Introduction

Introduction to Polynomials

Concept Introduction: Polynomials in Real Life

Have you ever wondered how companies calculate profit when they produce goods? Imagine a factory manufacturing notebooks. The total cost might be ₹100 for setup plus ₹5 per notebook. If they produce x notebooks, the cost is 100 + 5x. The revenue from selling them at ₹8 each is 8x. The profit is revenue minus cost: 8x − (100 + 5x) = 3x − 100.

This expression 3x − 100 is a polynomial — a mathematical expression involving variables and coefficients. When the factory asks "How many notebooks must we sell to break even?", they're finding the zero of the polynomial 3x − 100, which is x = 100/3 ≈ 34 notebooks.

Polynomials appear everywhere: calculating areas, predicting projectile motion, optimizing phone signal strength, and even in Google's search algorithms. Understanding polynomials unlocks the door to algebra, calculus, and real-world problem-solving.

{{FORMULA: expr=p(x) = axⁿ + bxⁿ⁻¹ + ... + cx + d | symbols=a,b,c,d:real coefficients, n:degree (highest power), x:variable}}

{{KEY: type=concept | title=What is a Zero of a Polynomial? | text=A real number k is a zero of polynomial p(x) if p(k) = 0. It represents the x-coordinate where the graph intersects the x-axis. For example, if p(x) = 2x + 3, then k = −3/2 is its zero because p(−3/2) = 0.}}

Definitions & Core Vocabulary

What Makes an Expression a Polynomial?

Polynomials must satisfy specific rules. Let's understand through logical reasoning:

Valid Polynomials (✅)

An expression is a polynomial if:

1. All exponents are whole numbers (0, 1, 2, 3, ...)

4x + 2  →  degree 1 ✅

x² − 3x + 5  →  degree 2 ✅

7u⁶ − u⁴ + 4u² − u + 8  →  degree 6 ✅

2. Coefficients are real numbers (including fractions, decimals)

(2/3)x² + (√5)x − π  →  valid ✅

3. No division by variables

3x² + 5  →  valid ✅

Invalid Expressions (❌)

An expression is NOT a polynomial if:

1. Negative or fractional exponents appear

x⁻¹ = 1/x  →  NOT a polynomial ❌

√x = x^(1/2)  →  NOT a polynomial ❌

2. Variable in the denominator

1/(x + 1)  →  NOT a polynomial ❌

(x + 2)/(x² + 2x + 3)  →  NOT a polynomial ❌

3. Trigonometric, exponential, or logarithmic functions

sin(x) + 2  →  NOT a polynomial ❌

The Standard Forms

Understanding the general form helps us identify relationships between coefficients and zeroes.

Linear Polynomial (Degree 1)

p(x) = ax + b,  where a ≠ 0

Zero: Solve ax + b = 0

x = −b/a

The zero is the negative ratio of constant term to coefficient of x.

Quadratic Polynomial (Degree 2)

p(x) = ax² + bx + c,  where a ≠ 0

Zeroes: Found by solving ax² + bx + c = 0 (we'll explore the relationship between zeroes and coefficients later).

Cubic Polynomial (Degree 3)

p(x) = ax³ + bx² + cx + d,  where a ≠ 0

How to Find the Value of a Polynomial at x = k

Logic in 3 Steps:

1. Write the polynomial expression clearly.

2. Substitute every occurrence of the variable with the value k.

3. Simplify using BODMAS (Brackets, Orders/powers, Division, Multiplication, Addition, Subtraction).

Let's see this with a detailed example.

Solved Examples

Example 1: Identifying Polynomial Type

Given: The expression 2x² − 3x + (1/5)

To Find: Is it a polynomial? If yes, state its degree and type.

Solution:

1. Check if all exponents are whole numbers.

Powers of x are 2 and 1 → both whole numbers ✅

2. Check if coefficients are real numbers.

Coefficients: 2, −3, 1/5 → all real ✅

3. Identify the highest power.

Degree = 2

4. Classify by degree.

Degree 2 → Quadratic polynomial

Final Answer: Yes, it is a quadratic polynomial of degree 2

Example 2: Finding Value of a Polynomial

Given: p(x) = x² − 3x − 4

To Find: p(2) and p(−1)

Solution:

1. Substitute x = 2 in the polynomial.

p(2) = (2)² − 3(2) − 4

2. Simplify step-by-step.

p(2) = 4 − 6 − 4 = −6

3. Now substitute x = −1.

p(−1) = (−1)² − 3(−1) − 4

4. Simplify carefully (watch the signs!).

p(−1) = 1 + 3 − 4 = 0

Final Answer: p(2) = −6 and p(−1) = 0

Example 3: Finding the Zero of a Linear Polynomial

Given: p(x) = 2x + 3

To Find: The zero of the polynomial

Solution:

