CG Board Class 10 Maths Chapter 1 Solutions: Polynomials (बहुपद) (2026-27)

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CG Board (CGBSE / SCERT Chhattisgarh) Class 10 Mathematics Solutions | 2026-27 Edition

📚 Chapter Overview:

  • Board: Chhattisgarh Board of Secondary Education (CGBSE) / CG SCERT Raipur
  • Subject & Unit: Class 10 Mathematics – Unit 1: Algebra (2026-27 Edition)
  • Chapter 1: Polynomials (बहुपद)
  • Medium: Bilingual Step-by-Step Solutions (English Medium + हिंदी माध्यम)
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  • *Polynomial: An algebraic expression in which the exponents of the variables are non-negative integers.
  • *Division Algorithm for Polynomials

Dividend = (Divisor × Quotient) + Remainder

  • *Remainder Theorem

If a polynomial f(x) is divided by (x – a), then the remainder is f(a). If the divisor is (x + a), the remainder is f(-a).

  • *Factor Theorem

If f(a) = 0 for a polynomial f(x), then (x – a) is a factor of f(x). Conversely, if (x – a) is a factor of f(x), then the remainder f(a) = 0.

  • *Relationship between Zeroes and Coefficients of a Quadratic Polynomial

For a quadratic polynomial ax2 + bx + c (a ≠ 0) having zeroes α and β:

  • *Sum of zeroes

α + β = -b/a = -(Coefficient of x)/(Coefficient of x2)

  • *Product of zeroes

α · β = c/a = (Constant term)/(Coefficient of x2)

  • *Forming Factors from Zeroes

If α and β are zeroes of a quadratic polynomial, its factors are (x – α) and (x – β).

Q1

Find the quotient and remainder on dividing polynomial…


📝 Question:

Find the quotient and remainder on dividing polynomial x2 – x + 1 by x + 1

💡 Solution:

Given Dividend = x2 – x + 1, Divisor = x + 1.

*Long Division Steps

  1. Divide first term of dividend by first term of divisor: x2/x = x (1st term of quotient).
  2. Multiply x(x + 1) = x2 + x. Subtract from x2 – x + 1

(x2 – x + 1) – (x2 + x) = -2x + 1

  1. Divide -2x by x: -2x/x = -2 (2nd term of quotient).
  2. Multiply -2(x + 1) = -2x – 2. Subtract from -2x + 1

(-2x + 1) – (-2x – 2) = 1 + 2 = 3

  • *Quotient = x – 2
  • *Remainder = 3

Q2

Find the quotient and remainder on dividing 6xsup2sup…


📝 Question:

Find the quotient and remainder on dividing 6x2 – 5x + 1 by 2x – 1

💡 Solution:

Given Dividend = 6x2 – 5x + 1, Divisor = 2x – 1.

*Steps

  1. (6x2)/(2x) = 3x. Multiply 3x(2x – 1) = 6x2 – 3x.
  2. Subtract: (6x2 – 5x + 1) – (6x2 – 3x) = -2x + 1.
  3. -2x/2x = -1. Multiply -1(2x – 1) = -2x + 1.
  4. Subtract: (-2x + 1) – (-2x + 1) = 0.
  • *Quotient = 3x – 1
  • *Remainder = 0

Q3

Find the quotient and remainder on dividing 2ysup3sup…


📝 Question:

Find the quotient and remainder on dividing 2y3 + 4y2 + 3y + 1 by y + 1

💡 Solution:

Given Dividend = 2y3 + 4y2 + 3y + 1, Divisor = y + 1.

*Steps

  1. (2y3)/(y) = 2y2. Multiply 2y2(y + 1) = 2y3 + 2y2.
  2. Subtract: (2y3 + 4y2 + 3y + 1) – (2y3 + 2y2) = 2y2 + 3y + 1.
  3. (2y2)/(y) = 2y. Multiply 2y(y + 1) = 2y2 + 2y.
  4. Subtract: (2y2 + 3y + 1) – (2y2 + 2y) = y + 1.
  5. y/y = 1. Multiply 1(y + 1) = y + 1.
  6. Subtract: (y + 1) – (y + 1) = 0.
  • *Quotient = 2y2 + 2y + 1
  • *Remainder = 0

Q4

Find the quotient and remainder on dividing xsup5sup…


📝 Question:

Find the quotient and remainder on dividing x5 + 5x + 3x2 + 5x3 + 3 by 4x + x2 + 2

💡 Solution:

First, arrange terms in descending order of exponents

  • Dividend = x5 + 0x4 + 5x3 + 3x2 + 5x + 3
  • Divisor = x2 + 4x + 2

*Steps

  1. x5/x2 = x3. Multiply x3(x2 + 4x + 2) = x5 + 4x4 + 2x3.

