Question 141543
Do the binomial expansion on {{{(x-2)^3}}}.  Using the result as the divisor and {{{x^4 - 4x^3 + 16x - 16 = 0}}} as the dividend, perform polynomial long division or synthetic division.  Remember to insert {{{0x^2}}} as a placeholder for the missing 2nd degree term.  If x = 2 is a root of multiplicity 3, you will get no remainder and the quotient will be different from {{{x-2}}}.


If the quotient IS {{{x-2}}}, then {{{(x-2)}}} is a root with multiplicity 4.


If there is a remainder, then {{{x-2}}} is not a root at all.


The quotient, {{{x-a}}} where {{{a<>2}}}, will be the 4th factor, and you can solve {{{x-a=0}}} to get the 4th root.