Solve $(x^2+3x+2)(x^2-2x-1)(x^2-7x+12)+24=0$

[anemone]

Find all real numbers that satisfy $(x^2+3x+2)(x^2-2x-1)(x^2-7x+12)+24=0$.

 

[mathbalarka]

An old trick :

Spoiler

We see that $x^2 + 3x + 2 = (x + 1)(x + 2)$ and $x^2 - 7x + 12 = (x - 3)(x - 4)$. Then, we have :

$$\begin{aligned}(x^2+3x+2)(x^2-2x-1)(x^2-7x+12)+24 &= (x + 1)(x + 2)(x^2 - 2x - 1)(x - 3)(x - 4) + 24 \\ &= \{(x + 1)(x - 3)\}\{x^2 - 2x - 1\}\{(x + 2)(x - 4)\} + 24 \\ &= (x^2 - 2x - 3)(x^2 - 2x - 1)(x^2 - 2x - 8) + 24\end{aligned}$$

Setting $t = x^2 - 2x$ gives the twisted cubic of the form

$$(t - 3)(t - 1)(t - 8) + 24 = t^3 - 12t^2 + 35t = t(t - 5)(t - 7)$$

Reversing the transformation gives

$$x(x - 2)(x^2 - 2x - 5)(x^2 - 2x - 7)$$

The real roots are then

$$ x = 0, 2, 1 - \sqrt{6}, 1 + \sqrt{6}, 1 - 2\sqrt{2}, 1 + 2\sqrt{2} $$

Balarka
.

 

[anemone]

Thanks for participating, Balarka and fyi, I used the same method to tackle the problem as well!(Smile)

 

[topsquark]

That. Was. Sooooooo. Cool! (heart)

-Dan

 

[mathbalarka]

Yeah, quite interesting, this trick. Whenever you find $ABC + K$, factorize $A$, $B$ and $C$ - see what you have :p

I discovered this one when I was 5, thought it was original then but a few years afterwards, I noticed that some textbooks using this method. Funny, no?