NCERT Solutions For Class 10 Maths Chapter 2 Ex 2.3 textbook focuses on the division algorithm of polynomials. It helps students learn how to divide one polynomial by another and understand the relationship between dividend, divisor, quotient, and remainder. The problems are based on applying the division algorithm and interpreting degrees of polynomials. These NCERT Solutions For Class 10 Maths Chapter 2 Ex 2.3 are designed to provide clear and step-by-step explanations, helping students gain confidence and accuracy in solving polynomial problems.
NCERT Solutions For Class 10 Maths Chapter 2 Ex 2.3 Textbook
Question 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
(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
We divide:
x - 3
________________
x² – 2 | x³ – 3x² + 5x – 3
x³ - 2x
________________
-3x² + 7x
-3x² + 6
________________
x – 9
Quotient: x – 3
Remainder: x – 9
(ii) Divide:
p(x) = x⁴ – 3x² + 4x + 5
g(x) = x² – x + 1
We divide:
x² + x – 3
______________________
x²–x+1 | x⁴ + 0x³ – 3x² + 4x + 5
x⁴ – x³ + x²
_______________________
x³ – 4x² + 4x
x³ – x² + x
_______________________
–3x² + 3x + 5
–3x² + 3x – 3
_______________________
0x + 8
Quotient: x² + x – 3
Remainder: 8
(iii) Divide:
p(x) = x⁴ – 5x + 6
g(x) = 2 – x² ⇒ write as –x² + 2
Arrange p(x):
x⁴ + 0x³ + 0x² – 5x + 6
We divide:
–x² – 2
______________________
–x²+2 | x⁴ + 0x³ + 0x² – 5x + 6
x⁴ – 2x²
_______________________
0x³ + 2x² – 5x
2x² – 4
_______________________
–5x + 10
Quotient: –x² – 2
Remainder: –5x + 10
Ex 2.3 Class 10 Maths Question 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
(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
(iii) x2 + 3x + 1, x5 – 4x3 + x2 + 3x + 1
Solution:
(i) Divide:
2t⁴ + 3t³ – 2t² – 9t – 12 ÷ t² – 3
2t² + 3t + 4
________________________
t² - 3 | 2t⁴ + 3t³ - 2t² - 9t - 12
- (2t⁴ - 6t²)
________________________
3t³ + 4t²
- (3t³ - 9t)
________________
4t² + 0t -12
- (4t² -12)
______________
0
Remainder = 0 → t² – 3 is a factor
(ii)
Divide:
3x⁴ + 5x³ – 7x² + 2x + 2 ÷ x² + 3x + 1
3x² - 4x + 2
___________________________
x²+3x+1 | 3x⁴ + 5x³ - 7x² + 2x + 2
- (3x⁴ + 9x³ + 3x²)
____________________
-4x³ -10x² + 2x
- (-4x³ -12x² -4x)
__________________
2x² + 6x + 2
- (2x² + 6x + 2)
__________________
0
Remainder = 0 → x² + 3x + 1 is a factor
(iii)
Divide:
x⁵ – 4x³ + x² + 3x + 1 ÷ x² + 3x + 1
x³ - 3x² + 4x - 8
_______________________________
x²+3x+1 | x⁵ + 0x⁴ - 4x³ + x² + 3x + 1
- (x⁵ + 3x⁴ + x³)
________________________
-3x⁴ - 5x³ + x²
- (-3x⁴ -9x³ -3x²)
___________________
4x³ + 4x² + 3x
- (4x³ +12x² + 4x)
__________________
-8x² - x + 1
- (-8x² -24x -8)
__________________
23x + 9
Remainder ≠ 0 → x² + 3x + 1 is NOT a factor
Question:
Obtain all other zeroes of the polynomial
3x⁴ + 6x³ – 2x² – 10x – 5,
if two of its zeroes are √5/3 and –√5/3.
Solution:
Two zeroes given are √5/3 and –√5/3.
These form the factor:
(x – √5/3)(x + √5/3) = x² – 5/9 = (9x² – 5)/9
So, the factor with integer coefficients is 9x² – 5
x² + 2x – 1
_______________________________
9x² - 5 | 3x⁴ + 6x³ – 2x² – 10x – 5
– (3x⁴ – 5x²)
______________________
6x³ + 3x²
– (6x³ – 10x)
__________________
3x² – 10x
– (3x² – 5)
________________
–10x – 0
Now divide:
3x⁴ + 6x³ – 2x² – 10x – 5 = (9x² – 5)(x² + 2x – 1)
Now solve:
x² + 2x – 1 = 0
Using the quadratic formula:
x = (–2 ± √8)/2 = –1 ± √2
Ex 2.3 Class 10 Maths Question 4.
