 is our constant and 1 is our leading coefficient the only possible rational roots are 1 divided by 1 or -1 which is the same as those terms we have already tested.
:
now synthetic division using the root 1 (x-1)
:
 
   |
  1|1 -4  2   1
   |___1_-3__-1___________________
    1 -3 -1   0
:

so when we factor out the root 1 we get
:
x%5E2-3x-1=0
:and using the quadratic formula we get 2 irrational roots from this depressed equation
3%2F2%2Bsqrt%2813%29%2F2 and 3%2F2-sqrt%2813%29%2F2
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2Cx%5E3-4x%5E2%2B2x%2B1%29
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-3x%2B-1+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-3%29%5E2-4%2A1%2A-1=13.

Discriminant d=13 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--3%2B-sqrt%28+13+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-3%29%2Bsqrt%28+13+%29%29%2F2%5C1+=+3.30277563773199
x%5B2%5D+=+%28-%28-3%29-sqrt%28+13+%29%29%2F2%5C1+=+-0.302775637731995

Quadratic expression 1x%5E2%2B-3x%2B-1 can be factored:
1x%5E2%2B-3x%2B-1+=+%28x-3.30277563773199%29%2A%28x--0.302775637731995%29
Again, the answer is: 3.30277563773199, -0.302775637731995. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-3%2Ax%2B-1+%29