Subtract: -4x4 + 3x3 + 3x2 + 5x + 3.

  1. (-4x4)/(x2) = -4x2. Multiply -4x2(x2 + 4x + 2) = -4x4 – 16x3 – 8x2.

Subtract: 19x3 + 11x2 + 5x + 3.

  1. (19x3)/(x2) = 19x. Multiply 19x(x2 + 4x + 2) = 19x3 + 76x2 + 38x.

Subtract: -65x2 – 33x + 3.

  1. (-65x2)/(x2) = -65. Multiply -65(x2 + 4x + 2) = -65x2 – 260x – 130.

Subtract: (-33x + 3) – (-260x – 130) = 227x + 133.

  • *Quotient = x3 – 4x2 + 19x – 65
  • *Remainder = 227x + 133

Q5

Find the quotient and remainder on dividing xsup2sup…


📝 Question:

Find the quotient and remainder on dividing x2 – 2xy + y2 by x – y

💡 Solution:

Since x2 – 2xy + y2 = (x – y)2

((x – y)2)/(x – y) = x – y

  • *Quotient = x – y
  • *Remainder = 0

Q6

Divide polynomial a by polynomial a b


📝 Question:

Divide polynomial a by polynomial a – b

💡 Solution:

Dividend = a, Divisor = a – b.

*Steps

  1. a/a = 1 (Quotient).
  2. Multiply 1(a – b) = a – b.
  3. Subtract: a – (a – b) = a – a + b = b.
  • *Quotient = 1
  • *Remainder = b

Q7

If divisor 3xsup2sup 2x 2 quotient x 1…


📝 Question:

If divisor = 3x2 – 2x + 2, quotient = x + 1, remainder = 3, then what is the dividend?

💡 Solution:

Using Division Algorithm

Dividend = (Divisor × Quotient) + Remainder

= (3x2 – 2x + 2)(x + 1) + 3

= 3x2(x + 1) – 2x(x + 1) + 2(x + 1) + 3

= (3x3 + 3x2) – (2x2 + 2x) + (2x + 2) + 3

= 3x3 + (3x2 – 2x2) + (-2x + 2x) + (2 + 3)

= 3x3 + x2 + 5

  • *Dividend = 3x3 + x2 + 5

Q8

If divisor 4x 7 quotient x 1 remainder…


📝 Question:

If divisor = 4x – 7, quotient = x + 1, remainder = 0, then what is the dividend?

💡 Solution:

Dividend = (Divisor × Quotient) + Remainder

= (4x – 7)(x + 1) + 0

= 4x(x + 1) – 7(x + 1) = 4x2 + 4x – 7x – 7 = 4x2 – 3x – 7

  • *Dividend = 4x2 – 3x – 7

Q9

Prove that on dividing the polynomial 4xsup3sup 3xsup2sup…


📝 Question:

. Prove that on dividing the polynomial 4x3 + 3x2 + 2x – 9 by x – 1, the remainder is zero.**

💡 Solution:

  • Proof /

Let p(x) = 4x3 + 3x2 + 2x – 9.

Divisor g(x) = x – 1.

By Remainder Theorem, when p(x) is divided by x – 1, remainder R = p(1).

p(1) = 4(1)3 + 3(1)2 + 2(1) – 9

= 4(1) + 3(1) + 2 – 9 = 4 + 3 + 2 – 9 = 9 – 9 = 0

Since remainder R = 0, it is proved that the remainder is zero on dividing by x – 1.

Q10

Verify whether on dividing polynomial xsup2sup 5x 3…


📝 Question:

Verify whether on dividing polynomial x2 – 5x + 3 by x – 3 the remainder is zero or not

💡 Solution:

Let p(x) = x2 – 5x + 3.

