# Thread: evaluating a cubic equation.

1. ## evaluating a cubic equation.

The question is as follows:

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.(Use the quadratic formula to solve the depressed equation.

A. 1
B. 1
C. $\frac{1}{1}, -\frac{1}{1}$
D. Through application of synthetic substitution I determined that 1 is the only rational root.
E. This is where I have problems...
How do I derive this depressed equation? I remember reading that the product of synthetic substitution is a depressed equation. But I'm coming up with strange answers. Where am I going wrong?
I can provide working on request, but it is quite time consuming to do on my evo so I won't put it up immediately.

Thank you!

2. Originally Posted by quikwerk
The question is as follows:

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.(Use the quadratic formula to solve the depressed equation.

A. 1
B. 1
C. $\frac{1}{1}, -\frac{1}{1}$
D. Through application of synthetic substitution I determined that 1 is the only rational root.
E. This is where I have problems...
How do I derive this depressed equation? I remember reading that the product of synthetic substitution is a depressed equation. But I'm coming up with strange answers. Where am I going wrong?
I can provide working on request, but it is quite time consuming to do on my evo so I won't put it up immediately.

Thank you!
A is wrong. You need to use Descarte's Rule of Sign.

x^3 is positive, squared -, x +, and constant +

+ - + +

From + to -, 1 change, and from - to positive, another change.

We have two sign changes so (+) = 2 or 0.

For c, you can just write $\pm 1$. You don't need them in fraction form.

What did you get from your synthetic division?

3. Originally Posted by dwsmith
A is wrong. You need to use Descarte's Rule of Sign.

x^3 is positive, squared -, x +, and constant +

+ - + +

From + to -, 1 change, and from - to positive, another change.

We have two sign changes so (+) = 2 or 0.

For c, you can just write $\pm 1$. You don't need them in fraction form.

What did you get from your synthetic division?
Before I post up my working, I have a question.
Wouldn't "a" being wrong make "c" wrong as well?

4. Originally Posted by quikwerk
Before I post up my working, I have a question.
Wouldn't "a" being wrong make "c" wrong as well?
No.

5. Originally Posted by quikwerk
The question is as follows:

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.(Use the quadratic formula to solve the depressed equation.

A. 1
B. 1
C. $\frac{1}{1}, -\frac{1}{1}$
D. Through application of synthetic substitution I determined that 1 is the only rational root.
E. This is where I have problems...
How do I derive this depressed equation? I remember reading that the product of synthetic substitution is a depressed equation. But I'm coming up with strange answers. Where am I going wrong?
I can provide working on request, but it is quite time consuming to do on my evo so I won't put it up immediately.

Thank you!
You will have established that $x=1$ is a rational root and from synthetic division that:

$x^3-4x^2+2x+1=(x-1)(x^2-3x-1)=0$

The quadratic in the middle set equal 0 is the depressed equation:

$x^2-3x-1=0$

CB