Answer :

(i) Let f (x) = x6-ax5+x4-ax3+3x-a+2 be the given polynomial

From factor theorem,


If (x – a) is a factor of f (x) then f (a) = 0 [Therefore, x – a = 0, x = a]


f (a) = 0


(a)6 – a (a)5 + (a)4 – a (a)3 + 3 (a) – a + 2 = 0


a6 – a6 + a4 – a4 + 3a – a + 2 = 0


2a + 2 = 0


a = -1


Hence, (x – a) is a factor f (x) when a = -1.


(ii) Let, f (x) = x5-a2x3+2x+a+1 be the given polynomial


From factor theorem,


If (x – a) is a factor of f (x) then f (a) = 0 [Therefore, x – a = 0, x = a]


f (a) = 0


(a)5 – a2 (a)3 + 2 (a) + a + 1 = 0


a5 – a5 + 2a + a + 1 = 0


3a + 1 = 0


3a = -1


a =


Hence, (x – a) is a factor f (x) when a = .


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