SOLUTION: 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? B. How many possible negative roots are there

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Question 216529: 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?
B. How many possible negative roots are there?
C. What are the possible rational roots?
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.
Answer by Alan3354(69443)   (Show Source): 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.
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