Assume that there is a polynomial $P(x)$
$$ P(4) = P(5) = P(6) = \cdots = P(10) = 10! $$ $$ P(3) = 11! $$ $$ \deg(P(x)) = 7 $$
What is the constant term of polynomial $P(x)$?
The answer is
$$ 1201\times 10! $$
Algebra Polynomials Problem 1
1
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algebra-precalculus
polynomials
1 Answers
7
Hint
We first shift your polynomial down by $10!$ to get a new polynomial $P^*$ $$P^*(4) = P^*(5) = P^*(6) = P^*(7)=P^*(8)=P^*(9)= P^*(10) = 0$$ We now have all $7$ roots of this polynomial, and they are all zero. Try to express your polynomial in terms of these roots.