Since $ p(1) = 0 $, $ x = 1 $ is indeed a root. To check the multiplicity, we compute the derivative $ p'(x) $:

February 22, 2026

$Since $ p(1) = 0 $, $ x = 1 $ is indeed a root. To check the multiplicity, we compute the derivative $ p'(x) $:$

Since \( p(1) = 0 \), \( x = 1 \) is indeed a root. To check the multiplicity, we compute the derivative \( p'(x) \):