Home / 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) $:

February 22, 2026

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

Solution:** To verify if $ x = 1 $ is a root of multiplicity greater than 1 for $ p(x) = x^4 - 4x^3 + 6x^2 - 4x + 1 $, we first check if $ p(1) = 0 $.
Calculate $ p(1) $:
p(1) = 1^4 - 4 imes 1^3 + 6 imes 1^2 - 4 imes 1 + 1 = 1 - 4 + 6 - 4 + 1 = 0
p'(x) = 4x^3 - 12x^2 + 12x - 4
Now, evaluate $ p'(1) $:
p'(1) = 4 imes 1^3 - 12 imes 1^2 + 12 imes 1 - 4 = 4 - 12 + 12 - 4 = 0