Answer :

Let, f(x) = x3+3x2+3x+1

(i) x + 1


Apply remainder theorem


x + 1 =0


x = - 1


Replace x by – 1 we get


x3+3x2 + 3x + 1


(-1)3 + 3(-1)2 + 3(-1) + 1


-1 + 3 - 3 + 1


0


Hence, the required remainder is 0.


(ii) x-


Apply remainder theorem


x – 1/2 =0


x = 1/2


Replace x by 1/2 we get


x3+3x2 + 3x + 1


(1/2)3 + 3(1/2)2 + 3(1/2) + 1


1/8 + 3/4 + 3/2 + 1


Add the fraction taking LCM of denominator we get


(1 + 6 + 12 + 8)/8


27/8


Hence, the required remainder is 27/8


(iii) x = x – 0


By remainder theorem required remainder is equal to f (0)


Now, f (x) = x3+3x2+3x+1


f (0) = (0)3 + 3 (0)2 + 3 (0) + 1


= 0 + 0 + 0 + 1


= 1


Hence, the required remainder is 1.


(iv) x+π = x – (-π)


By remainder theorem required remainder is equal to f (-π)


Now, f (x) = x3 + 3x2 + 3x + 1


f (- π) = (- π)3 + 3 (- π)2 + 3 (- π) + 1


= - π3 + 3π2 - 3π + 1


Hence, required remainder is - π3 + 3π2 - 3π + 1.


(v) 5 + 2x = 2 [x – ()]


By remainder theorem required remainder is equal to f ()


Now, f (x) = x3 + 3x2 + 3x + 1


f () = )3 + 3 ()2 + 3 () + 1


= + 3 * + 3 * + 1


= + - + 1


=


Hence, the required remainder is .

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