# Polynomial Challenge

#### anemone

##### MHB POTW Director
Staff member
Prove that there are only two real numbers such that $(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)=720$.

#### MarkFL

Staff member
My solution:

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

##### MHB POTW Director
Staff member
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

Staff member
I had a little help from the computer with the grunt work of dividing, differentiating and factoring... #### MarkFL

Staff member

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

##### MHB POTW Director
Staff member
$$\displaystyle f(x)=x^4-14x^3+77x^2-196x+252$$
$$\displaystyle \left(x^2-7x \right)^2+28\left(x^2-7x \right)+252=\left(x^2-7x+14 \right)^2+56$$
Hey MarkFL, I like this method even more! 