📐 Sum of Zeroes:

α + β = -b/a

📐 Product of Zeroes:

αβ = c/a

Part (i): x² – 2x – 8

Step 1: Identify coefficients

Comparing with ax² + bx + c:
a = 1, b = -2, c = -8

Step 2: Find zeroes by factorization

x² – 2x – 8 = 0
We need two numbers whose sum = -2 and product = -8
Numbers: -4 and +2 (because -4 + 2 = -2 and -4 × 2 = -8)

x² – 4x + 2x – 8 = 0
x(x – 4) + 2(x – 4) = 0
(x – 4)(x + 2) = 0

Therefore: x – 4 = 0 or x + 2 = 0
Zeroes: α = 4, β = -2

Step 3: Verify the relationship

Part (ii): 4s² – 4s + 1

Step 1: Identify coefficients

a = 4, b = -4, c = 1

Step 2: Recognize perfect square pattern

Notice: 4s² – 4s + 1 = (2s)² – 2(2s)(1) + (1)²
This is of the form a² – 2ab + b² = (a – b)²

4s² – 4s + 1 = (2s – 1)²
(2s – 1)(2s – 1) = 0

Therefore: 2s – 1 = 0 (twice)
2s = 1
s = 1/2
Zeroes: α = 1/2, β = 1/2 (repeated root)

Step 3: Verify the relationship

Part (iii): 6x² – 3 – 7x

Step 1: Write in standard form

Rearrange: 6x² – 7x – 3
a = 6, b = -7, c = -3

Step 2: Find zeroes by factorization

6x² – 7x – 3 = 0
We need to split -7x such that product = 6 × (-3) = -18
Numbers: -9 and +2 (because -9 + 2 = -7 and -9 × 2 = -18)

6x² – 9x + 2x – 3 = 0
3x(2x – 3) + 1(2x – 3) = 0
(2x – 3)(3x + 1) = 0

Therefore: 2x – 3 = 0 or 3x + 1 = 0
x = 3/2 or x = -1/3
Zeroes: α = 3/2, β = -1/3

Step 3: Verify the relationship

Part (iv): 4u² + 8u

Step 1: Identify coefficients

Notice: constant term c = 0
a = 4, b = 8, c = 0

Step 2: Find zeroes by taking common factor

4u² + 8u = 0
4u(u + 2) = 0

Therefore: 4u = 0 or u + 2 = 0
u = 0 or u = -2
Zeroes: α = 0, β = -2

Step 3: Verify the relationship

⚠️ Important: Don’t forget that 0 is also a zero! Many students miss this.

Part (v): t² – 15

Step 1: Identify coefficients

Notice: middle term b = 0
a = 1, b = 0, c = -15

Step 2: Use difference of squares formula

t² – 15 = 0
t² = 15
t = ±√15

Or using formula: a² – b² = (a + b)(a – b)
t² – (√15)² = (t + √15)(t – √15) = 0
Zeroes: α = √15, β = -√15

Step 3: Verify the relationship

Part (vi): 3x² – x – 4

Step 1: Identify coefficients

a = 3, b = -1, c = -4

Step 2: Find zeroes by factorization

3x² – x – 4 = 0
We need to split -x such that product = 3 × (-4) = -12
Numbers: -4 and +3 (because -4 + 3 = -1 and -4 × 3 = -12)

3x² – 4x + 3x – 4 = 0
x(3x – 4) + 1(3x – 4) = 0
(3x – 4)(x + 1) = 0

Therefore: 3x – 4 = 0 or x + 1 = 0
x = 4/3 or x = -1
Zeroes: α = 4/3, β = -1

Step 3: Verify the relationship

Quadratic Formula:

x = [-b ± √(b² – 4ac)] / 2a

Understanding Different Types of Polynomials

Type 1: Standard Form (ax² + bx + c with all terms)

Examples: x² – 2x – 8, 3x² – x – 4
Method: Factorization by splitting middle term
Key: Find two numbers whose sum = b/a and product = c

Type 2: Perfect Square Trinomials

Example: 4s² – 4s + 1 = (2s – 1)²
Pattern: a² ± 2ab + b² = (a ± b)²
Key: Check if first and last terms are perfect squares and middle term = 2√(first × last)

Type 3: Missing Constant Term (c = 0)

Example: 4u² + 8u
Method: Take common factor (one zero will always be 0)
Key: Factor out the variable completely

Type 4: Missing Middle Term (b = 0)

Example: t² – 15
Method: Difference of squares: a² – b² = (a + b)(a – b)
Key: Zeroes will be equal in magnitude but opposite in sign

Type 5: Not in Standard Form

Example: 6x² – 3 – 7x
First Step: Rearrange to standard form: 6x² – 7x – 3
Then: Apply appropriate method

Graphical Interpretation

The zeroes of a quadratic polynomial represent the x-coordinates where the parabola crosses or touches the x-axis.

