NCERT Solutions for Class 10 Maths Chapter 2 Ex 2.3 For Class 10 Polynomials is a perfect guide to boost your preparation during CBSE 10th Class Maths exams.
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Pre 2.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Chapter 2 Polynomial Class 10
Ex 2.3.
- Polynomials Class 10 Ex 2.1
- Polynomials Class 10 Ex 2.2
- Polynomials Class 10 Ex 2.3
- NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 10 |
| Subject | Maths |
| Chapter | Chapter 2 |
| Chapter Name | Polynomials |
| Exercise | Ex 2.3 |
| Number of Questions Solved | 5 |
| Category | NCERT Solutions |
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3
Page No: 36
Question 1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
Quotient = x – 3
Remainder = 7x – 9
Quotient = x2 + x – 3
Remainder = 8
Quotient = -x2 – 2
Remainder = -5x + 10
Concept insight: When dividing a polynomial by another polynomial, first arrange the polynomial in the descending powers of the variable. In the process of division, be mindful of the signs of the coefficients of the terms of the polynomials. After doing the division, one can check his answer obtained by the division algorithm which is given below:
Dividend = Divisor x Quotient + Remainder
Also remember that the quotient obtained is just a polynomial.
Question 2. Check whether the first polynomial is a factor of the second polynomial by dividing the
second polynomial by the first polynomial:
Solution:
The polynomial 2t4 + 3t3 – 2t2 – 9t – 12 can be divided by the polynomial t2 – 3 = t2 + 0.t – 3 as follows:
Since the remainder is 0, t² – 3 is a factor of 2t4 + 3t3 – 2t2 – 9t – 12 .
(ii) The polynomial 3x4 + 5x3 – 7x2 + 2x + 2 can be divided by the polynomial x2 + 3x + 1 as follows:
Since the remainder is 0, x² + 3x + 1 is a factor of 3x4 + 5x3 – 7x2 + 2x + 2
(iii) The polynomial x5 – 4x3 + x2 + 3x + 1 can be divided by the polynomial x3 – 3x + 1 as follows:
Since the remainder is not equal to 0, x3 – 3x + 1 is not a factor of x5 – 4x3 + x2 + 3x + 1.
Concept insight: A polynomial g(x) is a factor of another polynomial p(x) if the remainder obtained by dividing p(x) by g(x) is zero and not just a constant. When changing the sign, make sure that you do not change the sign of words that were not included in the previous operation. For example, don’t change the sign of 3x + 1 in the first step of (iii).
Question 3. Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are √(5/3) and – √(5/3).
Solution:
Let p(x) = 3x4 + 6x3 – 2x2 – 10x -5
Now, x² + 2x + 1 = (x + 1)2
So, the two zeroes of x² + 2x + 1 are -1 and -1.
Concept insight: Remember that if (x – a) and (x – b) are factors of a polynomial, then (x – a)(x – b) will also be a factor of that polynomial. Also, if a is the zero of a polynomial p(x), where the power of p(x) is greater than 1, then (x – a) will be a factor of p(x), when p(x) (x – a) ), the remainder obtained will be 0 and the quotient will be a factor of the polynomial p(x). To cross check your answer, the number of zeroes of the polynomial will be less than or equal to the power of the polynomial.
Question 4. On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and
-2x + 4, respectively. Find g(x).
Solution:
Divided, p(x) = x3 – 3x2 + x + 2
Quotient = (x – 2)
Remainder = (-2x + 4)
Let g(x) be the divisor.
According to the division algorithm,
Dividend = Divisor x Quotient + Remainder
Concept Insight: When a polynomial is divided by any other non-zero polynomial, then it satisfies the division algorithm which is as below:
Dividend = Divisor x Quotient + Remainder
Divisor x Quotient = Dividend – Remainder
So, from this relation, the divisor can be obtained by dividing the result of (Dividend – Remainder) by the quotient.
Question 5. Give examples of polynomial p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Solution:
According to the division algorithm, if p(x) and g(x) are two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that
p(x) = g(x) x q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of g(x).
(i) Degree of quotient will be equal to degree of dividend when divisor is constant.
Let us consider the division of 18x2 + 3x + 9 by 3.
Here, p(x) = 18x2 + 3x + 9 and g(x) = 3
q(x) = 6x2 + x + 3 and r(x) = 0
Here, degree of p(x) and q(x) is the same which is 2.
Checking:
p(x) = g(x) x q(x) + r(x)
Thus, the division algorithm is satisfied.
(ii) Let us consider the division of 2x4 + 2x by 2x3,
Here, p(x) = 2x4 + 2x and g(x) = 2x3
q(x) = x and r(x) = 2x
Clearly, the degree of q(x) and r(x) is the same which is 1.
Checking,
p(x) = g(x) x q(x) + r(x)
2x4 + 2x = (2x3 ) x x + 2x
2x4 + 2x = 2x4 + 2x
Thus, the division algorithm is satisfied.
(iii) Degree of remainder will be 0 when remainder obtained on division is a constant.
Let us consider the division of 10x3 + 3 by 5x2.
Here, p(x) = 10x3 + 3 and g(x) = 5x2
q(x) = 2x and r(x) = 3
Clearly, the degree of r(x) is 0.
Checking:
p(x) = g(x) x q(x) + r(x)
10x3 + 3 = (5x2 ) x 2x + 3
10x3 + 3 = 10x3 + 3
Thus, the division algorithm is satisfied.
Concept Insight: To answer these types of questions, one must remember the division algorithm. Also, remember the condition on the remainder of the polynomial r(x). The polynomial r(x) is either 0 or its power is absolutely less than g(x). The answer may not be unique in all cases because there may be many polynomials that satisfy the given conditions.
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