# Find the remainder when x^{3} + 3x^{2} + 3x + 1 is divided by

i) x + 1 ii) x - (1/2) iii) x iv) x + π v) 5 + 2x

**Solution:**

Let p(x) be any polynomial of degree greater than or equal to one and let 'a' be any real number.

If a polynomial p(x) is divided by x - a then the remainder is p(a).

Let p(x) = x^{3} + 3x^{2} + 3x + 1

(i) The root of x + 1 = 0 is -1

p(-1) = (-1)^{3} + 3(-1)^{2} + 3(-1) + 1

= -1 + 3 - 3 + 1

= 0

Hence by the remainder theorem, 0 is the remainder when x^{3} + 3x^{2} + 3x + 1 is divided by x + 1. We can also say that x + 1 is a factor of x^{3} + 3x^{2} + 3x + 1.

(ii) The root of x - (1/2) = 0 is 1/2.

p(1/2) = (1/2)^{3} + 3(1/2)^{2} + 3(1/2) + 1

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

= (1 + 6 + 12 + 8)/8 = 27/8

Hence by the remainder theorem, 27 / 8 is the remainder when x^{3} + 3x^{2} + 3x + 1 is divided by x.

(iii) The root of x = 0 is 0

p(0) = (0)^{3} + 3(0)^{2} + 3(0) +1

= 0 + 0 + 0 + 1

= 1

Hence by the remainder theorem, 1 is the remainder when x^{3} + 3x^{2} + 3x + 1 is divided by x.

(iv) The root of x + π = 0 is - π

p(-π) = (-π)^{3} + 3(-π)^{2} + 3(-π) + 1

= -π^{3} + 3π^{2} - 3π + 1

Hence by the remainder theorem, -π^{3} + 3π^{2} - 3π + 1 is the remainder when x^{3} + 3x^{2} + 3x +1 is divided by x + π.

(v) Now, the root of 5 + 2x = 0 is -5/2

p(-5/2) = [(-5/2)^{3} + 3(-5/2)^{2} + 3(-5/2) + 1]

= [(-125/8) + (75/4) + (-15/2) + 1]

= (-125 + 150 - 60 + 8) / 8

= (-185 + 158) / 8

= -27/8

Hence by remainder theorem, -27/8 is the remainder when x^{3} + 3x^{2} + 3x + 1 is divided by 5 + 2x.

**☛ Check: **Class 9 Maths NCERT Solutions Chapter 2

**Video Solution:**

## Find the remainder when x³ + 3x² + 3x + 1 is divided by i) x + 1 ii) x - (1/2) iii) x iv) x + π v) 5 + 2x.

NCERT Solutions Class 9 Maths Chapter 2 Exercise 2.3 Question 1:

**Summary:**

The remainder when x³ + 3x² + 3x + 1 is divided by x + 1, x - 1/2, x, x + π, and 5 + 2x respectively are 0, 27/8, 1, −π³ + 3π² - 3π + 1 , and -27/8.

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