polynomials

Polynomials – NCERT Solutions for Class 10 Maths Chapter 2

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Exercise 2.3 Polynomials

1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

(i) p(x) = x3-3x2+5x–3 , g(x) = x2–2

Solution:

Apologies for the formatting. Let’s perform the division again with proper formatting:

           x - 3
       _______________
x^2 - 2 | x^3 - 3x^2 + 5x - 3
           - (x^3 +0x^2  - 2x)
           _______________
                      - 3x^2 + 7x -3
                     -(-3x^2 -0x + 6)
              _______________
                                   7x - 9

The quotient is x – 3 and the remainder is 7x-9

Therefore, when dividing the polynomial p(x) = x^3 – 3x^2 + 5x – 3 by the polynomial g(x) = x^2 – 2, the quotient is x – 3 and the remainder is 7x-9.

(ii) p(x) = x4-3x2+4x+5 , g(x) = x2+1-x

Solution:

To divide the polynomial p(x) = x^4 – 3x^2 + 4x + 5 by the polynomial g(x) = x^2 + 1 – x, we can perform polynomial long division. Let’s do the division:

                x^2 + x -3
          _____________________
x^2 - x +1 | x^4 - 3x^2 + 4x + 5
          - (x^4 - x^3 + x^2)
          _____________________
                  -x^3 - 4x^2 + 4x + 5
                 +(x^3 - x^2 + x)
                 _____________________
                        3x^2 + 3x + 5
                      - (3x^2 +3x - 3)
                          _____________________
                                    8

The quotient isx^2 + x -3 and the remainder is 8.

Therefore, when dividing the polynomial p(x) = x^4 – 3x^2 + 4x + 5 by the polynomial g(x) = x^2 + 1 – x, the quotient isx^2 + x -3 and the remainder is 8.

(iii) p(x) =x4–5x+6, g(x) = 2–x2

Solution:

To divide the polynomial p(x) = x^4 – 5x + 6 by the polynomial g(x) = 2 – x^2, we can perform polynomial long division. Let’s do the division:

              -x^2 - 2
        ____________________
-x^2 +2  | x^4 + 0x^3 - 5x + 6
        - (x^4 +0x^3-2x^2)
        ____________________
                   2x^2 - 5x + 6
                - (2x^2 +0x  - 4)
              ____________________
                         -5x + 10

The quotient is -x^2 - 2, and the remainder is -5x + 10.

Therefore, when dividing the polynomial p(x) = x^4 – 5x + 6 by the polynomial g(x) = 2 – x^2, the quotient is -x^2 - 2, and the remainder is -5x + 10.

2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

(i) t2-3, 2t4 +3t3-2t2-9t-12

Solution:

First polynomial = t2-3

Second polynomial = 2t4 +3t3-2t2 -9t-12


Apologies for the confusion. You are correct.

Let’s perform the polynomial division again:

                  2t^2 + 3t + 4
          ___________________________
t^2 - 3 | 2t^4 + 3t^3 - 2t^2 - 9t - 12
        -(2t^4 + 0t^3 - 6t^2)
          ___________________________
                   3t^3 + 4t^2 - 9t - 12
                - (3t^3 + 0t^2 - 9t)
                  ___________________________
                          4t^2 - 0t - 12
                         -(4t^2 -0t - 12)
                      -----------------------------
                                                                             0

As we can see, the remainder is left as 0. Therefore, we say that, t2-3 is a factor of 2t4 +3t3-2t2 -9t-12.

(ii)x2+3x+1 , 3x4+5x3-7x2+2x+2

Solution:

The first polynomial is: x^2 + 3x + 1 The second polynomial is: 3x^4 + 5x^3 – 7x^2 + 2x + 2

To perform the division, we’ll use long division. Here’s the step-by-step process

x^2 + 3x + 1 | 3x^4 + 5x^3 – 7x^2 + 2x + 2

– (3x^4 + 9x^3 + 3x^2)

_________________________

-4x^3 – 10x^2 + 2x

– (-4x^3 – 12x^2 – 4x)

___________________________

2x^2 + 6x + 2

– (2x^2 + 6x + 2)

___________________

0

After performing the division, we see that the remainder is zero. Therefore, the first polynomial, x^2 + 3x + 1, is a factor of the second polynomial, 3x^4 + 5x^3 – 7x^2 + 2x + 2.

(iii) x3-3x+1, x5-4x3+x2+3x+1

Solution:

Given,

First polynomial = x3-3x+1

Second polynomial = x5-4x3+x2+3x+1

As we can see, the remainder is not equal to 0. Therefore, we say that, x3-3x+1 is not a factor of x5-4x3+x2+3x+1 .

3. Obtain all other zeroes of 3x4+6x3-2x2-10x-5, if two of its zeroes are √(5/3) and – √(5/3).

Solutions:

From the factorization (3x^2 – 5)(x^2 + 2x + 1), we have two quadratic equations:

  1. 3x^2 – 5 = 0
  2. x^2 + 2x + 1 = 0

Let’s solve these equations separately:

  1. 3x^2 – 5 = 0 By adding 5 to both sides and dividing by 3, we get: 3x^2 = 5 x^2 = 5/3 x = ±√(5/3)
  2. x^2 + 2x + 1 = 0 This quadratic equation can be factored as (x + 1)(x + 1) = 0, which gives us a repeated root of x = -1.

Therefore, the four zeroes of the given polynomial equation, 3x^4 + 6x^3 – 2x^2 – 10x – 5, are: x = √(5/3), -√(5/3), -1, and -1.

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:

5. Give examples of polynomials 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

Solutions:

To satisfy the given conditions of the division algorithm, we can choose the following examples of polynomials:

(i) deg p(x) = deg q(x):
Let’s consider:
p(x) = 2x^3 + 3x^2 – 4x + 1
q(x) = x^2 + 2x – 3

In this example, both p(x) and q(x) are polynomials of degree 3.

(ii) deg q(x) = deg r(x):
Let’s consider:
q(x) = 3x^2 + 5x – 2
r(x) = x – 1

In this example, both q(x) and r(x) are polynomials of degree 1.

(iii) deg r(x) = 0:
Let’s consider:
r(x) = 4

In this example, r(x) is a constant polynomial of degree 0.

These examples satisfy the conditions of the division algorithm as specified.

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