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, 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, 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, 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, required remainder is .