<s>[INST] Roots of the equation $(x – 1)(x – 2)(x – 3) = 24$ The equation $(x – 1)(x – 2)(x – 3) = 24$ has the real root equal to 'a' and the complex roots 'b' and 'c'. Then find the value of $\frac{bc}{a}$
My approach is as follow $y=f(x)=(x – 1)(x – 2)(x – 3) - 24=0$
$y'=3x^2-12x+11=0$
Solving we get $x=2\pm\sqrt{\frac{1}{3}}$
$f(2+\sqrt{\frac{1}{3}})<0$ & $ f(2-\sqrt{\frac{1}{3}})<0$
It is Local Minimum at $2+\sqrt{\frac{1}{3}}$ and Local Maximum at $2-\sqrt{\frac{1}{3}}$
By hit and trial I got $f(5)=0$ viz a=5
Given abc=30, therefore bc=6.
Hence the answer is $\frac{6}{5}$ which is correct.
My only concern is to find the real value without using any HIT and TRIAL.
 [/INST] There are actual computation methods for cubics but we can poke some number theory style fun into this.
$(x-1)(x-2)(x-3)=24$ are numbers whose product is 24. If we focused on integers, these numbers would be consecutive.
Oh what luck befalls us today. Watch this:
$(x-1)(x-2)(x-3)=4 \cdot 3 \cdot 2$
Let each factor pick a number and the value of $x$ remains the same.
Hence one answer is $x=5$
We have one answer. Perform synthetic division or long division of the full cubic by $x-5$ in order to represent it as $(x-5)P(x)=0$ where P has degree 2. 
Doing this would yield $(x-5)(x^2-x+6)$ and the quadratic has no real solutions.
A better way of playing this guessing game would be to try out all factors of the $-30$ in the cubic and see if they work. This would be the rational root theorem.
</s>