On dividing x3 – 3x2 + x + 2bya polynomial g(x), the quotient and remainder were x – 2 and -2x + 4 respectively. Find g(x).
Solution:
Let’s write it as: Dividend = Divisor × Quotient + Remainder
g(x) × (x – 2) + (–2x + 4) = x³ – 3x² + x + 2
We want to isolate g(x)
Let’s move remainder to the left side:
g(x) × (x – 2) = x³ – 3x² + x + 2 – (–2x + 4)
= x³ – 3x² + x + 2 + 2x – 4
= x³ – 3x² + 3x – 2
Now divide x³ – 3x² + 3x – 2 by x – 2 to find g(x).
x² – x + 1
_________________________
x – 2 | x³ – 3x² + 3x – 2
–(x³ – 2x²)
______________
–x² + 3x
–(–x² + 2x)
___________
x – 2
–(x – 2)
________
0
g(x) = x² – x + 1
Ex 2.3 Class 10 Maths Question 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
Solution:
(i) deg p(x) = deg q(x)
Let:
p(x) = x + 2
g(x) = 1
q(x) = x + 2
r(x) = 0
Then:
p(x) = g(x) × q(x) + r(x)
p(x) = 1 × (x + 2) + 0 = x + 2
Degree of p(x) = 1
Degree of q(x) = 1
(ii) deg q(x) = deg r(x)
Let:
g(x) = x
q(x) = x + 1
r(x) = 2x + 3
Then:
p(x) = g(x) × q(x) + r(x)
p(x) = x(x + 1) + (2x + 3)
p(x) = x² + x + 2x + 3 = x² + 3x + 3
Degree of q(x) = 1
Degree of r(x) = 1
(iii) deg r(x) = 0
Let:
g(x) = x + 2
q(x) = x + 3
r(x) = 4
Then:
p(x) = g(x) × q(x) + r(x)
p(x) = (x + 2)(x + 3) + 4
p(x) = x² + 5x + 6 + 4 = x² + 5x + 10
Degree of r(x) = 0
New Syllabus – NCERT Solutions For Class 10 Maths Chapter 2 Ex 2.3
Question 1:
Verify that the numbers given alongside the polynomials are zeroes of the respective polynomials. Also, verify the relationship between the zeroes and the coefficients in each case.
(i) Polynomial: x² – 2x – 3; Zeroes: 3, –1
Factor: (x – 3)(x + 1)
Zeroes: 3 and –1
Sum = 3 + (–1) = 2
Product = 3 × (–1) = –3
Check: –b/a = –(–2)/1 = 2, c/a = –3
✔ Verified
(ii) Polynomial: x² + 4x + 3; Zeroes: –1, –3
Factor: (x + 1)(x + 3)
Zeroes: –1 and –3
Sum = –1 + (–3) = –4
Product = –1 × (–3) = 3
Check: –b/a = –4, c/a = 3
✔ Verified
(iii) Polynomial: x² + x – 6; Zeroes: 2, –3
Factor: (x – 2)(x + 3)
Zeroes: 2 and –3
Sum = 2 + (–3) = –1
Product = 2 × (–3) = –6
Check: –b/a = –1, c/a = –6
✔ Verified
(iv) Polynomial: x² + 2x – 8; Zeroes: 2, –4
Factor: (x – 2)(x + 4)
Zeroes: 2 and –4
Sum = 2 + (–4) = –2
Product = 2 × (–4) = –8
Check: –b/a = –2, c/a = –8
✔ Verified
(v) Polynomial: x² – 4; Zeroes: 2, –2
Factor: (x – 2)(x + 2)
Zeroes: 2 and –2
Sum = 2 + (–2) = 0
Product = 2 × (–2) = –4
Check: –b/a = 0, c/a = –4
✔ Verified
(vi) Polynomial: x² – 5x + 6; Zeroes: 2, 3
Factor: (x – 2)(x – 3)
Zeroes: 2 and 3
Sum = 2 + 3 = 5
Product = 2 × 3 = 6
Check: –b/a = 5, c/a = 6
✔ Verified
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NCERT Solutions For Class 10 Maths Chapter 2 Ex 2.3 help students understand the concept of polynomial division thoroughly. By solving problems step by step, students develop confidence in handling remainders and finding missing polynomials.
NCERT Solutions For Class 10 Maths Chapter 2 Ex 2.3 strengthen the foundation of algebra by applying the division algorithm. Practicing these questions ensures better performance in exams and competitive tests. Hence, students are encouraged to revise
NCERT Solutions For Class 10 Maths Chapter 2 Ex 2.3 regularly. For complete mastery, go through NCERT Solutions For Class 10 Maths Chapter 2 Ex 2.3 multiple times and test your understanding by revisiting tricky questions.