RD Sharma Chapter 6 Class 9 Maths Exercise 6.3 Solutions

RD Sharma Chapter 6 Class 9 Maths Exercise 6.3 Solutions is based on the Remainder Theorem of Factorization of Polynomial. In this article, we will provide complete detailed information about the remainder theorem and terms related to the Factorization of Polynomial. The remainder theorem of polynomials provides us a connection between the remainder and its dividend.

Moreover, download the RD Sharma Chapter 6 Class 9 Maths Excercise 6.3 Solution PDF for practicing to score well in the exam. Our experts prepare the PDF for the students, which helps them understand this topic of Chapter 6 Class 9. Go down to the article for the detailed information about the im[portant definition of the Remainder Theorem with examples.

Click Here to know about the Chapter 6 Factorization of Polynomial

Download RD Sharma Chapter 6 Class 9 Maths Exercise 6.3 Solutions PDF

 


Solutions for Class 9 Maths Chapter 6 Factorization of Polynomials Exercise 6.3

Important Definitions RD Sharma Chapter 6 Class 9 Maths Exercise 6.3 Solutions

The remainder theorem of polynomials provides us a link between the remainder and its dividend. Let f(p) be any polynomial of degree greater than or equal to one and ‘a’ be any real number. If f(p) is divided by the linear polynomial p – a, then the remainder is f(a).

So fundamentally, p -a is the divisor of f(p) if and only if f(a) = 0. It is applied to factorize polynomials of each degree in a simple manner.

Examples of Remainder Theorem

Ques- Find the root of the polynomial x2− 3x– 4.

Solution- x2– 3x– 4

= f(4) = 42– 3(4)– 4

= f(4)=16–16=0

So,  (x-4) must be a factor of x2– 3x– 4

Ques- 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

Ques- 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

Frequently Asked Questions (FAQs) of RD Sharma Chapter 6 Class 9 Maths Exercise 6.3 Solutions

Ques 1- How does remainder theorem work?

Ans-If you divide a polynomial, f(x), by linear identity, x-A, the remainder will be similar as f(A). For example, the remainder when is divided by x-3 is.

Ques 2- What happens if the remainder is zero?

Ans- When the remainder is zero, both the quotient and divisor are the factors of dividend. But, when the remainder is not zero, neither the quotient nor the divisor is the factor of the dividend.

Ques- How do you find the quotient and remainder?

Ans- Finding Dividend, divisor, quotient, and the remainder from the following expressions-

a/n = q + r/n

  1. a is the first number to divide, known as the dividend.
  2. n is the number divide by; it is known as the divisor.
  3. q is the result of division turned down to the nearest integer; it is known as the quotient.
  4. r is the remainder of the mathematical operation.

Know about CBSE Board

Leave a Comment