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The polynomial $$ (x-1)(x-2)(x-3) + 1 $$ has one root (I have seen it by a plot in LaTeX), $0 < x_{root} < 1$. So I presume that the polynomial can be rewritten as : $$ (x-1)(x-2)(x-3) + 1 = (x - a)^{3} $$ but this is not possible, since $$ (x-1)(x-2)(x-3)+1 = x^{3} - 6x^{2} + 11x -5 $$ $$ (x-a)^{3} = x^{3} - 3ax^{2} + 3a^{2}x -a^{3} $$

How to find the form of the polynomial without the remainder? (also manually, without numerical method)

Thanks before.

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  • $\begingroup$ Note that : $(x-1)(x-2)(x-3)=0 $has 3 roots ,but $$(x-1)(x-2)(x-3)+1=0$$ has one root. It is not like $(x-a)^3$. $\endgroup$ – Khosrotash Sep 7 '17 at 5:34
  • $\begingroup$ It cannot necessarily be written like $(x-a)^3$ but it could be written like $(x-a)(x-b)(x-c)$ but it could be that only $a$ is real and $b$ and $c$ are both complex non-reals. $\endgroup$ – JMoravitz Sep 7 '17 at 5:37
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    $\begingroup$ This has one real root and two complex roots. $\endgroup$ – StephenG Sep 7 '17 at 5:38
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    $\begingroup$ In terms of real factorizations, incidentally, the polynomial factors as $(x-a)(x^2+bx+c)$ for $b,c$ with $b^2\lt 4c$ - that is, the one real root $a$ and two complex conjugate roots (which lead to a real quadratic factor). $\endgroup$ – Steven Stadnicki Sep 7 '17 at 6:20
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    $\begingroup$ in fact, this function has one real root and two complex roots link. $\endgroup$ – GAVD Sep 7 '17 at 6:21
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We can solve $ (x-1)(x-2)(x-3) + 1 = 0 $ for $x \in \mathbb{R}$ by manipulating it in a quadratic equation with some convenient variable substitutions as follows:

Expand the polynomial $$ x^3-6 x^2+11 x-5 = 0 $$

Substitute $ y = x - 2 $ to get rid of the quadratic term $$ (y+2)^3 - 6(y+2)^2 + 11(y+2) - 5 = y^3 - y + 1 = 0 $$

Let $c \in \mathbb{R}$ be a constant we yet don't know its value but will determine a suitable value later. Substitute $ y = z + \frac{c}{z} $ $$ (z + \frac{c}{z})^3 - (z + \frac{c}{z}) + 1 = 0 $$

Multiply both sides by $z^3$ and group by $z$ $$ z^6 + z^4(3c-1) + z^3 + z^2c(3c-1)+c^3 = 0 $$

Note that by setting $ c = \frac{1}{3} $ we get rid of both the quadratic and the quartic terms yielding $$ z^6 + z^3 + c^3 = 0 $$

To make computations easier make $ k = c^3$. Now substitute $ u = z^3 $ and we get a quadratic equation in $u$ $$ u^2 + u + k = 0 $$

Can you carry on from here? Let me know if you get stuck.

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  • $\begingroup$ I had to jump to other things. Thanks @mucciolo this is interesting. $\endgroup$ – Arief Anbiya Oct 31 '17 at 18:00
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Short answer:

Factoring a polynomial in linear or quadratic factors and finding its roots are equivalent tasks, one solves the other.

Your polynomial has no simple form and you can't escape Cardano/Tartaglia's formulas, which are a little heavy, though tractable by hand https://en.wikipedia.org/wiki/Cubic_function#General_solution_to_the_cubic_equation_with_real_coefficients

https://www.wolframalpha.com/input/?i=roots+of+(x-1)(x-2)(x-3)%2B1

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