Question

using the remainder therom, find the remainder, when p(x) is divided by g(x), where

p(x) =x cube - ax square +6x-a
g(x)= x-a


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Answer 1

Apply remainder theorem 
=>x – a =0 
=> x =  a
Replace x by  a we get 
=> x3 – ax2 + 6x – a 
=>( a)3 -a(a)2 + 6(a) - a
=>  a3sup> – a3 + 6a – a
=>  5a

Answer 2

$p(x) = {x}^{3} - a {x}^{2} + 6x - a$

g (x)

g (x) = x-a = 0

x = a

putting the value of x in p (x)

${a}^{3} - a \times {a}^{2} + 6a - a \\ \\ {a}^{3} - {a}^{3} + 6a - a \\ \\ 6a - a \\ \\ = 5a$

Remainder = 5a

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