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Given the polynomial $-2x^3+23x^2-59x+24$, how do I find any non-real zeros? If there aren't, how can I explain why? I have tried factoring it by various methods, but have failed.

Pomponazzo
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Andrew Liu
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    Hi, welcome. What are the various methods you tried? Have you tried, for instance, the rational root theorem? – Matthew Leingang Nov 13 '22 at 21:47
  • I have tried it. I don't think the RR Theorem shows any imaginary roots though. – Andrew Liu Nov 13 '22 at 21:48
  • If there's non real solutions then it can be factored as $$(-2x^2+Bx+C)(x+D)$$such that $$B^2+8C\lt 0$$ – MaximusFastidiousIrreverence Nov 13 '22 at 21:52
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    Does the rational root theorem fail to show imaginary roots because all the roots are rational? – Matthew Leingang Nov 13 '22 at 21:54
  • @lonestudent correct, but I am trying to figure out how I can find the possible imaginary roots. – Andrew Liu Nov 13 '22 at 21:56
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    @Hi! Welcome. Please add context and also show attempt. –  Nov 13 '22 at 21:58
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    You say you've tried rational root theorem. If so, what roots have you found? – Sil Nov 13 '22 at 22:02
  • The possible roots are positive/negative (1, 1/2, 2, 3, 3/2, 4, 6, 8, 12, and 24). – Andrew Liu Nov 13 '22 at 22:04
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    Those are only candidates, you need to check if any of those are actual roots of the polynomial. (Spoiler: three of them are) – Sil Nov 13 '22 at 22:05
  • That is correct @Sil. The three roots are 0.5, 3, and 8. As my answer states, a cubic function can only have either 1 or 3 roots. As the three roots mentioned fulfill that requirement, we can conclude that there are no more real or non-real roots. – Andrew Liu Nov 13 '22 at 22:07
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    Exactly, rational root theorem solves the problem simply. Keep in mind that what you said is true only about real roots, cubics all have 3 (complex) roots including multiplicity. – Sil Nov 13 '22 at 22:13

1 Answers1

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Factoring $2 \, x^3 - 23 \, x^2 + 59 \, x - 24 = 0$ can be done by considering a factor of the form $2 \, x + a$ multiplied by a quadratic. In this view it can be seen as $$ 2 \, x^3 - 23 \, x^2 + 59 \, x - 24 = (2 x -1) (x-3) (x-8) = 0.$$ This shows that the roots are $$ x \in \left\{ \frac{1}{2}, 3, 8 \right\}$$ and does not contain any non-real values.

Leucippus
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    How did you get (2x-1) as a factor? – Andrew Liu Nov 13 '22 at 22:18
  • @AndrewLiu The first term in the cubic, if it can be factored, has a form of either $(2 x) (x^2)$ or $ (2 x^2) (x)$. The last term is negative so either the resulting binomial has a negative constant term of the factor does. The remaining is experience, guessing, or division. – Leucippus Nov 14 '22 at 09:29