Polynomials
Theory, Identities, Factorisation & Exercises — Class 9 Mathematics
📋 Contents
- Introduction to Polynomials
- Terms and Their Coefficients
- Degree of a Polynomial
- Types of Polynomials
- Zeroes of a Polynomial
- Algebraic Identities
- Factors of a Polynomial
- Remainder & Factor Theorem
- Exercise 1 (MCQ)
- Exercise 2 (Short Answer)
- Exercise 3 (Long Answer)
1. Introduction to Polynomials
A polynomial is an algebraic expression having one or more terms involving powers of a variable with non-negative integer exponents.
In general, a polynomial in variable x is written as:
where a₀, a₁, a₂, …, aₙ are real numbers and n is a non-negative integer.
Classification by Number of Terms
| Name | Number of Terms | Examples |
|---|---|---|
Monomial | 1 | 3x², –5y, 7 |
Binomial | 2 | x + 1, 2x² – 3 |
Trinomial | 3 | x² + 2x + 1, 3t³ – 8 + 5t |
Polynomial | 4 or more | x⁴ – x³ + 2x – 1 |
EXAMPLE
Which of the following are polynomials?
(i) 5x³ – 2x + 1 (ii) y² + 3y – 4 (iii) x^(1/2) + 2 (iv) 4t³ – t + 7 (v) y^(–2) + 1
Solution:
- (i) All powers of x are non-negative integers → Polynomial ✓
- (ii) All powers of y are non-negative integers → Polynomial ✓
- (iii) The term x1/2 has a rational (fractional) power → Not a polynomial ✗
- (iv) All powers of t are non-negative integers → Polynomial ✓
- (v) The term y–2 has a negative exponent → Not a polynomial ✗
Want more solved examples on identifying polynomials? Visit Myclass24.com for practice problems and video solutions.
2. Terms and Their Coefficients
In a polynomial p(x) = aₙxⁿ + … + a₁x + a₀, each expression aₙxⁿ, …, a₁x, a₀ is called a term. The constants aₙ, aₙ₋₁, …, a₀ are called coefficients, and a₀ is the constant term.
EXAMPLE
Write the coefficients of x² in each of the following:
(i) x² + 3x – 4 (ii) –2x² + 5x + 1 (iii) (a + d)x² + bx + c
Solution:
- (i) Coefficient of x² is 1
- (ii) Coefficient of x² is –2
- (iii) Coefficient of x² is (a + d)
3. Degree of a Polynomial
The degree of a polynomial is the highest power of the variable present in any term.
| Degree | Type | General Form | Example |
|---|---|---|---|
| 0 | Constant / Zero-degree polynomial | f(x) = a, a ≠ 0 | f(x) = 7 |
| 1 | Linear Polynomial | ax + b, a ≠ 0 | 3x – 5 |
| 2 | Quadratic Polynomial | ax² + bx + c, a ≠ 0 | 2x² – x + 4 |
| 3 | Cubic Polynomial | ax³ + bx² + cx + d, a ≠ 0 | x³ – 3x + 2 |
| 4 | Biquadratic Polynomial | ax⁴ + bx³ + cx² + dx + e, a ≠ 0 | x⁴ – 5x² + 4 |
• Zero degree polynomial: Any non-zero constant, e.g. f(x) = 7. It can be written as 7x⁰, so its degree is 0.
• Zero polynomial: f(x) = 0. Its degree is not defined.
• Polynomials of degree 5 or more have no particular name; they are simply called "polynomial of degree 5 (or 6, …)".
EXAMPLE
Classify as linear, quadratic, or cubic:
(i) x³ – 3x² + 2 (ii) 4t (iii) 1 – x + 5x² – 3x²
Solution:
- (i) Degree 3 → Cubic polynomial
- (ii) Degree 1 → Linear polynomial
- (iii) Simplify first: 1 – x + (5 – 3)x² = 1 – x + 2x² → Degree 2 → Quadratic polynomial
EXAMPLE
Give one example each of: (i) a binomial of degree 53, and (ii) a monomial of degree 90.
Solution:
- (i) A binomial of degree 53: x⁵³ + 1 (has two terms, highest power = 53)
- (ii) A monomial of degree 90: x⁹⁰ (single term, highest power = 90)
4. Types of Polynomials — Summary
(A) Based on Degree
See the table in Section 3 above for linear, quadratic, cubic, and biquadratic types.
(B) Based on Number of Terms
| Name | Terms | Examples |
|---|---|---|
Monomial | 1 | x, 9x², 5x³ |
Binomial | 2 | 2x² + 3x, x + 5x³, –8x³ + 3 |
Trinomial | 3 | 3x³ – 8 + 8x⁴ – 3x², 5 – 7x + 8x⁹ |
5. Zeroes (Roots) of a Polynomial
A value x = a is called a zero of the polynomial p(x) if p(a) = 0. To find a zero, set the polynomial equal to zero and solve.
