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Solutions ⇒ Class 10th ⇒ Mathematics ⇒ 2. Polynomials

Exercise 2.3 Class 10 maths 2. Polynomials - ncert solutions - Toppers Study

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Exercise 2.3 Class 10 maths 2. Polynomials - ncert solutions - Toppers Study

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Exercise 2.3 class 10 Mathematics Chapter 2. Polynomials

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  • Exercise 2.3 Class 10 Maths 2. Polynomials - Ncert Solutions - Toppers Study
  • Class 10 Ncert Solutions
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  • Chapter 2. Polynomials Exercise 2.3 Class 10

2. Polynomials

| Exercise 2.3 |

Exercise 2.3 Class 10 maths 2. Polynomials - ncert solutions - Toppers Study


Exercise 2.3 class 10 maths chapter 2. Polynomials


Q1. 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

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

(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2

Solution: (i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2

Quotients q(x) = x –  3 and Remainder = 7x –  9

Solution: (ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x

Quotients q(x) = x2 + x –  3 and Remainder = 8

Solution: (iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2

Quotients q(x) = – x2 –  2 and Remainder = –  5x + 10  

Q2. 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

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

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

Solution: (i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12

Hence Remainder r(x) is 0

Therefore, t2 – 3 is the factor of 2t4 + 3t3 – 2t2 – 9t – 12

Solution: (ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2

Hence Remainder r(x) is 0

Therefore,  x2 + 3x + 1 is the factor of 3x4 + 5x3 – 7x2 + 2x + 2

Solution: (iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1

Hence Remainder r(x) = 2

Therefore, x3 – 3x + 1, is not a factor of x5 – 4x3 + x2 + 3x + 1

Q3. Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10– 5, if two of its zeroes are

Solution:

Given that : p(x) = 3x4 + 6x3 – 2x2 – 10– 5

Or 3x2‚Äč - 5 = 0

Therefore, 3x2 - 5 is the factor of p(x)

Now Dividing 3x4 + 6x3 - 2x2 - 10x - 5 by 3x2 - 5

Therefore,  p(x) = (3x2 –  5) (x2 + 2x + 1)

Now, factorizing and getting zeroes x2 + 2x + 1 -

= x2 + x + x + 1 = 0

= x(x + 1) + 1(x + 1) = 0

= (x + 1) (x + 1) = 0 

Or x + 1 = 0, x + 1 = 0

Or x = – 1, x = – 1

Therefore, two zeroes are – 1 and – 1.

Q4. On dividing x3 – 3x2 + + 2 by a polynomial g(x), the quotient and remainder were – 2 and –2+ 4, respectively. Find g(x).

Solution:

Given that: Dividend p(x) = x3 – 3x2 + x + 2

Quotient q(x) = x – 2,

Remainder r(x) = – 2x + 4

Divisor g(x) =?

Dividend = divisor × quotient + remainder

p(x) = g(x) × q(x) + r(x)

x3  3x2 + x + 2 = g(x) (x  2) + ( 2x + 4)

x3  3x2 + x + 2 + 2x  4 = g(x) (x  2)

g(x) (x  2) = x3  3x2 + 3x  2

Dividing x3  3x2 + 3x  2 by x - 2 we obtain g(x)-

Therefore, Divisor g(x) = x2  x + 1

Q5. 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

Solution:

Using Euclid’s Division algorithm:

p(x) = g(x) × q(x) + r(x)  where q(x)  0

(i) deg p(x) = deg q(x)

The deg of dividend p(x) and quotient q(x) can be equal when deg of divisor is 0 or any number.

Example : Let p(x) = 2x2 - 6x + 3

And let g(x) = 2

On dividing  

p(x) = 2x2 - 6x + 2 + 1

     = 2(x2 - 3x + 1) + 1

Now comparing 2(x2 - 3x + 1) + 1 by p(x) = g(x) × q(x) + r(x) we get:

So, q(x) = x2 - 3x + 1and r(x) = 1

By which we obtain deg p(x) = deg q(x)

(ii) deg q(x) = deg r(x)

Solution: This situation comes when deg p(x) and deg g(x) is equal-

Let p(x) = 2x2 + 6x + 7 and g(x) = x2 + 3x + 2

On dividing: q(x) = 2 and r(x) = 3

Therefore, deg q(x) = deg r(x)

(iii) deg r(x) = 0

Solution:

r(x) = 0 is obtained when p(x) is completely divisible by g(x):

Let p(x) = x2 – 1 and g(x) = x + 1

On dividing we obtain:

q(x) = x – 1 and r(x) = 0

 

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Mathematics Class - 11th

NCERT Maths book for CBSE Students.

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Study Materials List:

Solutions ⇒ Class 10th ⇒ Mathematics
1. Real Numbers
2. Polynomials
3. Pair of Linear Equations in Two Variables
4. Quadratic Equations
5. Arithmetic Progressions
6. Triangles
7. Coordinate Geometry
8. Introduction to Trigonometry
9. Some Applications of Trigonometry
10. Circles
11. Constructions
12. Areas Related to Circles
13. Surface Areas and Volumes
14. Statistics
15. Probability

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