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
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) = x3 + 3x2 + 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 x3 + 3x2 + 3x + 1 is divided by x + 1. We can also say that x + 1 is a factor of x3 + 3x2 + 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 x3 + 3x2 + 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 x3 + 3x2 + 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
= -π3 + 3π2 - 3π + 1
Hence by the remainder theorem, -π3 + 3π2 - 3π + 1 is the remainder when x3 + 3x2 + 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 x3 + 3x2 + 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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