Polynomials form one of the most essential topics in CBSE Class 10 mathematics. They provide the foundation for understanding equations, algebraic expressions, and the relationship between zeroes and coefficients. A clear grasp of polynomial formulas not only helps in solving academic problems efficiently but also builds a strong base for higher mathematics.

In this guide, we cover all major formulas for linear, quadratic, and cubic polynomials, their practical applications, tips to memorize them effectively, and step-by-step examples to help students prepare confidently for their CBSE exams.

Understanding Polynomials and Their Types

Polynomials are algebraic expressions that involve variables raised to whole-number powers. Depending on their degree:

• Linear Polynomial: Degree 1, e.g., ax+bax + bax+b
• Quadratic Polynomial: Degree 2, e.g., ax2+bx+cax^2 + bx + cax2+bx+c
• Cubic Polynomial: Degree 3, e.g., ax3+bx2+cx+dax^3 + bx^2 + cx + dax3+bx2+cx+d

Key Concept: The degree of a polynomial determines the maximum number of roots it can have. Linear polynomials have 1 root, quadratic polynomials have 2, and cubic polynomials have 3.

Essential Polynomial Formulas – Class 10

Quadratic Polynomial

For a quadratic polynomial ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0, with roots α\alphaα and β\betaβ:

• Sum of roots: α+β=−ba\alpha + \beta = -\frac{b}{a}α+β=−ab​
• Product of roots: αβ=ca\alpha \beta = \frac{c}{a}αβ=ac​

Cubic Polynomials

For a cubic polynomial ax3+bx2+cx+d=0ax^3 + bx^2 + cx + d = 0ax3+bx2+cx+d=0, with roots α,β,γ\alpha, \beta, \gammaα,β,γ:

• Sum of roots: α+β+γ=−ba\alpha + \beta + \gamma = -\frac{b}{a}α+β+γ=−ab​
• Sum of product of roots taken two at a time: αβ+βγ+γα=ca\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}αβ+βγ+γα=ac​
• Product of roots: αβγ=−da\alpha\beta\gamma = -\frac{d}{a}αβγ=−ad​
  

Additional Formulas

• The relation between coefficients and roots helps in constructing polynomials from given roots.
• Factorization formula: (x−α)(x−β)=x2−(α+β)x+αβ(x – \alpha)(x – \beta) = x^2 – (\alpha + \beta)x + \alpha\beta(x−α)(x−β)=x2−(α+β)x+αβ
• HCF/LCM of polynomials: Useful in simplifying expressions or solving algebraic fractions.
  

Applications of Polynomial Formulas

Polynomial formulas are not limited to classroom problems. They are widely applied in various fields:

• Algebra & Equations: Solve quadratic and cubic equations efficiently.
• Engineering: Modeling forces, motion, and mechanical systems.
• Economics & Finance: Analyze cost, revenue, and profit patterns using polynomial functions.
• Science & Research: Represent growth patterns, chemical reaction rates, or population studies mathematically.
• Exam-Focused Use: Helps students simplify computations, avoid mistakes, and save time while solving numerical problems in CBSE exams.
  

Tips to Memorize Polynomials Formulas

1. Logical Understanding: Learn formulas by understanding how roots relate to coefficients.
2. Practice Extensively: Solve varied problems, including factorization, sum/product of roots, and polynomial equations.
3. Use Flashcards: Write formulas on one side and examples on the other for quick revision.
4. Group Formulas: Categorize formulas for linear, quadratic, and cubic polynomials separately.
5. Visual Aids: Diagrams showing the relation of roots on the number line can reinforce memory.
6. Daily Revision: Repetition ensures formulas are retained for exams.
   

Examples of Polynomial Formulas in Action

Example 1: Construct a Quadratic Polynomial

Problem: Form a quadratic equation whose roots are 4 and -3.

Solution:

1. Factor form: (x−4)(x+3)=0(x – 4)(x + 3) = 0(x−4)(x+3)=0
2. Expand: x2+3x−4x−12=x2−x−12x^2 + 3x – 4x – 12 = x^2 – x – 12×2+3x−4x−12=x2−x−12
3. Verification:
   • Sum of roots = 4+(−3)=1=−ba4 + (-3) = 1 = -\frac{b}{a}4+(−3)=1=−ab​
   • Product of roots = 4×(−3)=−12=ca4 \times (-3) = -12 = \frac{c}{a}4×(−3)=−12=ac​

Answer: x2−x−12=0x^2 – x – 12 = 0x2−x−12=0

Example 2: Construct a Cubic Polynomial

Problem: Find a cubic polynomial with roots 1, 2, -3.

Solution:

1. Factor form: (x−1)(x−2)(x+3)(x – 1)(x – 2)(x + 3)(x−1)(x−2)(x+3)
2. Multiply step by step:
   (x2−3x+2)(x+3)=x3+0x2−7x+6(x^2 – 3x + 2)(x + 3) = x^3 + 0x^2 – 7x + 6(x2−3x+2)(x+3)=x3+0x2−7x+6
3. Verification:
   • Sum of roots = 1+2−3=0=−ba1 + 2 – 3 = 0 = -\frac{b}{a}1+2−3=0=−ab​
   • Product of roots = 1×2×(−3)=−6=−da1 \times 2 \times (-3) = -6 = -\frac{d}{a}1×2×(−3)=−6=−ad​

Answer: x3−7x+6=0x^3 – 7x + 6 = 0x3−7x+6=0

Example 3: Find Roots Using Sum & Product

Problem: A quadratic polynomial has sum of roots = 5, product of roots = 6. Form the polynomial.

Solution:

x2−(sum of roots)x+(product of roots)=x2−5x+6x^2 – (\text{sum of roots})x + (\text{product of roots}) = x^2 – 5x + 6×2−(sum of roots)x+(product of roots)=x2−5x+6

Answer: x2−5x+6=0x^2 – 5x + 6 = 0x2−5x+6=0