$$x^4 + 2x^3 -13x^2 -14x + 24 = 0$$
I tried factoring, but it didn't seem to work out. I tried pulling out an $x^3$, but no progress. Please help.
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Does this answer your question? How to solve a quartic equation? – an4s Jan 20 '20 at 20:36
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Are you actually trying to find the roots of this polynomial or, as you wrote in the title, only trying to determine their nature? You can do the latter with Descartes’ rule of signs and, if that gives you an ambiguous result, Sturm’s theorem.. – amd Jan 20 '20 at 21:44
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$x=1$ is a root so divide by $x-1$ and try to factor the resulting cubic. $x=3$ is also a root so divide by $x-3$ and solve the resulting quadratic.
Mohammad Riazi-Kermani
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$$x^4 + 2x^3 -13x^2 -14x + 24=$$ $$=x^4-x^3+3x^3-3x^2-10x^2+10x-24x+24=(x-1)(x^3+3x^2-10x-24)=$$ $$=(x-1)(x^3+2x^2+x^2+2x-12x-24)=(x-1)(x+2)(x^2+x-12).$$ Can you end it now?
Michael Rozenberg
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@Bob Smith We know that $1$ is a root of the equation, which by the Bézout's theorem says that there is a factor $x-1$ and I made this factoring. – Michael Rozenberg Jan 20 '20 at 20:39
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You can start looking for the rational roots by the rational root theorem. Because the polynomial is monic, they'll need to be integers. We start with $\pm1$: $1$ is a root, but $-1$ isn't. We eventually find the roots are $1,\,-2,\,3,\,-4$.
J.G.
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