You can
put this solution on YOUR website! Hi, I need some help with my American School Algebra 2 work.
Given x^3-4x^2+2x+1=0
A. How many possible positive roots are there?
Using Decartes' rule, there are either 2 or 0 positive roots.
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B. How many possible negative roots are there?
Descartes again, 1 negative root.
You can find Descartes on google.
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C. What are the possible rational roots?
+1 is a root. I found it by graphing the function.
Then dividing it out, you're left with
x^2-3x-1 = 0
| Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc) |
Quadratic equation  (in our case  ) has the following solutons:
For these solutions to exist, the discriminant  should not be a negative number.
First, we need to compute the discriminant :  .
Discriminant d=13 is greater than zero. That means that there are two solutions:  .
Quadratic expression  can be factored:
Again, the answer is: 3.30277563773199, -0.302775637731995.
Here's your graph:
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x = (3+sqrt(13))/2
x = (3-sqrt(13))/2
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D. Using synthetic substitution, which of the possible rational roots is actually a root of the equation?
E. Find the irrational roots of the equation(hint:ust the quadratic formula to solve the depressed equation.
