RD Sharma Solutions for Class 9 Mathematics Chapter 6 Exercise 6.3 Factorization of Polynomials are provided here. This exercise deals with the remainder theorem, along with the definition of polynomials and their terms. The remainder theorem of polynomials gives us a relationship between the remainder and its dividend. Let p(x) be any polynomial of degree greater than or equal to one and ‘a’ be any real number. If p(x) is divided by the linear polynomial x – a, then the remainder is p(a).
For a better understanding of concepts, students can practise RD Sharma Solutions Class 9 Chapter 6.
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Exercise 6.3 Page No: 6.14
In each of the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division : (1 – 8)
Question 1: f(x) = x3 + 4x2 – 3x + 10, g(x) = x + 4
Solution:
f(x) = x3 + 4x2 – 3x + 10, g(x) = x + 4
Put g(x) =0
=> x + 4 = 0 or x = -4
Remainder = f(-4)
Now,
f(-4) = (-4)3 + 4(-4)2 – 3(-4) + 10 = -64 + 64 + 12 + 10 = 22
Actual Division:
Question 2: f(x) = 4x4 – 3x3 – 2x2 + x – 7, g(x) = x – 1
Solution:
f(x) = 4x4 – 3x3 – 2x2 + x – 7
Put g(x) =0
=> x – 1 = 0 or x = 1
Remainder = f(1)
Now,
f(1) = 4(1)4 – 3(1)3 – 2(1)2 + (1) – 7 = 4 – 3 – 2 + 1 – 7 = -7
Actual Division:
Question 3: f(x) = 2x4 – 6X3 + 2x2 – x + 2, g(x) = x + 2
Solution:
f(x) = 2x4 – 6X3 + 2x2 – x + 2, g(x) = x + 2
Put g(x) = 0
=> x + 2 = 0 or x = -2
Remainder = f(-2)
Now,
f(-2) = 2(-2)4 – 6(-2)3 + 2(-2)2 – (-2) + 2 = 32 + 48 + 8 + 2 + 2 = 92
Actual Division:
Question 4: f(x) = 4x3 – 12x2 + 14x – 3, g(x) = 2x – 1
Solution:
f(x) = 4x3 – 12x2 + 14x – 3, g(x) = 2x – 1
Put g(x) =0
=> 2x -1 =0 or x = 1/2
Remainder = f(1/2)
Now,
f(1/2) = 4(1/2)3 – 12(1/2)2 + 14(1/2) – 3 = ½ – 3 + 7 – 3 = 3/2
Actual Division:
Question 5: f(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – 2x
Solution:
f(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – 2x
Put g(x) = 0
=> 1 – 2x = 0 or x = 1/2
Remainder = f(1/2)
Now,
f(1/2) = (1/2)3 – 6(1/2)2 + 2(1/2) – 4 = 1 + 1/8 – 4 – 3/2 = -35/8
Actual Division:
Question 6: f(x) = x4 – 3x2 + 4, g(x) = x – 2
Solution:
f(x) = x4 – 3x2 + 4, g(x) = x – 2
Put g(x) = 0
=> x – 2 = 0 or x = 2
Remainder = f(2)
Now,
f(2) = (2)4 – 3(2)2 + 4 = 16 – 12 + 4 = 8
Actual Division:
Question 7: f(x) = 9x3 – 3x2 + x – 5, g(x) = x – 2/3
Solution:
f(x) = 9x3 – 3x2 + x – 5, g(x) = x – 2/3
Put g(x) = 0
=> x – 2/3 = 0 or x = 2/3
Remainder = f(2/3)
Now,
f(2/3) = 9(2/3)3 – 3(2/3)2 + (2/3) – 5 = 8/3 – 4/3 + 2/3 – 5/1 = -3
Actual Division:
RD Sharma Solutions for Class 9 Maths Chapter 6 Factorization of Polynomials Exercise 6.3
RD Sharma Solutions for Class 9 Maths Chapter 6 Factorization of Polynomials Exercise 6.3 are based on the Remainder Theorem. RD Sharma Solutions for Class 9 are prepared by subject experts to help students grasp the concepts better and score well in examinations.
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