1.

Using remainder theorem x³-6x²+9x+3 is divided by x-1

Answer» PUT
x-1=0
hence x=1
BY REMAINDER THEOREM
WE GET REMAINDER IF WE PUT X=1 IN GIVEN EQUATION
{x}^{3} - 6 {x}^{2} + 9x + 3 \\ {1}^{ 3} - 6 \times {1}^{2} +9 \times 1 + 3 \\ = 1 - 6 + 9 + 3 \\ = <klux>7</klux>
Remainder is 7


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