# Find the remainde

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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