• Two distinct zeroes: Parabola crosses x-axis at two points (e.g., x² – 2x – 8 has zeroes at x = 4 and x = -2)
• Two equal zeroes: Parabola touches x-axis at one point (e.g., 4s² – 4s + 1 touches at s = 1/2)
• No real zeroes: Parabola doesn’t touch x-axis (discriminant < 0)

💡 Fun Fact: The sum of zeroes tells you about the axis of symmetry of the parabola! The axis is at x = (α + β)/2 = -b/2a.

❌ Mistake 1: Sign Errors in -b/a Formula

Wrong: For x² – 2x – 8, writing sum = -2/1 = -2
Right: Sum = -b/a = -(-2)/1 = +2
Tip: Pay attention to the negative sign in the formula!

❌ Mistake 2: Forgetting to Write Polynomial in Standard Form

Wrong: Trying to factorize 6x² – 3 – 7x directly
Right: First rearrange to 6x² – 7x – 3, then factorize
Tip: Always write in ax² + bx + c form first!

❌ Mistake 3: Missing Zero as a Zero

Wrong: For 4u² + 8u, writing only u = -2
Right: Zeroes are u = 0 and u = -2
Tip: When constant term is 0, one zero is always 0!

❌ Mistake 4: Calculation Errors in Splitting Middle Term

Wrong: For 6x² – 7x – 3, splitting as 6x² – 6x – x – 3
Right: Split as 6x² – 9x + 2x – 3 (product must equal 6 × -3 = -18)
Tip: Always verify: sum of split terms = middle term, product = a × c

❌ Mistake 5: Not Simplifying Fractions

Wrong: Leaving answer as 8/6
Right: Simplify to 4/3
Tip: Always simplify final answers!

❌ Mistake 6: Incomplete Verification

Wrong: Verifying only sum or only product
Right: Verify BOTH sum and product relationships
Tip: Both conditions must be satisfied for full marks!

❌ Mistake 7: Forgetting ± Sign for Square Roots

Wrong: For t² = 15, writing only t = √15
Right: t = ±√15 (both +√15 and -√15)
Tip: Square root always has two values: positive and negative!

🎯 Trick 1: “Negative Boy” for Sum Formula

Remember: “Sum is NEGATIVE b over a”
α + β = -b/a (the negative sign is crucial!)
Think: “Negative boy (-b) divided by a”

🎯 Trick 2: “Positive Cat” for Product Formula

Remember: “Product is POSITIVE c over a”
αβ = c/a (no negative sign here!)
Think: “Positive cat (c) divided by a”

🎯 Trick 3: ABC Method for Standard Form

A = coefficient of x² (Always first)
B = coefficient of x (Before constant)
C = constant term (Comes last)
Remember: “ABC – Always Before Constant”

🎯 Trick 4: Quick Recognition of Perfect Squares

Check if: Middle term = 2 × √(first term × last term)
Example: 4s² – 4s + 1
√(4s²) = 2s, √1 = 1, and 2 × 2s × 1 = 4s ✓
So it’s (2s – 1)²

🎯 Trick 5: Factor-Pair Method for Splitting

For ax² + bx + c, find factors of (a × c) that add up to b
Quick tip: List factor pairs of (a × c) systematically:
For -18: (1,-18), (2,-9), (3,-6)… stop when you find pair that adds to b

🎯 Trick 6: Zero Detector

If constant term c = 0: One zero is always 0
If middle term b = 0: Zeroes are equal and opposite (±√something)
Remember: “No constant? Zero’s there. No middle? Opposites pair!”

🎯 Trick 7: Verification Shortcut

Don’t recalculate everything! Use your zeroes:
Sum: Just add the two zeroes you found
Product: Just multiply them
Then compare with formulas. Saves time! ⏱️

📌 Exercise 2.1 Questions

Questions asking to find zeroes graphically or verify if a given value is a zero – uses same concepts of finding zeroes.