1. Set the polynomial equal to zero.

2x + 3 = 0

2. Isolate x by subtracting 3 from both sides.

2x = −3

3. Divide both sides by 2.

x = −3/2

4. Verify by substituting back.

p(−3/2) = 2(−3/2) + 3 = −3 + 3 = 0 ✅

Final Answer: x = −3/2

Example 4: Verifying Zeroes of a Quadratic Polynomial (Tricky)

Given: p(x) = x² − 3x − 4, and claims that −1 and 4 are its zeroes

To Find: Verify if both values are indeed zeroes

Solution:

1. Check x = −1 by substituting.

p(−1) = (−1)² − 3(−1) − 4

2. Simplify.

p(−1) = 1 + 3 − 4 = 0 ✅

3. Check x = 4 by substituting.

p(4) = (4)² − 3(4) − 4

4. Simplify.

p(4) = 16 − 12 − 4 = 0 ✅

5. Both values satisfy p(k) = 0.

Final Answer: Yes, both −1 and 4 are zeroes of the polynomial

Tips & Tricks for Quick Calculations

Common Mistakes Students Make

Brain-Teaser Questions

Q1: If p(x) = kx² − 4x + k and p(2) = 0, find the value of k.

💡 Answer: Substitute x = 2: k(2)² − 4(2) + k = 0 → 4k − 8 + k = 0 → 5k = 8 → k = 8/5

Q2: Can a polynomial have infinitely many zeroes? If yes, give an example.

💡 Answer: Yes, the zero polynomial p(x) = 0 (all coefficients are zero) satisfies p(k) = 0 for every real number k, so it has infinitely many zeroes.

Q3: If p(x) = x² − (a + b)x + ab, what are its zeroes in terms of a and b?

💡 Answer: Zeroes are x = a and x = b. Verify: p(a) = a² − (a+b)a + ab = a² − a² − ab + ab = 0 ✅. Similarly p(b) = 0 ✅.

Mini Cheatsheet: Key Formulas & Concepts

Next Page Preview: We'll explore the geometrical meaning of zeroes — why the zero of a polynomial corresponds to where its graph crosses the x-axis, and how the shape of parabolas reveals the number of zeroes a quadratic can have.

Geometrical Meaning of the Zeroes of a Polynomial — Part 1

Page 2 of 5: Geometrical Meaning of the Zeroes of a Polynomial — Part 1

{{FORMULA: expr=y = ax² + bx + c | symbols=y:value of polynomial, x:variable, a:leading coefficient, b:linear coefficient, c:constant}}

Concept Introduction

Imagine you're playing basketball. When you shoot the ball, it flies through the air in a beautiful arc before (hopefully!) going through the hoop. This curved path is a perfect real-world example of a parabola, which is the shape of the graph of a quadratic polynomial. Now, think about the ball's journey. It starts from your hands (a certain height), reaches a maximum height, and then comes down.

If we imagine a coordinate plane with the ground as the x-axis, the points where the ball would hit the ground are the "zeroes" of its path. These are the points where the height (y) is zero. This connection between a real-world path and a graph is exactly what we're exploring. The zeroes of a polynomial aren't just abstract numbers; they are the specific points where the polynomial's value is zero, which geometrically means they are the points where its graph crosses the horizontal axis.

{{VISUAL: diagram: A basketball player shooting a ball. A dotted parabolic arc shows the ball's path from the player's hands, reaching a peak, and then going down towards a hoop. The ground is labeled as the x-axis, and the point where the parabolic path would intersect the ground is labeled "Zero of the polynomial path".}}

Definitions & Key Terms

Before we dive deep, let's clarify the essential terms we'll be using.

The Logic: Connecting Algebra to Geometry

The link between the zeroes of a polynomial and its graph is a cornerstone of algebra. It's a simple but powerful idea that turns abstract equations into visual shapes. Let's walk through the logic.

1. The Core Definition of a Zero By definition, a number k is a zero of a polynomial p(x) if substituting x = k into the polynomial makes the entire expression equal to zero.

   p(k) = 0
   

2. The Role of a Graph To draw the graph of a polynomial, we set it equal to y. We plot the graph of the equation y = p(x). This means for any point (x, y) on the graph, the y-coordinate is simply the value of the polynomial at that specific x-coordinate.

3. The Identity of the X-axis The x-axis is a special line. What do all points lying on the x-axis have in common? Their y-coordinate is always zero. Any point on the x-axis can be written in the form (x, 0).

4. The Point of Intersection When the graph of y = p(x) intersects the x-axis, it's touching a point where the y-coordinate is zero. Let's say this happens at an x-value of k. The coordinates of this intersection point must therefore be (k, 0).

