MHB POTW Director
- Feb 14, 2012
Prove that there are only two real numbers such that $(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)=720$.
Hey MarkFL, I like this method even more!Addendum to my solution:
After dividing, we found:
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.