Solve Polynomial Challenge: Prove $(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)=720$

[anemone]

Prove that there are only two real numbers such that $(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)=720$.

 

[MarkFL]

My solution:

Spoiler

Given that $6!=720$ we can see that:

\(\displaystyle x=0,\,7\)

And so, we may look at the number of real roots for:

\(\displaystyle f(x)=\frac{(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)-720}{x(x-7)}=x^4-14x^3+77x^2-196x+252\)

\(\displaystyle f'(x)=2(2x-7)\left(x^2-7x+14 \right)\)

The discriminant of the quadratic factor is negative, hence there is only 1 critical value:

\(\displaystyle x=\frac{7}{2}\)

\(\displaystyle f''(x)=12x^2-84x+154=3(2x-7)^2+7\)

Thus, the function is concave up for all $x$, and we find:

\(\displaystyle f\left(\frac{7}{2} \right)=\frac{945}{16}\)

And therefore \(\displaystyle 0<f(x)\) for all $x$, hence the original equation has no real roots beyond the two found by inspection.

 

[anemone]

Thanks MarkFL for participating and your solution is awesome and...you're FAST! I used essentially the same method as yours to solve this problem too...

 

[MarkFL]

I had a little help from the computer with the grunt work of dividing, differentiating and factoring...(Bandit)

 

[MarkFL]

Addendum to my solution:

Spoiler

After dividing, we found:

\(\displaystyle f(x)=x^4-14x^3+77x^2-196x+252\)

We may write this as:

\(\displaystyle \left(x^2-7x \right)^2+28\left(x^2-7x \right)+252=\left(x^2-7x+14 \right)^2+56\)

This is enough to show that this function as no real roots.

 

[anemone]

Addendum to my solution:

Spoiler

After dividing, we found:

\(\displaystyle f(x)=x^4-14x^3+77x^2-196x+252\)

We may write this as:

\(\displaystyle \left(x^2-7x \right)^2+28\left(x^2-7x \right)+252=\left(x^2-7x+14 \right)^2+56\)

This is enough to show that this function as no real roots.

Hey MarkFL, I like this method even more!(Nerd)