SOLUTION: Given x^3 - 4x^2 + 2x + 1 =0. b. How many possible negative roots are there? c. What are the possible rational roots? d. Using synthetic substitution, which of the possible ra

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Given x^3 - 4x^2 + 2x + 1 =0. b. How many possible negative roots are there? c. What are the possible rational roots? d. Using synthetic substitution, which of the possible ra      Log On


   

Question 176377: Given x^3 - 4x^2 + 2x + 1 =0.
b. How many possible negative roots are there?
c. What are the possible rational roots?
d. Using synthetic substitution, which of the possible rational toots is actually a root of the equation?
e. Find the irrational roots of the equation. (Hint: Use the quadratic formula to the the depressed equation.)
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
If the coefficients of a polynomial are integers, it is natural to look for roots which are also integers. Any such root must divide the constant term.
:
Given+x%5E3+-+4x%5E2+%2B+2x+%2B+1+=0.
:
our constant term is 1 ...so 1 and -1 are the only possibles
:
lets try 1:%281%29%5E3-4%281%29%5E2%2B2%281%29%2B1=0so 1 is a factor
:
lets try -1:%28-1%29%5E3-4%28-1%29%5E2%2B2%28-1%29%2B1=-6so -1 is not a factor
:
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.
:
 
   |
  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%2Bsqrt%2813%29%2F2 and 3-sqrt%2813%29%2F2
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