Roots of the equation $(x – 1)(x – 2)(x – 3) = 24$

Question

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.

Answer 1

Rewrite $$(x – 1)(x – 2)(x – 3) = 24$$

as

$$x^3-6x^2+11x-30=0$$

which factorizes as

$$(x-5)(x^2-x+6)=0$$

Thus,

$$\frac{bc}a= \frac 65$$

Answer 2

Noticing that $24=2\cdot3\cdot4$, $5$ must be a root. Then after long division by $x-5$, $x^2-x+6=0$.

Using Vieta,

$$\frac{bc}a=\frac65.$$

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Now for a "general" solution, you first deplete the cubic by setting $z:=x-2$ and the equation is

$$(z+1)z(z-1)=z^3-z=24.$$

Then with $z:=\dfrac2{\sqrt 3}\cosh u$,

$$4\cosh^3u-3\cosh u=\cosh3u=36\sqrt3,$$

finally giving

$$x=\frac2{\sqrt3}\cosh\frac{\text{arcosh }36\sqrt3}3+2.$$

Needless to say, this is $5$.

Answer 3

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.