📌 Exercise 2.2 Question 2

Finding a quadratic polynomial when zeroes are given – reverse application of sum and product formulas!

📌 Exercise 2.3 Questions

Division algorithm problems – you’ll need to find zeroes to verify division results.

📌 Exercise 2.4 Word Problems

Real-life problems that form quadratic equations – you’ll use these same methods to find solutions.

📌 Chapter 4: Quadratic Equations

The entire chapter builds on finding zeroes – same methods, different context. Master this, master Chapter 4!

🚀 1. Projectile Motion (Physics)

When you throw a ball, its height follows a quadratic equation. Finding zeroes tells you when the ball hits the ground!
Example: h(t) = -5t² + 20t (height in meters at time t seconds)
Zeroes: t = 0 (start) and t = 4 (lands after 4 seconds)

💰 2. Profit Maximization (Business)

Companies use quadratic functions to model profit. Zeroes represent break-even points (no profit, no loss).
Between the two zeroes is where profit exists!

🏗️ 3. Architecture & Engineering

Parabolic arches in bridges and buildings use quadratic equations. Zeroes determine where the arch meets the ground.

📱 4. Computer Graphics & Gaming

Curved paths of objects in games (like Angry Birds trajectory) use quadratic equations. Zeroes help calculate landing positions.

🌾 5. Agriculture

Farmers use quadratic models for crop yield vs. fertilizer amount. Zeroes indicate minimum fertilizer needed for any yield.

🎯 6. Sports Analytics

Basketball shot trajectories, soccer ball kicks – all follow parabolic paths. Zeroes help calculate shooting angles and distances.

✅ DO:

• Show all factorization steps clearly
• Write “Therefore, zeroes are α = ___, β = ___”
• For verification, write both LHS and RHS separately
• Use ∴ (therefore) symbol to conclude verification
• Box or underline final answers
• Write “Verified ✓” after each verification

❌ DON’T:

• Skip steps in factorization (even if you can do mentally)
• Write just the final zeroes without showing method
• Forget to verify BOTH sum and product
• Leave answers in unsimplified form (like 8/6 instead of 4/3)
• Use pencil for final answers (use pen)

📊 Word Count: ~200 words per polynomial × 6 = ~1200 words total (can be condensed)

⏱️ Time Required: 6-8 minutes

✅ Expected Marks: 4/4

Question: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

Solution:

(i) x² – 2x – 8

x² – 2x – 8 = 0
x² – 4x + 2x – 8 = 0
x(x – 4) + 2(x – 4) = 0
(x – 4)(x + 2) = 0
∴ x = 4 or x = -2

Zeroes: α = 4, β = -2

Verification:
Here, a = 1, b = -2, c = -8
Sum of zeroes: α + β = 4 + (-2) = 2 = -(-2)/1 = -b/a ✓
Product of zeroes: αβ = 4 × (-2) = -8 = -8/1 = c/a ✓
Hence verified.

(ii) 4s² – 4s + 1

4s² – 4s + 1 = (2s)² – 2(2s)(1) + (1)²
= (2s – 1)²
(2s – 1)(2s – 1) = 0
∴ s = 1/2, 1/2

Zeroes: α = 1/2, β = 1/2

Verification:
Here, a = 4, b = -4, c = 1
Sum of zeroes: α + β = 1/2 + 1/2 = 1 = -(-4)/4 = -b/a ✓
Product of zeroes: αβ = 1/2 × 1/2 = 1/4 = 1/4 = c/a ✓
Hence verified.

(iii) 6x² – 3 – 7x

Rearranging: 6x² – 7x – 3 = 0
6x² – 9x + 2x – 3 = 0
3x(2x – 3) + 1(2x – 3) = 0
(2x – 3)(3x + 1) = 0
∴ x = 3/2 or x = -1/3

Zeroes: α = 3/2, β = -1/3

Verification:
Here, a = 6, b = -7, c = -3
Sum of zeroes: α + β = 3/2 + (-1/3) = 9/6 – 2/6 = 7/6 = -(-7)/6 = -b/a ✓
Product of zeroes: αβ = 3/2 × (-1/3) = -1/2 = -3/6 = c/a ✓
Hence verified.

(iv) 4u² + 8u

4u² + 8u = 0
4u(u + 2) = 0
∴ u = 0 or u = -2

Zeroes: α = 0, β = -2

Verification:
Here, a = 4, b = 8, c = 0
Sum of zeroes: α + β = 0 + (-2) = -2 = -8/4 = -b/a ✓
Product of zeroes: αβ = 0 × (-2) = 0 = 0/4 = c/a ✓
Hence verified.

