synthetic division for function F(x)=2x^3-x^2+2x-1

[ochocki]

For the function F(x)=2x^3-x^2+2x-1

I need to use Descartes rule of signs as well as the rational zero test. Through Descartes I find that there are no negative zeroes and 3 or 1 positive zero.

with the rational zero test I find that possible factors are + or - 1 & + or - 1/2

through synthetic division I found that 1 is not a factor, and since there are no negative zeroes, 1/2 must be a factor. The problem arises when I plug 1/2 into a synthetic division problem, it leaves a remainder for me. As far as I know for a number to be a factor it musnt leave a remainder. Can anyone spot what I am possibly doing wrong?

 

[stapel]

I need to use Descartes rule of signs as well as the rational zero test.

You need to use these to do what...? Are you factoring? Finding the real zeroes? The complex zeroes? Or something else?

Can anyone spot what I am possibly doing wrong?

Since you have described some of your work, rather than showing all of it, I'm afraid "spotting" anything will likely be impossible.

Please reply with the full and exact statement of the exercise, the complete instructions, and a clear listing of your steps.

Thank you.

Eliz.

 

[ochocki]

The directions were this:

Using Descartes rule of signs and the rational zero test, find the factors of the function stated above.

 

[stapel]

Thank you for providing the instructions. Now, in order for us to be able to check your work, please reply with that information as well.

Thank you.

Eliz.

 

[soroban]

Hello, ochocki!

Find the factors of: \(\displaystyle \:F(x)\:=\:2x^3\,-\,x^2\,+\,2x\,-\,1\)

\(\displaystyle \frac{1}{2}\) must be a factor.
When I plug \(\displaystyle \frac{1}{2}\) into a synthetic division problem, it leaves a remainder.

Can anyone spot what I am possibly doing wrong?

It's hard to see your work from here
but it looks like you transposed to \(\displaystyle B\flat\) minor instead of playing the \(\displaystyle J\diamond\).


How about just factoring it?

By grouping: \(\displaystyle \:x^2(2x\,-\,1)\,+\,(2x\,-\,1)\)

Factor again: \(\displaystyle \2x\,-\,1)(x^2\,+\,1)\)

 

[Mrspi]

Here's the synthetic division to test whether 1/2 is a root:

1/2  |  2    -1      2    -1
              1      0     1
        --------------------
        2     0      2     0

remainder is 0. The remaining coefficients give this reduced polynomial:

2x<SUP>2</SUP> + 2

The polynomial can be factored as

(x - (1/2))(2x<SUP>2</SUP> + 2)
or
2(x - (1/2))(x^2 + 1)

or,
(2x - 1)(x^2 + 1)

But Sorobano's method is "neater"......

 

[ochocki]

Thanks everyone, i see what I did. I apologize for not typing it up completely. It was a simple clerical thing, I checked it numerous time, funny how that happens.