Find the remainder when $x^{3}$ −2 $x^{2}$ +6x is divided by x−2.

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Solution

Let p(x)= $x^{3}$ −2 $x^{2}$ +6x and q(x)=(x−2)

According to remainder theorem, remainder is equal to p(a) when p(x) is divided by (x−a).

p(2)= $2^{3}$ −2 ${(2)}^{2}$ +6(2) = $2^{3}$ − $2^{3}$ +12 = 12

Therefore, remainder is equal to 12 when p(x)= $x^{3}$ −2 $x^{2}$ +6x is divided by (x−2).

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