Taxicab Rooted Polynomial

$x(x+6)(x+12)(x+18) = 1729$

$\Rightarrow x(x+18)(x+6)(x+12) = 1729$

$\Rightarrow (x^2+18x)(x^2 + 18x + 72) = 1729$

Let $x^2 + 18x = a$

$\Rightarrow a(a + 72) = 1729$

$\Rightarrow a^2 + 72a - 1729 = 0$

$\Rightarrow (a-19)(a+91) = 0 \Rightarrow a = 19 \quad or \quad -91$

When $a = -91 \Rightarrow x^2 + 18x + 91 = 0 \Rightarrow x = \dfrac{-18 \pm \sqrt{-40}}{2}$ which is not possible since the question asks for real roots.

When $a = 19 \Rightarrow x^2 + 18x = 19 \Rightarrow x^2 + 18x - 19 = 0 \Rightarrow (x + 19)(x - 1) = 0$

$\Rightarrow x = 1 , -19$

Thus , sum real roots $\huge=1-19 = \color{#D61F06}{ \boxed{-18}}$

Alternative trial and error:

We can easily see that $1729 = 1 \times 7 \times 13 \times 19$ , where $1 , 7 , 13 , 19$ form an Arithmetic progression with difference $6$ .

Thus $x = 1 , -19$

However , it is better to check whether there are other real roots existing or not

Using Vieta's Formula
Find the first two coefficients of the expanded polynomial. They are $1$ and $6 + 12 + 18 = 36.$ By Vieta's formula, the sum of all roots is $\frac{-36}{1} = -36$ but by looking at the graph, two roots aren't real.

By the complex conjugate root theorem, if $a + bi$ is a root, then so is $a - bi.$ Since the polynomial is symmetric about $x = -9,$ the real roots are equidistant from $-9,$ and the imaginary roots are equidistant from $-9,$ and also equidistant from the real roots. Thus they are of the form $-9 + bi$ and $-9 - bi.$ Subtract their sum from $-36:$ $-36 - (-18) = \boxed{-18}$

Using number theory
Looking at the graph, there are two real roots. Luckily, $1729$ 's prime factors differ by the differences in the polynomial's factors: including $1,$ they are $\{1,7,13,19\}.$ And since the polynomial is of even degree, we just need to find the case where the factors are all negative and the case where they're all positive. These are respectively $x = -19$ and $x = 1.$ $1 - 19 = \boxed{-18}$

$\begin{aligned} 1729 & = \color{#3D99F6}{(1)(7) (13) (19)} = \color{#D61F06}{(-1)(-7) (-13) (-19)} \\ & = \color{#3D99F6}{1}(\color{#3D99F6}{1}+6)(\color{#3D99F6}{1}+12)(\color{#3D99F6}{1}+18) = (\color{#D61F06}{-19}+18) (\color{#D61F06}{-19}+12) (\color{#D61F06}{-19}+6) (\color{#D61F06}{-19}) \end{aligned}$

The sum of real roots $= 1 - 19 = \boxed{-18}$

Let $y=x+9$ , then we have $x(x+6)(x+12)(x+18)=(y-9)(y-3)(y+3)(y+9)=(y^2-81)(y^2-9)=y^4-90y^2+729=1729$ $\Rightarrow y^4-90y^2-1000=0$ $\Rightarrow (y^2-100)(y^2+10)=0$ $\Rightarrow (y-10)(y+10)(y^2+10)=0$ Since $y^2+10=0$ don't have real roots, then we have $y_1=10$ and $y_2=-10$ . So, we have $x_1=10-9=1$ and $x_2=-10-9=-19$ , and the sum of all real roots of the equation above is $x_1+x_2=-18$ .

Notice the prime factorization of 1729 = (7)(13)(19)= (91)(19) This makes it obvious x=1 is a solution

Using synthetic division with x=1, you find that the new cubic polynomial is x^3+37x^2+433x+1729 = 0

You can factor this further to be

(x+19) (x (x+18)+91) = 0

The number of imaginary solutions to any polynomial must be an even number. Since, (x(x+18)+91) = 0 only has imaginary solutions because the discriminant (b^2-4ac) < 0, you can conclude there are 2 real solutions and 2 complex solutions to the original quartic equation.

So, x=1,-19 are the only real solutions, so you have 1+ -19 = -18 as the answer.

$1729 \text{factors into} 1729=7\times 13\times 19 \\ \text{hence the only possible solutions according to the above expression are}\\ 1729=1\times 7\times 13\times 19 \\ \text{OR}\\ 1729=-1\times -7\times -13\times -19\\ \text{and } x=1 \text{ or } -19 \\ 1-19=-18.$

1729 = 7x13x19

Se x = 1

Logo, 1x7x13x19 = 1729

Induz o x+18 = -1, logo x = -19

x(x+6)(x+12)(x+18) = -19x(-13)x(-7)x(-1) = 1729

Foi pensado nessa possibilidade, pois são 4 raízes. Caso a função tenha números de raízes pares, logo o conjunto de valores negativos resultará num valor positivo.

Então se "x" pode ser -19 e 1. Somando, -19 + 1 = -18

$x(x+6)(x+12)(x+18)=1729$

$x(x+18)(x+6)(x+12)=1729$

$(x^2+18x)(x^2+18x+72)=1729$

Let $n=(x^2+18x)$

$n(n+72)=1729$

$n^2+72n-1729=0$

$(n-19)(n+91)=0$ $\Rightarrow$ $n=19$ or $n=-91$

When $n=-91$ $\Rightarrow$ $x^2+18x+91=0\Rightarrow$ $x=\dfrac{-18\pm\sqrt{-40}}{2}$ . But the question asks for real roots, not complex ones, so this solution is irrelevant.

However, when $n=19\Rightarrow$ $x^2+18x=19\Rightarrow x^2+18x-19=0\Rightarrow (x+19)(x-1)=0$

$\Rightarrow x=-19, x=1$

$-19+1=18$

Let's say $y=x+9,\quad z=y^{ 2 }-45\\ x(x+6)(x+12)(x+18)=1729\\ (y-9)(y-3)(y+3)(y+9)=1729\\ (y^{ 2 }-81)(y^{ 2 }-9)=1729\\ (z-36)(z+36)=1729\\ z^{ 2 }-1296=1729\\ z^{ 2 }=3025\\ y^{ 2 }=\pm 55+45,\quad y^{ 2 }=-10\quad or \quad y^{ 2 }=100\\ y=\pm 10\\ x=1\quad or\quad x=-19\\ 1-19=\boxed{ -18 }\\$

I'm horrible at editing -_-

A bit of graph sketching will sort this one. The LHS values at x= (0,-6,-12,-18) are all 0. The polynomial is a quartic (with a +ve x^4 coeff) so will have 'w' form and there will 2 real roots at x>0 and x< -18 and maybe 2 more. The symmetry of the four zero crossing coordinates means that roots will have the form of a $\pm$ b and a $\pm$ c where a = -9 (-9 being the mid-way between 0 and -18). x = -9 is the maximum between the two minima and its value is 729 which is less than 1729 this means that there are only two real roots. Since the roots are -9+b and -9-b so the sum is 2 x -9 = -18.

A bit of sketching and looking can save a great deal of maths.

Put X=x+9 Then the given equation reduces to (X-9)(X-3)(X+3)(X+9)=1729 => (X^2 - 9^2)(X^2 - 3^2)=1729 Put X^2=t => (t - 81)(t - 9)=1729 => t^2 - 90t - 1000=0 By quadratic formula t=100,-10=(x+9)^2 =>x=1,-19

So, the sum of all(2) real roots is -18.