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Let $f(t)=36t^3-19t+5$ be a cubic polynomial. How we can factor $f$ to its roots?

Mathematica says that $f(t)=(-1+2 t) (-1+3 t) (5+6 t)$. How?

SKMohammadi
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4 Answers4

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The Rational root theorem can be used here to find all roots of this polynomial

Belgi
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  • is there another way? – SKMohammadi Jan 02 '15 at 20:07
  • @MathMan I can't think of any better way, maybe other answers will. Anyway, you can use this theorem without having to justify it - one you get the roots you can prove those are the roots by evaluating the polynomial at those points – Belgi Jan 02 '15 at 20:09
  • How can you know all roots are rational, without looking at the mathematica result? – Dasherman Jan 02 '15 at 20:13
  • @MathMan Yes, guess the roots. Or use Cardano's formulas if you want to get messy. – user2345215 Jan 02 '15 at 20:13
  • If you check t=1/2 f(0.5)=0 so divide f(t) by t-0.5 to find others – Khosrotash Jan 02 '15 at 20:15
  • @MathMan - This method finds all rational roots – Belgi Jan 02 '15 at 20:15
  • You can't know all roots are rational except by testing all possible values. Cardano's formula won't work here as there are 3 real roots. You also can try a trigonometric method, but you'll have to use inverse trigonometric function. – Bernard Jan 02 '15 at 22:28
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Hint $\ 6f(t) = (6t)^3\! -19(6t) + 30 = x^3\! - 19x + 30 =: g(x)\,$ for $\,x = 6t.$ By the Rational Root Test the only possible rational roots of $g$ are integer factors of $30.\,$

Remark $\ $ This is a generalization of the high-school "AC method" and it works generally to reduce finding rational roots to finding integer roots. See here for much further discussion, including more general ring-theoretic viewpoints

Community
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Bill Dubuque
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  • The test does not require that the polynomial will be monic, why the extra step ? – Belgi Jan 02 '15 at 22:32
  • @Belgi Because often reduction to the monic special case yields simplifications, both conceptually and computationally. Follow the link for more. – Bill Dubuque Jan 02 '15 at 22:40
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Looks like using the rational root theorem is a good bet for this problem. If you feel it is tedious to use the rational root theorem to find all three roots, then use it to find just one root (or just plain guess and check until you find one root, like $t=\frac{1}{2}$) and then use polynomial division. Since $t=\frac{1}{2}$ is a root, you can divide your polynomial by $\left(t-\frac{1}{2}\right)$ and will get $$36t^3-19t+5 =\left(t-\frac{1}{2}\right)(36t^2+18t-10)$$ From here you should be able to factor the quadratic portion $(36t^2+18t-10)$ by hand. The rest of the algebra should follow as: $$\left(t-\frac{1}{2}\right)(36t^2+18t-10) = \left(t-\frac{1}{2}\right)2\cdot(18t^2+9t-5) \\ = \left(2\left[t-\frac{1}{2}\right]\right)(6t+5)(3t-1) \\ = (2t-1)(6t+5)(3t-1)$$ This technique will work for whichever of the three roots you begin with.

graydad
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  • Better than solving a quadratic equation, you can use the hint in the previous answer: $g(x)$ also has rootequal to $2$, hence the root $\dfrac{2}{6}=\dfrac{1}{3}$ for $f(x)$. You'll find the last root using relations between coefficients and roots. – Bernard Jan 02 '15 at 22:32
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And without using Rational root theorem: $$36t^3-10t+5=36t^3-30t^2+30t^2-25t+6t+5$$ $$36t^3-10t+5=6t(6t^2-5t+1)+5(6t^2-5t+1)$$ $$\,\,\,\,\,\,\,=(6t^2-5t+1)(6t+5)$$ $$\,\,\,\,\,\,\,\,\,\,=(2t-1)(3t-1)(6t+5)$$

SKMohammadi
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