Divisor = x – 3.

By Remainder Theorem, remainder R = p(3).

p(3) = (3)2 – 5(3) + 3 = 9 – 15 + 3 = 12 – 15 = -3

Since p(3) = -3 ≠ 0, the remainder is not zero.

Q11

If the area of a rectangle is 45xsup2sup…


📝 Question:

If the area of a rectangle is 45x2 + 30x square meter and its breadth is 15x meter, then what will be the length?

💡 Solution:

  • Area of rectangle = 45x2 + 30x m2
  • Breadth = 15x m
  • Length = (Area)/(Breadth) = (45x2 + 30x)/(15x)

Length = (45x2)/(15x) + 30x/15x = 3x + 2

  • *Length = (3x + 2) meters

Q12

A line segment AB whose length is 28x…


📝 Question:

A line segment AB whose length is 28x unit is to be divided into two equal parts. What will be the length of each part?

💡 Solution:

  • Total length of line segment AB = 28x units
  • Length of each part = 28x/2 = 14x units
  • *Length of each part = 14x units

#### *Exercise – 2

Q1

If px xsup3sup 3xsup2sup 5x 8 is divided…


📝 Question:

If p(x) = x3 + 3x2 – 5x + 8 is divided by the following, then find the remainder with the help of remainder theorem

💡 Solution:

(i) x + 1**

  • Divisor x + 1 = 0 ⇒ x = -1.
  • Remainder R = p(-1) = (-1)3 + 3(-1)2 – 5(-1) + 8 = -1 + 3(1) + 5 + 8 = -1 + 3 + 5 + 8 = 15.
  • (ii) 2x – 1
  • Divisor 2x – 1 = 0 ⇒ x = 1/2.
  • Remainder R = p(1/2) = (1/2)3 + 3(1/2)2 – 5(1/2) + 8 = 1/8 + 3/4 – 5/2 + 8.
  • Taking LCM = 8

R = (1 + 6 – 20 + 64)/(8) = 51/8

  • (iii) x + 2
  • Divisor x + 2 = 0 ⇒ x = -2.
  • Remainder R = p(-2) = (-2)3 + 3(-2)2 – 5(-2) + 8 = -8 + 3(4) + 10 + 8 = -8 + 12 + 10 + 8 = 22.
  • (iv) x – 4
  • Divisor x – 4 = 0 ⇒ x = 4.
  • Remainder R = p(4) = (4)3 + 3(4)2 – 5(4) + 8 = 64 + 3(16) – 20 + 8 = 64 + 48 – 20 + 8 = 100.
  • (v) x + 1/3
  • Divisor x + 1/3 = 0 ⇒ x = -1/3.
  • Remainder R = p(-1/3) = (-1/3)3 + 3(-1/3)2 – 5(-1/3) + 8 = -1/27 + 3(1/9) + 5/3 + 8 = -1/27 + 1/3 + 5/3 + 8 = -1/27 + 2 + 8 = 10 – 1/27 = 269/27.

Q2

Verify in the following whether gx is a…


📝 Question:

Verify in the following whether g(x) is a factor of p(x)

💡 Solution:

(i) g(x) = x – 3, p(x) = x3 – 4x2 + x + 6**

  • Put x = 3 in p(x)

p(3) = (3)3 – 4(3)2 + 3 + 6 = 27 – 4(9) + 9 = 27 – 36 + 9 = 0

  • Since remainder p(3) = 0, by Factor Theorem, g(x) is a factor of p(x).
  • (ii) g(x) = x + 1, p(x) = 2x3 + x2 – 2x + 1
  • Put x = -1 in p(x)

p(-1) = 2(-1)3 + (-1)2 – 2(-1) + 1 = 2(-1) + 1 + 2 + 1 = -2 + 4 = 2

  • Since p(-1) = 2 ≠ 0, g(x) is NOT a factor of p(x).
  • (iii) g(x) = x – 2, p(x) = x4 – x3 – x2 – x – 2
  • Put x = 2 in p(x)

