Question

Find the remainder when x³ + 3x² + 3x + 1 is divided by

(i) x + 1

(ii) x - \(\frac{1}{2}\)

(iii) x

(iv) x + π

(v) 5 + 2x

Answer 1

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.

x - \(\frac{1}{2}\)

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 – \((\frac{-5}{2})]\)

By remainder theorem required remainder is equal to f \((\frac{-5}{2})\)

Now, f (x) = x³ + 3x² + 3x + 1

Hence, the required remainder is \(\frac{-27}{8}.\)