Analyzing the Setup
Imagine you are sitting in the exam hall, and you see this polynomial:
f(x)=x9−6x8−2x7+12x6+x4−7x3+6x2+x−3
Your heart might skip a beat. It looks like a monster, doesn't it? A degree-nine polynomial!
The immediate, panicked reaction is to start calculating 69, 68, and so on. But stop. Take a deep breath.
In the world of JEE Advanced, whenever you see a high-degree polynomial and a simple integer evaluation, you are not being asked to calculate; you are being asked to observe.
The Brute Force Fallacy
If you try to calculate 69, you are falling right into the trap. The examiners want to see if you have the mathematical maturity to recognize structure.
Let us substitute
x=6 into our expression:
f(6)=69−6(68)−2(67)+12(66)+64−7(63)+6(62)+6−3
Now, look at the first two terms: 69−6(68). This is just 69−69, which equals 0. The monster has already started to shrink.
The Dance of Cancellation
Let us look at the next pair: −2(67)+12(66). This looks intimidating, but remember that 12=2⋅6.
So, we can rewrite this as:
−2(67)+2⋅6(66)
Since 6⋅66=67, this becomes −2(67)+2(67), which is also zero! The pattern is emerging. We are not doing arithmetic; we are performing a dance of cancellations.
Moving to the middle terms: 64−7(63)+6(62). We know that 6(62)=63.
So, the expression becomes:
64−7(63)+63=64−6(63)
Since 6(63)=64, we have 64−64, which is yet again zero!
The Grand Finale
After all that, what remains? We have systematically eliminated almost every term.
We are left with the very last part of the polynomial:
x−3. Substituting
x=6 into this final, simple expression gives us:
6−3=3
The entire, terrifying degree-nine polynomial collapses into the number 3.
This is the beauty of mathematics—hidden order beneath apparent chaos. Never fear the monster; look for the pattern, and you will find that the solution is often simpler than you ever imagined.