EXAMPLE
Find the zero of the polynomial p(x) = 2x – 3r.
Solution:
Set p(x) = 0:
2x = 3r
x = 3r / 2
∴ The zero of p(x) is x = 3r/2.
EXAMPLE
Find the zero of q(t) = 3t – 4.
Solution:
∴ The zero is t = 4/3.
EXAMPLE
Find the value of each polynomial at the given point:
(i) p(x) = 5x³ – 4x + 3 at x = –1 (ii) q(y) = 3y³ – 4y + 11 at y = 2 (iii) p(t) = 4t⁴ + 5t³ – t² + 6 at t = a
Solution:
- (i) p(–1) = 5(–1)³ – 4(–1) + 3 = –5 + 4 + 3 = 2
- (ii) q(2) = 3(8) – 4(2) + 11 = 24 – 8 + 11 = 27
- (iii) p(a) = 4a⁴ + 5a³ – a² + 6
EXAMPLE
Find the value of:
(i) 36x² + 49y² + 84xy, when x = 3, y = 6
(ii) 25x² + 16y² – 40xy, when x = 6, y = 7
Solution:
- (i) Recognise the pattern: 36x² + 49y² + 84xy = (6x)² + (7y)² + 2·(6x)·(7y) = (6x + 7y)²
At x = 3, y = 6: (6×3 + 7×6)² = (18 + 42)² = (60)² = 3600 - (ii) 25x² + 16y² – 40xy = (5x)² + (4y)² – 2·(5x)·(4y) = (5x – 4y)²
At x = 6, y = 7: (5×6 – 4×7)² = (30 – 28)² = (2)² = 4
6. Algebraic Identities
An identity is an equality that holds true for all values of the variables involved. Practice these identities with interactive exercises at Myclass24.com.
Standard Identities
| # | Identity |
|---|---|
| (i) | (a + b)² = a² + 2ab + b² |
| (ii) | (a – b)² = a² – 2ab + b² |
| (iii) | a² – b² = (a + b)(a – b) |
| (iv) | a³ + b³ = (a + b)(a² – ab + b²) |
| (v) | a³ – b³ = (a – b)(a² + ab + b²) |
| (vi) | (a + b)³ = a³ + b³ + 3ab(a + b) |
| (vii) | (a – b)³ = a³ – b³ – 3ab(a – b) |
| (viii) | a⁴ + a²b² + b⁴ = (a² + ab + b²)(a² – ab + b²) |
| (ix) | a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca) |
Special Case: If a + b + c = 0, then a³ + b³ + c³ = 3abc. | |
Derived Value Forms
These are useful when certain sums or products are known:
(b) a² + b² = (a – b)² + 2ab [use when a–b and ab are given]
(c) a + b = √[(a – b)² + 4ab]
(d) a – b = √[(a + b)² – 4ab]
(e) a³ + b³ = (a + b)³ – 3ab(a + b)
(f) a³ – b³ = (a – b)³ + 3ab(a – b)
(g) a⁴ + b⁴ = (a² + b²)² – 2a²b² = [(a + b)² – 2ab]² – 2a²b²
(h) a⁵ + b⁵ = (a³ + b³)(a² + b²) – a²b²(a + b)
7. Factors of a Polynomial
If a polynomial f(x) can be expressed as a product of two or more polynomials f₁(x) · f₂(x) · …, then each of them is called a factor of f(x). The process of finding factors is called factorisation.
(a) Factorisation by Making a Trinomial a Perfect Square
Recognise an expression of the form A² – 2AB + B² = (A – B)² or A² + 2AB + B² = (A + B)².
EXAMPLE
Factorise: 81a²b²c² + 64a⁶b² – 144a⁴b²c
Solution:
= (9abc – 8a³b)²
= a²b²(9c – 8a²)²
(b) Factorisation Using a² – b² = (a + b)(a – b)
EXAMPLE
Factorise: 4(2a + 3b – 4c)² – (a – 4b + 5c)²
Solution:
Let A = 2(2a + 3b – 4c) = 4a + 6b – 8c, B = a – 4b + 5c
= (A + B)(A – B)
A + B = (4a + 6b – 8c) + (a – 4b + 5c) = 5a + 2b – 3c
A – B = (4a + 6b – 8c) – (a – 4b + 5c) = 3a + 10b – 13c
∴ = (5a + 2b – 3c)(3a + 10b – 13c)
EXAMPLE
Factorise: x⁴ + x²y² + y⁴
Solution:
= (x² + y²)² – (xy)²
= (x² + y² + xy)(x² + y² – xy)
(c) Factorisation Using a³ ± b³ Formulas
a³ – b³ = (a – b)(a² + ab + b²)
EXAMPLE
Factorise: 64a¹³b + 343ab¹³
Solution:
= ab[(4a⁴)³ + (7b⁴)³]
= ab(4a⁴ + 7b⁴)[(4a⁴)² – (4a⁴)(7b⁴) + (7b⁴)²]
= ab(4a⁴ + 7b⁴)(16a⁸ – 28a⁴b⁴ + 49b⁸)
8. Remainder Theorem & Factor Theorem
These theorems are fundamental for testing factors and finding roots of polynomials. Get step-by-step solutions and video lectures on Myclass24.com.
