2023 AMC 12B Problem 6

Question

When the roots of the polynomial

\begin{gathered} P(x)=(x-1)^1(x-2)^2 \\ {}\cdot (x-3)^3\cdots \\ {}\cdot (x-10)^{10} \end{gathered}

are removed from the number line, what remains is the union of

11

disjoint open intervals. On how many of these intervals is

P(x)

positive?

3

7

6

4

5

The sign of

P(x)

flips at a root only when that root’s exponent is odd

For

x\gt 10

the value is positive; move left, flipping sign only at the odd roots

1,3,5,7,9

The exponent of the factor

(x-k)

is

k,

so the sign of

P

changes at

x=k

only when

k

is odd, i.e. at

1,3,5,7,9.

For

x\gt 10

every factor is positive, so

P\gt 0.

Sweeping left and flipping at each odd root, the positive intervals are

(10,\infty),

(9,10),

(6,7),

(5,6),

(2,3),

and

(1,2)

— six intervals in all.

Thus, the correct answer is C.

Problem 6 in other years