(v) t² – 15

t² – 15 = 0
t² = 15
∴ t = ±√15

Zeroes: α = √15, β = -√15

Verification:
Here, a = 1, b = 0, c = -15
Sum of zeroes: α + β = √15 + (-√15) = 0 = -0/1 = -b/a ✓
Product of zeroes: αβ = √15 × (-√15) = -15 = -15/1 = c/a ✓
Hence verified.

(vi) 3x² – x – 4

3x² – x – 4 = 0
3x² – 4x + 3x – 4 = 0
x(3x – 4) + 1(3x – 4) = 0
(3x – 4)(x + 1) = 0
∴ x = 4/3 or x = -1

Zeroes: α = 4/3, β = -1

Verification:
Here, a = 3, b = -1, c = -4
Sum of zeroes: α + β = 4/3 + (-1) = 4/3 – 3/3 = 1/3 = -(-1)/3 = -b/a ✓
Product of zeroes: αβ = 4/3 × (-1) = -4/3 = -4/3 = c/a ✓
Hence verified.

✅ Examiner’s Checklist – Did you include:

• ✓ All factorization steps shown clearly
• ✓ Both zeroes stated explicitly
• ✓ Values of a, b, c identified
• ✓ Both sum and product verified
• ✓ “Hence verified” written at end
• ✓ All fractions simplified

🎯 Key Takeaways:

• Zeroes are values of x where polynomial = 0
• For ax² + bx + c: Sum = -b/a, Product = c/a
• Always verify BOTH relationships for full marks
• Factorization is fastest, quadratic formula is most reliable
• Special cases: c = 0 (one zero is 0), b = 0 (zeroes are opposites)
• Show all steps – don’t skip even if you can do mentally!

🚀 Quick Revision Points:

• Standard form: ax² + bx + c (always rearrange first)
• Factorization: Find two numbers with sum = b, product = a×c
• Perfect squares: Check if middle term = 2√(first × last)
• Verification formula: α + β = -b/a, αβ = c/a
• Time management: 6-8 minutes total, ~1 min per polynomial

Q1: What is the relationship between zeroes and coefficients of a quadratic polynomial?

Answer: For a quadratic polynomial ax² + bx + c with zeroes α and β: Sum of zeroes (α + β) = -b/a, and Product of zeroes (αβ) = c/a. These relationships are derived from the factored form a(x – α)(x – β) = ax² + bx + c.

Q2: How do I find zeroes of a quadratic polynomial?

Answer: Methods to find zeroes: 1) Factorization (split middle term) – fastest for simple polynomials, 2) Quadratic formula: x = [-b ± √(b² – 4ac)] / 2a – works for all polynomials, 3) Completing the square method. Choose the easiest method based on the polynomial structure. For exam, factorization is preferred if possible.

Q3: What is the easiest method to solve Exercise 2.2 Question 1?

Answer: For most polynomials in this question, factorization by splitting the middle term is easiest. For polynomials like t² – 15 (no middle term), use the difference of squares formula. For 4s² – 4s + 1, recognize it as a perfect square (2s – 1)². Practice identifying patterns to solve quickly!

Q4: How do I verify the relationship between zeroes and coefficients?

Answer: After finding zeroes α and β: 1) Calculate sum (α + β) and check if it equals -b/a, 2) Calculate product (αβ) and check if it equals c/a. Both conditions must be satisfied. Show all calculations clearly in your answer with proper steps for full marks.

Q5: What are common mistakes in finding zeroes of polynomials?

Answer: Common mistakes: 1) Sign errors when applying -b/a formula, 2) Forgetting to write polynomial in standard form (ax² + bx + c), 3) Calculation errors in factorization, 4) Not simplifying final answers, 5) Missing one zero (especially zero itself in polynomials like 4u² + 8u), 6) Incomplete verification (verifying only sum or only product).

Q6: How many marks is this question worth in board exams?

Answer: This is typically a 4-mark question in CBSE board exams. Mark distribution: Finding zeroes (2 marks), Verifying sum relationship (1 mark), Verifying product relationship (1 mark). Showing all steps is crucial for full marks. Partial marks may be awarded for correct method even with calculation errors.

Q7: What if I can’t factorize the polynomial easily?

