# Thread: Factor 2x^3 - 3x + 1

1. ## Factor 2x^3 - 3x + 1

Please show me how to factor 2x^3 - 3x + 1

2. ## Re: Factor 2x^3 - 3x + 1

Please show me how to factor 2x^3 - 3x + 1

It is easy to see that expression =0 when x=1 It follows that x-1 is a factor. The 2nd bracket then must begin with 2x^2 and end in -1
SO 2x^3-3x+1= (x-1)(2x^2 _1) Multiplying these out as they are would give a term -2x^2 which we dont want. Putting =2x in the middle of the 2nd bracket will get rid of this. So 2x^3-3x+1=(x-1)(2x^2+2x-1) Notice this also gives us the required -3x term.

3. ## Re: Factor 2x^3 - 3x + 1

Thank you, but I was wanting to break it into three roots. The term being factored is a derivative for which I am trying to find the critical values. I got as far as you did but was not able to break it down further. Feel free to speak in calculus terms; what do I do next?

4. ## Re: Factor 2x^3 - 3x + 1

Please show me how to factor 2x^3 - 3x + 1

Following on from my earlier post the 2nd bracket breaks down to (2x+1)(x-1) so final answer is
2x^3-3x+1= (x-1)(2x+1)(x-1)=(2x+1)(x-1)^2

5. ## Re: Factor 2x^3 - 3x + 1

How did you get that 2x^2+2x-1 breaks down to (2x+1)(x-1) ? When I multiply them together I reach 2x^2-x-1, not 2x^2+2x-1

6. ## Re: Factor 2x^3 - 3x + 1

You will have to use the Quadratic formula, to find the factors, because 2x^2+2x-1 cannot be factored into integers.

7. ## Re: Factor 2x^3 - 3x + 1

Please show me how to factor 2x^3 - 3x + 1
$\displaystyle \left( {x - 1} \right)\left( {x - \left[ {\frac{{ - 1}}{2} + \frac{{\sqrt 3 }}{2}} \right]} \right)\left( {x - \left[ {\frac{{ - 1}}{2} - \frac{{\sqrt 3 }}{2}} \right]} \right)$

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# factor 2x^3 3x-6

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