# How to find the one real root of $(x-1)(x-2)(x-3) + 1$, manually?

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.

• 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 $(x-a)^3$. – Khosrotash Sep 7 '17 at 5:34
• 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. – JMoravitz Sep 7 '17 at 5:37
• This has one real root and two complex roots. – StephenG Sep 7 '17 at 5:38
• 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). – Steven Stadnicki Sep 7 '17 at 6:20
• in fact, this function has one real root and two complex roots link. – GAVD Sep 7 '17 at 6:21

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.

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