EXAMPLE
Determine whether (x – 3) is a factor of p(x) = x³ – 3x² + 4x – 12.
Solution:
By the Factor Theorem, check p(3):
Since p(3) = 0, (x – 3) is a factor of p(x). ✓
EXAMPLE
Using the Factor Theorem, prove that p(x) = 4x⁴ + 5x³ – 12x² – 11x + 5 is divisible by g(x) = 4x + 5.
Solution:
If g(x) = 4x + 5 is a factor, then x = –5/4 should give p(–5/4) = 0.
= 4·(625/256) + 5·(–125/64) – 12·(25/16) + 55/4 + 5
= 625/64 – 625/64 – 300/16 + 55/4 + 5
= 0 – 75/4 + 55/4 + 5 = –20/4 + 5 = –5 + 5 = 0 ✓
Hence, (4x + 5) is a factor of p(x).
Exercise 1 — Multiple Choice Questions
- The product of (x + a)(x + b) is:
- The value of 150 × 98 is:
- The expansion of (x + y – z)² is:
- The value of (x + 2y + 2z)² + (x – 2y – 2z)² is:
- The value of 25x² + 16y² + 40xy at x = 1, y = –1 is:
- On simplifying (a + b)³ + (a – b)³ + 6a(a² – b²) we get:
- Find the value when a = –5, b = –6, c = 10 (using a³ + b³ + c³ – 3abc formula):
- If (x + y + z) = 1, xy + yz + zx = –1, xyz = –1, then x³ + y³ + z³ =
- In the method of factorisation, which statement is false?
- Factors of (a + b)³ – (a – b)³ are:
- Degree of the zero polynomial is:
- Factors of (42 – x – x²) are:
- Factors of 6x² + x – 1 are:
- Factors of x³ – 3x² – 10x + 24 are:
- If (x + a) is a factor of both x² + px + q and x² + mx + n, then a =
- Which one is NOT a polynomial?
- The polynomial 4x⁴ + 2x³ – x + 5 is of type:
- Degree of the zero polynomial is:
- The zero of the polynomial 5x – 10 is:
- The number of zeros of p(x) = x² – 4x + 4 is:
- The polynomial p(x) = x is of type:
- The value of k if (x – 1) is a factor of kx² – 3x + k is:
- Degree of polynomial x + x² + 3 is:
- If x + 1/x = 6, the value of x³ + 1/x³ is:
- Value of 104 × 96 is:
- The value of 5.63² + 11.26×2.37 + 2.37² is:
- If x – y = 5 and xy = 84, the value of x³ – y³ is:
- If x + y = 3, xy = 2, then x³ + y³ is:
- If (x – 2) is a factor of x³ – ax² + ax – 2a, the value of a is:
- If one factor of x⁴ – 5x² + 4 is (x – 1), find the other:
- If x = 5 + 2√6, value of x² + 1/x² is:
- When x³ – 3x² + 3x – 1 is divided by x, the remainder is:
- Degree of the polynomial (x² + 2)(x + 3) is:
- Which one is a polynomial in one variable?
- If p(x) = x + 3, then p(x) + p(–x) is:
Answers to Exercise 1
Exercise 2 — Short Answer Problems
- If a + b = 6 and a – b = 2, find the value of a² – b².
- If x = 152, y = –91, find the value of 9x² + 30xy + 25y².
- Evaluate: (i) (5x + 4y)² (ii) (4x – 5y)² (iii) (2x – 1/x)²
- If x + y = 3 and xy = –18, find the value of x³ + y³.
- If a + 1/a = 8, find the value of a³ + 1/a³.
- Evaluate:
- Find the product:
- Evaluate: (i) 102 × 106 (ii) 994 × 1006 (iii) 34 × 36
- Factorise: 4x⁴ + (7a)⁴
- Factorise: x¹² – 1
- Evaluate: ∛(216) × ∛(–125) (or equivalent cube-root expression)
- Write in standard form:
- Factorise: (x + 1)(x + 2)(x + 3)(x + 4) – 3
- Factorise: 64a³ – 27b³ – 144a²b + 108ab²
- Factorise: x⁴ + 2x³y – 2xy³ – y⁴
- Factorise: 8x³ + 16 – 9 (note: verify original expression)
- Factorise: x⁴ + x³ – 7x² – x + 6
- Factorise: 9z³ – 27z² – 100z + 300
- Determine whether (x – 3) is a factor of p(x) = x³ – 3x² + 4x – 12.