p(2) = (2)4 – (2)3 – (2)2 – 2 – 2 = 16 – 8 – 4 – 2 – 2 = 16 – 16 = 0

  • Since p(2) = 0, g(x) is a factor of p(x).
  • (iv) g(x) = x – 1, p(x) = x3 + 5x2 – 5x + 1
  • Put x = 1 in p(x)

p(1) = (1)3 + 5(1)2 – 5(1) + 1 = 1 + 5 – 5 + 1 = 2

  • Since p(1) = 2 ≠ 0, g(x) is NOT a factor of p(x).
  • (v) g(y) = y + 4, p(y) = y2 + 2y – 1
  • Put y = -4 in p(y)

p(-4) = (-4)2 + 2(-4) – 1 = 16 – 8 – 1 = 7

  • Since p(-4) = 7 ≠ 0, g(y) is NOT a factor of p(y).

Q3

Find the value of a when gx is…


📝 Question:

Find the value of a when g(x) is a factor of p(x)

💡 Solution:

(i) g(x) = x + 1, p(x) = x2 + ax + 2**

  • Since g(x) is a factor, p(-1) = 0.
  • (-1)2 + a(-1) + 2 = 0 ⇒ 1 – a + 2 = 0 ⇒ 3 – a = 0 ⇒ a = 3.
  • a = 3
  • (ii) g(x) = x – 1, p(x) = ax2 – 5x + 3
  • Since g(x) is a factor, p(1) = 0.
  • a(1)2 – 5(1) + 3 = 0 ⇒ a – 5 + 3 = 0 ⇒ a – 2 = 0 ⇒ a = 2.
  • a = 2
  • (iii) g(x) = x + 2, p(x) = 2x2 + 6x + a
  • Since g(x) is a factor, p(-2) = 0.
  • 2(-2)2 + 6(-2) + a = 0 ⇒ 2(4) – 12 + a = 0 ⇒ 8 – 12 + a = 0 ⇒ -4 + a = 0 ⇒ a = 4.
  • a = 4
  • (iv) g(t) = t – 3, p(t) = t2 + 2at – 2a + 3
  • Since g(t) is a factor, p(3) = 0.
  • (3)2 + 2a(3) – 2a + 3 = 0 ⇒ 9 + 6a – 2a + 3 = 0 ⇒ 4a + 12 = 0 ⇒ 4a = -12 ⇒ a = -3.
  • a = -3
  • (v) g(y) = y + 5, p(y) = y2 – 2y + a
  • Since g(y) is a factor, p(-5) = 0.
  • (-5)2 – 2(-5) + a = 0 ⇒ 25 + 10 + a = 0 ⇒ 35 + a = 0 ⇒ a = -35.
  • a = -35

Q4

If a polynomial fx is divided by xsup2sup…


📝 Question:

If a polynomial f(x) is divided by x2 – 9 then remainder is 3x + 2, what will be the remainder when the same polynomial is divided by (x – 3)?

💡 Solution:

By Division Algorithm

f(x) = (x2 – 9) · q(x) + (3x + 2)

Since x2 – 9 = (x – 3)(x + 3)

f(x) = (x – 3)(x + 3) · q(x) + (3x + 2)

By Remainder Theorem, when f(x) is divided by (x – 3), remainder R = f(3)

f(3) = (3 – 3)(3 + 3) · q(3) + [3(3) + 2]

f(3) = 0 · q(3) + (9 + 2) = 0 + 11 = 11

  • *Remainder = 11

Q5

If a polynomial fx is divided by xsup2sup…


📝 Question:

If a polynomial f(x) is divided by x2 – 16 then remainder is 5x + 3, what will be the remainder when the same polynomial is divided by (x + 4)?

💡 Solution:

(x – 4)(x + 1)

  • 2. x2 + 2x + 1
  • Product A · C = 1, sum B = 2. Numbers are +1 and +1.
  • x2 + x + x + 1 = x(x + 1) + 1(x + 1) = (x + 1)(x + 1).

(x + 1)(x + 1)

  • 3. x2 + x – 12
  • Product A · C = -12, sum B = +1. Numbers are +4 and -3.
  • x2 + 4x – 3x – 12 = x(x + 4) – 3(x + 4) = (x + 4)(x – 3).