Answer: Use the quadratic formula: x = [-b ± √(b² – 4ac)] / 2a. This works for ALL quadratic polynomials. First, ensure the polynomial is in standard form ax² + bx + c, then substitute values of a, b, and c into the formula. It’s a guaranteed method but takes slightly more time than factorization.

Q8: How do I remember the sum and product formulas?

Answer: Memory trick: “Sum is NEGATIVE b over a” (α + β = -b/a), “Product is POSITIVE c over a” (αβ = c/a). Remember: coefficient of x² is ‘a’, coefficient of x is ‘b’, constant term is ‘c’. The negative sign appears only in the sum formula. Use the mnemonic “Negative Boy, Positive Cat” to remember!

Q9: Can a quadratic polynomial have more than 2 zeroes?

Answer: No, a quadratic polynomial (degree 2) can have at most 2 zeroes. It may have: 2 distinct real zeroes, 2 equal real zeroes (repeated root like in 4s² – 4s + 1), or no real zeroes (complex roots when discriminant < 0). The number of zeroes cannot exceed the degree of the polynomial.

Q10: What is the difference between roots and zeroes?

Answer: Roots and zeroes mean the same thing – values of x that make the polynomial equal to zero. ‘Zeroes’ is used when referring to polynomials (like p(x) = x² – 5x + 6), while ‘roots’ is used for equations (like x² – 5x + 6 = 0). For p(x) = 0, the solutions are called both roots and zeroes interchangeably.

Q11: How much time should I spend on this question in the exam?

Answer: For a 4-mark question with 6 parts, allocate 6-8 minutes maximum. Spend ~1 minute per polynomial finding zeroes, then 2 minutes for verification. Practice to increase speed. Don’t spend more than 8 minutes as time management is crucial in board exams. If stuck on one part, move to the next and return later.

Q12: Are there similar questions in other exercises?

Answer: Yes! Similar questions appear in Exercise 2.1 (finding zeroes), Exercise 2.3 (division algorithm – uses zero concept), and Exercise 2.4 (word problems on polynomials). The concept of sum and product of zeroes is fundamental and appears throughout Chapter 2. Also crucial for Chapter 4: Quadratic Equations!

“Exercise 2.2 Question 1 is one of the most important questions in Chapter 2 – Polynomials. It tests two critical skills: finding zeroes and understanding the relationship between zeroes and coefficients. This question appears in almost every board exam in some form.”

Key Insights from an Examiner’s Perspective:

• Step-by-step approach is crucial: Even if you can solve mentally, write all steps. Examiners award marks for method, not just final answer.
• Verification is mandatory: Many students lose 2 marks by skipping verification. Always verify BOTH sum and product relationships.
• Common error – sign mistakes: The negative sign in -b/a is where 60% of students make mistakes. Double-check this!
• Pattern recognition saves time: Quickly identify perfect squares (like 4s² – 4s + 1) and polynomials with missing terms.
• Practice variation: This question type appears with different coefficients. Practice 10-15 similar problems to build speed.

Examiner’s Tip: In board exams, if you’re running short on time, prioritize showing the factorization steps clearly. Even if verification is incomplete, you can still secure 2-3 marks out of 4.

Exercise 2.2 Question 2

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

Connection: Reverse application of the same formulas used in Question 1!

Exercise 2.1 Question 1

The graphs of y = p(x) are given. Find the number of zeroes of p(x).

Connection: Graphical interpretation of zeroes – where parabola crosses x-axis.

Exercise 2.3 Question 1

Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder.

Connection: Uses division algorithm; zeroes help verify division results.

Exercise 2.4 Question 1

Verify that the numbers given alongside of the cubic polynomials are their zeroes.

Connection: Extends zero concept to cubic polynomials (degree 3).

Chapter 4: Quadratic Equations

All questions on solving quadratic equations use the same methods.

Connection: Finding roots of equations = finding zeroes of polynomials!

Problem 1:

Find the zeroes of 2x² + 7x + 3 and verify the relationship.

Hint: Split 7x as 6x + x

Problem 2:

Find the zeroes of 9x² – 6x + 1 and verify the relationship.

Hint: This is a perfect square trinomial!

Problem 3:

Find the zeroes of x² – 7 and verify the relationship.

Hint: Use difference of squares formula

Problem 4:

Find the zeroes of 5x² – 20x and verify the relationship.

Hint: Take common factor; one zero will be 0