- Using Factor Theorem, prove 4x + 5 divides 4x⁴ + 5x³ – 12x² – 11x + 5.
- Determine λ if (x + 1) is a factor of x³ – x² – (2 – λ)x + λ.
- Factorise: x³ – 23x² + 142x – 120
- Factorise: x³ + 13x² + 32x + 20
- Factorise: 2y³ + y² – 2y – 1
- Factorise: 4z³ + 20z² + 33z + 18
- Factorise: x⁴ + 5x² + 4
- Factorise: x³ – 10x² – 53x – 42
Answers to Exercise 2
- 36
- 1
- (i) 25x² + 40xy + 16y²; (ii) 16x² – 40xy + 25y²; (iii) 4x² – 4 + 1/x²
- 189
- 364
- (i) –281250; (ii) 9/10; (iii) –0.018
- (i) x² + 11x + 28; (ii) y² – y/2 – 15/4; (iii) P⁶ + 4096
- (i) 10812; (ii) 999964; (iii) 1224
- (2x² + 49a² + 14ax)(2x² + 49a² – 14ax)
- (x – 1)(x + 1)(x² + 1)(x² + x + 1)(x² – x + 1)(x⁴ – x² + 1)
- 3 (–6 × ½ = –3, take absolute value appropriate to expression)
- (i) x⁶ + 7x⁵ – 3x⁴ + 5x² + x + 4; (ii) m⁷ + 4m⁶ + 8m⁵ – 3m² + 6m – 11
- (x² + 5x + 3)(x² + 5x + 7)
- (4a – 3b)³
- (x – y)(x + y)³
- (2x – 1)(4x² + 2x + 9)
- (x + 1)(x – 1)(x + 3)(x – 2)
- (3z + 10)(z – 3)(3z – 10)
- Yes
- Proven (p(–5/4) = 0)
- λ = 0
- (x – 1)(x – 10)(x – 12)
- (x + 1)(x + 2)(x + 10)
- (y – 1)(y + 1)(2y + 1)
- (z + 2)(2z + 3)²
- (x – 1)(x + 1)(x – 2)(x + 2)
- (x + 1)(x – 14)(x + 3)
Exercise 3 — Long Answer / Higher-Order Problems
- Factorise each of the following:
- For each of the following figures, write an algebraic expression in factorised form for the shaded area. (Refer to diagrams in the original textbook or at Myclass24.com for visual representations of these area problems.)
- If p(x) = x³ – ax² + bx – c is divisible by (x – 1) and (x – 3), find the values of a and b.
- If x + y = 6 and xy = 8, find the value of x³ + y³. Ans: 18, 18
- If a – b = 5 and ab = 84, find the value of a³ – b³. Ans: 370
- If both (x – 1) and (x – 2) are factors of p(x) = x³ + ax² + bx + c, show that the remaining factor can be determined.
- Using the Factor Theorem, factorise:
- The polynomials p(x) = x³ – ax + 3 and q(x) = 2x³ + ax² – 4 when divided by (x – 2) leave remainders R₁ and R₂ respectively. Find a in each case if:
- Let R₁ and R₂ be the remainders when x³ + ax² – 5x + 6 and x³ + 2x² – ax + 8 are divided by (x – 1) and (x + 1) respectively. If R₁ = R₂, find a.
- Simplify: (a + b)³ + (b + c)³ + (c + a)³ – 3(a + b)(b + c)(c + a), given a + b + c = 0.
Answers to Exercise 3
- (i) (a + 1 + b)(a + 1 – b) (ii) (x + y)(x – y + 4) (iii) (7x + y)(–x + 7y) (iv) y(2x – y)(2x² – 2xy + y²) (v) (a – b + 2c)(a – b – 2c) (vi) 2(2x + y)(2x – y)(4x² + y²) [Note: using sum of cubes and simplification] (vii) (x – y)(x + y)(x² + xy + y²)(x² – xy + y²)
- (i) (R² – r²)π = π(R + r)(R – r); (ii) expression depends on diagram — see textbook.
- a = 3, b = –3 (from both p(1)=0 and p(3)=0)
- 18, 18
- 370
- —
- (i) (x – 1)(x – 2)(x – 3) (ii) (x – 2)(x + 3)(2x – 7)
- (i) a = 1; (ii) a = 7/3; (iii) a = 5/3
- a = 3
- 0 (since the three sums (a+b), (b+c), (c+a) add to 2(a+b+c) = 0)
📝 Make Your Own Notes
Chapter 02 – Polynomials | Class 9 Mathematics
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