(x + 4)(x – 3)

  • 4. x2 – 8x + 15
  • Product A · C = 15, sum B = -8. Numbers are -5 and -3.
  • x2 – 5x – 3x + 15 = x(x – 5) – 3(x – 5) = (x – 5)(x – 3).

(x – 5)(x – 3)

  • 5. t2 – 4t – 21
  • Product A · C = -21, sum B = -4. Numbers are -7 and +3.
  • t2 – 7t + 3t – 21 = t(t – 7) + 3(t – 7) = (t – 7)(t + 3).

(t – 7)(t + 3)

  • 6. -y2 + 35y + 156 (given in textbook as 156 + 35y – y2)
  • Product A · C = -1 × 156 = -156, sum B = 35. Numbers are +39 and -4.
  • -y2 + 39y – 4y + 156 = -y(y – 39) – 4(y – 39) = (39 – y)(y + 4) = -(y – 39)(y + 4).

(39 – y)(y + 4)

  • 7. 7x2 – 2x – 5
  • Product A · C = 7 × (-5) = -35, sum B = -2. Numbers are -7 and +5.
  • 7x2 – 7x + 5x – 5 = 7x(x – 1) + 5(x – 1) = (7x + 5)(x – 1).

(7x + 5)(x – 1)

  • 8. 12x2 – 24x + 12
  • Taking common factor 12: 12(x2 – 2x + 1) = 12(x – 1)(x – 1) = (12x – 12)(x – 1).

(12x – 12)(x – 1)

  • 9. 6x2 – 7x – 3
  • Product A · C = 6 × (-3) = -18, sum B = -7. Numbers are -9 and +2.
  • 6x2 – 9x + 2x – 3 = 3x(2x – 3) + 1(2x – 3) = (2x – 3)(3x + 1).

(2x – 3)(3x + 1)

  • 10. 14y2 + 19y – 3
  • Product A · C = 14 × (-3) = -42, sum B = 19. Numbers are +21 and -2.
  • 14y2 + 21y – 2y – 3 = 7y(2y + 3) – 1(2y + 3) = (2y + 3)(7y – 1).

(2y + 3)(7y – 1)

  • 11. √3y2 + 9y + 6√3
  • Product A · C = √3 × 6√3 = 18, sum B = 9. Numbers are +6 and +3.
  • √3y2 + 3y + 6y + 6√3 = √3y(y + √3) + 6(y + √3) = (y + √3)(√3y + 6).

(y + √3)(√3y + 6)

  • 12. 144x2 + 24x + 1
  • Product A · C = 144, sum B = 24. Numbers are +12 and +12.
  • 144x2 + 12x + 12x + 1 = 12x(12x + 1) + 1(12x + 1) = (12x + 1)2.

(12x + 1)2

Q1

Zeroes of some polynomials of the form axsup2sup…


📝 Question:

Zeroes of some polynomials of the form ax2 + bx + c are given below. Find the factors of the polynomials: Formula*: If zeroes are α and β, factors are (x – α) and (x – β)

💡 Solution:

(i) (3, 4)**

  • Factors = (x – 3)(x – 4)
  • (ii) (-2, -3)
  • Factors = (x – (-2))(x – (-3)) = (x + 2)(x + 3)
  • (iii) (1/2, -1/2)
  • Factors = (x – 1/2)(x + 1/2)
  • (iv) (15, 17)
  • Factors = (x – 15)(x – 17)
  • (v) (-18, 12)
  • Factors = (x – (-18))(x – 12) = (x + 18)(x – 12)

Q2

Find the sum and product of zeroes of…


📝 Question:

Find the sum and product of zeroes of the following polynomials: Formula*: For ax2 + bx + c, Sum (α + β) = -b/a, Product (α · β) = c/a

💡 Solution:

(i) x2 + 10x + 24**

  • a = 1, b = 10, c = 24.
  • (ii) 2x2 – 7x – 9
  • a = 2, b = -7, c = -9.
  • (iii) x2 + 11x + 30
  • a = 1, b = 11, c = 30.
  • (iv) -5x2 + 3x + 4
  • a = -5, b = 3, c = 4.
  • (v) x2 + x – 12
  • a = 1, b = 1, c = -12.

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