Find all roots of $\,(x + 1)(x + 2)(x + 3)^2(x + 4)(x + 5) = 360$

Question

The question is to find all complex roots of

$$(x + 1)(x + 2)(x + 3)^2(x + 4)(x + 5) = 360$$ and it is meant to be solved by hand.

Is there any quick way to solve this using some trick that I'm not aware of? The solution given by Wolfram used a lot of multiplying out and factoring that would be very difficult to think of or write. Also, just by observation $x=0$ and $x=-6$ are solutions.

Answer 1

$360 = 2^3 \times3^2 \times 5 = 1 \times 2 \times 3^2 \times 4 \times 5 = (-5) \times (-4) \times (-3)^2 \times (-2) \times (-1)$. We see that $x=0$ and $-6$ are solutions.

Let $(x+3) = y$. Then we get $$(x+1)(x+2)(x+3)^2(x+4)(x+5) = y^2 (y^2-1)(y^2-4) = 360$$ Note that since $x=0$ and $-6$ are solutions, we have $y = \pm 3$ to be solutions. Calling $y^2=t$, we then get that $$t(t-1)(t-4) = 360$$ Since $y=\pm3$ is a solution, $t=9$ is a solution for the above equation. Hence, $$t^3 - 5t^2 + 4t = 360 \implies t^3 - 5t^2 + 4t - 360 = (t-9)(t^2+at+b)$$ Hence, $a-9=-5$ and $9b = 360$. This gives us $a = 4$ and $b=40$. Hence, the solutions are $$t = 9, -2 \pm 6i$$ Hence, $$y = \pm3 , \pm \sqrt{-2\pm6i}$$ Hence, $$x = 0,-6, -3 \pm \sqrt{-2\pm6i}$$

Answer 2

This is another approach but with a lot more computation required than that of Marvis. It could still be done by hand as stated in the original post with some patience. Expanding the factors we get $$x^6 + 18x^5 + 130x^4 + 480x^3 + 949x^2 + 942x + 360=360$$ Then subtract $360$ from each side to get $$x^6 + 18x^5 + 130x^4 + 480x^3 + 949x^2 + 942x=0$$ Because it was noted by observation that $0$ and $-6$ are solutions let’s divide the polynomial by $x+6$. So

$$\frac{ x^6 + 18x^5 + 130x^4 + 480x^3 + 949x^2 + 942x}{x+6}=x^5+12x^4+58x^3+132x^2+157x$$ Because $0$ is also a solution we can factor out an $x$ for free to get $$x^4+12x^3+58x^2+132x+157=0$$ This is a quartic which we can convert to a depressed quartic by setting $$x=u-\frac{12}{4}$$ and expanding the following as seen in Quartic function

$$\left(u-\frac{12}{4}\right)^4+\frac{12}{4}\left(u-\frac{12}{4}\right)^3+ \frac{58}{4}\left(u-\frac{12}{4}\right)^2+ \frac{132}{4}\left(u-\frac{12}{4}\right)+\frac{157}{4}=0$$

This expands to $$u^4 + 4u^2 + 40=0$$ which is a biquadratic equation. Solving this quadratic for $u^2$ we get solutions

$$u=\pm \sqrt{-2\pm6i}$$ so all the complex solutions to the original quartic and also the original polynomial are

$$x=-3\pm \sqrt{-2\pm6i}$$ getting the $-3$ above from the $-\frac{12}{4}$ term used to depress the original quartic.