CG Board Class 9 Mathematics Chapter 4 Solutions: Polynomials (बहुपद) – 2026-27

(i) S – 3**

Polynomial. The power of variable S is 1, which is a whole number.

• (ii) 5y^-3

Not a polynomial. The power of variable y is -3, which is a negative integer (not a whole number).

• (iii) p + 1/p

Not a polynomial. Can be written as p + p^-1. The power of p in the second term is -1.

• (iv) ax² + b

Polynomial. The power of variable x is 2, which is a whole number.

• (v) (1)/(x²) + 1

Not a polynomial. Can be written as x^-2 + 1. The power of variable x is -2.

• (vi) 5p² + 2p + 1

Polynomial. Powers of variable p are 2 and 1, which are whole numbers.

A polynomial can have any number of terms, but the number of terms in any given polynomial is always finite. / , (finite)

• Q.2 Is 2p + p a monomial or a binomial?

Simplifying 2p + p gives 3p. Since it contains only 1 term after simplification, it is a monomial. / 2p + p 3p 1 , (monomial)

• Q.3 How many terms does the expression (x + 2)² have?

Expanding (x + 2)² = x² + 4x + 4. It has 3 terms (trinomial). / (x + 2)² = x² + 4x + 4 3

(i) 8**

• (ii) xyz
• (iii) uv + uv² + v³

No. A linear polynomial in one variable can be written in the form ax + b. On collecting similar terms, it can have at most 2 terms. / ax + b 2

(i) x² + 5x + 6, (ii) 2y² – 3y + 1, (iii) t² – 9, (iv) 3u² + 4u, (v) m² + 7.

• Q.2 What is the largest number of terms possible in a quadratic polynomial?

3 terms (in the general form ax² + bx + c). / 3

• Q.3 What is the minimum number of terms possible in a quadratic polynomial?

1 term (e.g., x² or 3x²). / 1

(i) p(y) = y² + 5y, q(y) = 3y – 5**

• (ii) p(r) = 5r² – 9, s(r) = 9r² – 4
• (iii) p(y) = 15y⁴ – 5y² + 27, s(y) = 15y⁴ – 9

(i) p(x) = x² + 3x + 2; q(x) = x² + 3x + 1**

• (ii) p(v) = v² – 3v + 2; q(v) = v + 1
• (iii) p(x) = 2x² + 7x + 3; q(x) = 5x² – 3x
• (iv) p(y) = y³ – y² + y – 1; q(y) = y + 1
• (v) p(u) = 3u² – 12u + 4; q(u) = u² – 2u + 1
• (i) 4x² – 3x + 5
• (ii) z + 3/z
• (iii) √y + 2y + 3 (or y^1/2 + 2y + 3)
• (iv) x² + 2/3x + 2
• (v) x¹⁰ + y³ + t⁵⁰
• (i) 3x³ + 2x² + 3x + 2 (Textbook variation 3x² + 2x² + 3x + 2)
• (ii) 3x² + 1
• (iii) 2 – 5x² + 1/2x³
• (iv) 1/2x² + 2
• (v) x⁴ + x³ + 1/4x²
• (i) x² + 1/5x + 5
• (ii) x² + 7
• (iii) x² + 2x
• (i) Binomial of degree 4 / 4 :
• (ii) Trinomial of degree 6 / 6 :
• (iii) Monomial of degree 5 / 5 :
• (i) x³ – 6x² + x + 1 → Degree = 3
• (ii) y⁹ – 3y⁷ + 3/2y² + 4 → Degree = 9
• (iii) 3 – y³z → Sum of powers = 3 + 1 = 4. Degree = 4
• (iv) x²y – 2x + 1 → Sum of powers = 2 + 1 = 3. Degree = 3
• (v) 5t – 11 → Degree = 1
• (vi) 7 → Constant polynomial 7 = 7x⁰. Degree = 0
• (i) x = 0
• (ii) x = 1
• (iii) x = -2
• (i) p(x) = 4x³ + 2x² – 3x + 2
• (ii) p(r) = (r – 1)(r + 1)
• (iii) p(t) = 2/3t² – 1/3t + 1/3
• (iv) p(y) = (y² – y + 1)(y + 1)
• (v) p(x) = x + 2
• (i) p(x) = 3x + 1; x = -1/3
• (ii) p(x) = x + 2; x = -2
• (iii) p(x) = 5x – 4; x = 5/4
• (iv) p(y) = y² – 1; y = 1, -1
• (v) p(t) = (t + 1)(t – 2); t = +1, -2
• (vi) p(x) = lx + m; x = -m/l
• (vii) p(r) = 3r² – 1; r = -1/√3, 1/√3
• (i) p(x) = x + 6 ⇒ x + 6 = 0 ⇒ x = -6
• (ii) p(x) = x – 6 ⇒ x – 6 = 0 ⇒ x = 6
• (iii) p(y) = 5y ⇒ 5y = 0 ⇒ y = 0
• (iv) p(t) = at, a ≠ 0 ⇒ at = 0 ⇒ t = 0
• (v) p(r) = cr + d, c ≠ 0 ⇒ cr + d = 0 ⇒ r = -d/c
• (vi) p(u) = 3u – 6 ⇒ 3u = 6 ⇒ u = 2
• (vii) r(s) = 2s + 3 ⇒ 2s = -3 ⇒ s = -3/2
• (viii) p(x) = 5 – x ⇒ 5 – x = 0 ⇒ x = 5
• (ix) q(t) = 1/2t – 2/3 ⇒ 1/2t = 2/3 ⇒ t = 4/3
• (i) 2x² + x + 1 and 3x² + 4x + 5
• (ii) 8p² – 3p + 4 and 3p³ – 4p + 7
• (iii) -5x³ + 9x² – 5x + 7 and -2x² + 7x³ – 3x – 8
• (i) 3y² + 2y – 5; 2y² + 5 + 8y and -y² – y
• (ii) 5 + 7r – 3r²; r² + 7 and r² – 3r + 5
• (iii) 4x + 7 – 3x² + 5x³; 7x² – 2x + 1 and -2x³ – 2x
• (i) t² – 5t + 2 from 7t³ – 3t² + 2
• (ii) 3p – 5p² + 7 + 3p³ from 2p² – 5 + 11p – p³
• (iii) 5z³ + 7z² + 2z – 4 from -3z² + 11z + 12z³ + 13
• (i) (3x + 4) by (7x² + 2x + 1)
• (ii) (5x³ + 2x) by (3x² – 9x + 6)
• (iii) (p⁴ – 5p² + 3) by (p³ + 1)