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While solving a problem on sequences and series, I got the following cubic equation

$$8x^3-16x-85=0$$

I cannot figure out how to solve it. I have tried to factor the L.H.S., but did not succeed. Please help. Are these type of equations solvable by factorization or is using a particular formula the only way?

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Hint. Find a rational zero $p/q$ such that $p$ divides $85$ and $q$ divides $8$. Note also that $f(2)=-53<0$ and $f(3)=83>0$.

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  • $\begingroup$ Is there any general rule that helps in factorizing polynomials like these? $\endgroup$ – MrAP Sep 10 '17 at 7:56
  • $\begingroup$ No a general rules, but the following criterion is useful: if a polynomial $P(x)=a_nx^n+\dots a_0\in\mathbb{Z}[x]$ has a rational root $p/q$ then $p$ divides $a_0$ and $q$ divides $a_n$. $\endgroup$ – Robert Z Sep 10 '17 at 8:07
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Hint: Assume that $8x^3-16x-85=(ax+b)(cx^2+dx+f)$ and compare coefficients. Then we see that $$ 8x^3-16x-85=(2x-5)(4x^2+10x+17). $$

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