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.

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.
