Mastering Factoring 3rd Degree Polynomials: x^3+6x^2-12x+8=0

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The discussion focuses on factoring the third-degree polynomial equation x^3 + 6x^2 - 12x + 8 = 0. A participant identifies that 2 is a root of the polynomial and suggests using synthetic division to simplify the equation by dividing it by (x - 2). The conversation highlights the connection between this polynomial and an eigenvalue problem, indicating its relevance in more advanced mathematical contexts. Participants seek methods to recall factoring techniques for cubic equations. The discussion emphasizes the importance of identifying roots and applying synthetic division for polynomial simplification.
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-x^3+6x^2-12x+8=0




The Attempt at a Solution



- this is actually part of an eigenvalue problem, but I can't seem to remember how to factor 3rd degree polynomials. Any thing to help me remember would be great.
 
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well by investigation we can see that 2 is a root of that equation, now can u perform synthetic division, or divide that polynomial by (x-2).
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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