Polynomial Question: 4x^3 - 9x^2 +6x -1

debased

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So I ran into this polynomial while doing some optimization for Calculus: 4x^3 - 9x^2 +6x -1 and I've had no trouble with Algebra, factoring, and so on until I realized that this thing can't be factored by grouping, and has imaginary roots when you get it down to a quadratic. When I punch it into WolframAlpha, it somehow converts it into (x-1)^2(4x-1) and I'm not sure how you get there. I feel like I'm missing some really basic Algebra thing that I've forgotten, any tips?
 
So I ran into this polynomial while doing some optimization for Calculus: 4x^3 - 9x^2 +6x -1 and I've had no trouble with Algebra, factoring, and so on until I realized that this thing can't be factored by grouping, and has imaginary roots when you get it down to a quadratic. When I punch it into WolframAlpha, it somehow converts it into (x-1)^2(4x-1) and I'm not sure how you get there. I feel like I'm missing some really basic Algebra thing that I've forgotten, any tips?
Have you covered the Rational Root Theorem?

Otherwise you have a cubic. Thus we know it factors as \(\displaystyle (x^2 + ax + b)(x + c)\) for some a, b, and c. Multiply it out and solve for a, b, and c.

-Dan
 
So I ran into this polynomial while doing some optimization for Calculus: 4x^3 - 9x^2 +6x -1 and I've had no trouble with Algebra, factoring, and so on until I realized that this thing can't be factored by grouping, and has imaginary roots when you get it down to a quadratic. When I punch it into WolframAlpha, it somehow converts it into (x-1)^2(4x-1) and I'm not sure how you get there. I feel like I'm missing some really basic Algebra thing that I've forgotten, any tips?

When you say it "has imaginary roots when you get it down to a quadratic", it sounds like you have used the rational root theorem, reducing the degree by dividing by a rational root. But what you say is not true: there are no imaginary roots, as clearly shown by the factoring you found. All the roots are rational.

So, what did you do that led you to your conclusion? I would guess that you divided incorrectly.
 
When you say it "has imaginary roots when you get it down to a quadratic", it sounds like you have used the rational root theorem, reducing the degree by dividing by a rational root. But what you say is not true: there are no imaginary roots, as clearly shown by the factoring you found. All the roots are rational.

So, what did you do that led you to your conclusion? I would guess that you divided incorrectly.

I misspoke, I assumed there were imaginary roots because of my miscalculation. My algebra is rusty so my real question is how DO you factor this. The other post mentioned using some method but where did you learn that? I did not see that covered in my algebra classes, all I remember is the quadratic formula, grouping, and the method where you multiply a and c and figure out two factors that multiply to their product and sum to b.
 
I misspoke, I assumed there were imaginary roots because of my miscalculation. My algebra is rusty so my real question is how DO you factor this. The other post mentioned using some method but where did you learn that? I did not see that covered in my algebra classes, all I remember is the quadratic formula, grouping, and the method where you multiply a and c and figure out two factors that multiply to their product and sum to b.

Upon reviewing my old algebra notes, I realize you can use synthetic division to get it into this form. I was so used to using quad form, grouping, and the other method I didn't even consider this. Thanks for the help.
 
I misspoke, I assumed there were imaginary roots because of my miscalculation. My algebra is rusty so my real question is how DO you factor this. The other post mentioned using some method but where did you learn that? I did not see that covered in my algebra classes, all I remember is the quadratic formula, grouping, and the method where you multiply a and c and figure out two factors that multiply to their product and sum to b.

Upon reviewing my old algebra notes, I realize you can use synthetic division to get it into this form. I was so used to using quad form, grouping, and the other method I didn't even consider this. Thanks for the help.

So you did something else entirely to "get it down to a quadratic", probably something illegal for a cubic. Sounds like you have it now; the rational root theorem tells you what to divide by, which then reduces it to a quadratic quotient.
 
So you did something else entirely to "get it down to a quadratic", probably something illegal for a cubic. Sounds like you have it now; the rational root theorem tells you what to divide by, which then reduces it to a quadratic quotient.

Precisely, thanks for the different perspective; I'll be sure to review some Algebra.
 
So I ran into this polynomial while doing some optimization for Calculus: 4x^3 - 9x^2 +6x -1 and I've had no trouble with Algebra, factoring, and so on until I realized that this thing can't be factored by grouping, and has imaginary roots when you get it down to a quadratic. When I punch it into WolframAlpha, it somehow converts it into (x-1)^2(4x-1) and I'm not sure how you get there. I feel like I'm missing some really basic Algebra thing that I've forgotten, any tips?
I think that by a quick inspection you can see that x=1 is a root so then x-1 must be a factor. Now do long division or synthetic division. Otherwise you should use the rational root test. My advise is always to spend a few seconds to see if you can see a root.
 
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