   {{VISUAL: diagram: A coordinate plane with a simple upward-opening parabola representing y = p(x). The parabola cuts the x-axis at two distinct points, labeled A and B. The coordinates are marked as A=(α, 0) and B=(β, 0). Arrows point to α and β on the x-axis, with the label "Zeroes of the polynomial".}}

5. The Final Connection Since the point (k, 0) is on the graph of y = p(x), it must satisfy the equation y = p(x). Let's substitute y = 0 and x = k into the equation.

   0 = p(k)
   

6. The Unavoidable Conclusion This result, p(k) = 0, is the exact same condition as our initial definition of a zero! This proves a fundamental truth:

{{KEY: type=concept | title=The Geometrical Meaning of a Zero | text=The zeroes of a polynomial p(x) are precisely the x-coordinates of the points where the graph of y = p(x) intersects the x-axis.}}

Solved Examples

Let's apply this concept to some problems, starting from the basics and moving to more complex scenarios.

Example 1: The Linear Case (Easy)

Given: The linear polynomial p(x) = 2x + 3.

To Find: The zero of the polynomial and the coordinates of the point where its graph intersects the x-axis.

Solution:

1. To find the zero, we set p(x) = 0.

   2x + 3 = 0
   

2. Now, we solve for x.

   2x = -3
   

   x = -3/2
   

3. The zero of the polynomial is -3/2.

4. Based on our key concept, the graph of y = 2x + 3 must intersect the x-axis at the point where the x-coordinate is the zero. The y-coordinate on the x-axis is always 0.

   Intersection Point = (-3/2, 0)
   

Final Answer: The zero is -3/2, and the graph intersects the x-axis at the point (-3/2, 0).

Example 2: Reading a Graph (Medium)

Given: The graph of a quadratic polynomial y = p(x).

{{VISUAL: diagram: A coordinate plane showing a downward-opening parabola. The parabola intersects the x-axis at x = -1 and x = 4. The vertex of the parabola is in the first quadrant.}}

To Find: The zeroes of the polynomial p(x).

Solution:

1. Recall that the zeroes of a polynomial are the x-coordinates of the points where its graph intersects the x-axis.

2. Observe the provided graph. The curve is a parabola that intersects the x-axis at two distinct points.

3. Identify the x-coordinates of these intersection points from the graph.

   • The first point of intersection is at x = -1.
   • The second point of intersection is at x = 4.

4. Therefore, these x-coordinates are the zeroes of the polynomial p(x).

Final Answer: The zeroes of the polynomial are -1 and 4.

Example 3: From Equation to Graph Insight (Hard)

Given: The quadratic polynomial p(x) = x² - 3x - 4.

To Find: The zeroes of p(x) and the number of times its graph will intersect the x-axis.

Solution:

1. First, let's find the zeroes by setting p(x) = 0.

   x² - 3x - 4 = 0
   

2. We can solve this quadratic equation by factoring (splitting the middle term). We need two numbers that multiply to -4 and add to -3. These numbers are -4 and +1.

   x² - 4x + 1x - 4 = 0
   

3. Factor by grouping.

   x(x - 4) + 1(x - 4) = 0
   

   (x + 1)(x - 4) = 0
   

4. This gives us two possible solutions for x.

   • x + 1 = 0 → x = -1
   • x - 4 = 0 → x = 4

5. The polynomial has two distinct zeroes: -1 and 4.

6. Since there are two distinct zeroes, the graph of y = x² - 3x - 4 will intersect the x-axis at two distinct points, (-1, 0) and (4, 0).

Final Answer: The zeroes are -1 and 4. The graph will intersect the x-axis at two distinct points.

Example 4: Interpreting the "a" Coefficient (Tricky)

Given: A quadratic polynomial p(x) = -2x² + 8x - 6.

To Find: The zeroes of p(x) and determine if its parabolic graph opens upwards or downwards.

Solution:

1. First, let's determine the shape of the parabola. The general form is ax² + bx + c. In our case, a = -2.

2. The rule is: if a > 0, the parabola opens upwards (like ∪). If a < 0, it opens downwards (like ∩).

3. Since a = -2, which is less than 0, the parabola opens downwards.

4. Now, find the zeroes by setting p(x) = 0.

   -2x² + 8x - 6 = 0
   

5. To simplify, we can divide the entire equation by -2.

   x² - 4x + 3 = 0
   

6. Factor the simplified quadratic equation. We need two numbers that multiply to 3 and add to -4. These are -1 and -3.

   (x - 1)(x - 3) = 0
   

7. This gives us two distinct zeroes.

   • x - 1 = 0 → x = 1
   • x - 3 = 0 → x = 3

Final Answer: The zeroes are 1 and 3. The graph is a parabola that opens downwards.

Tips & Tricks

Common Mistakes to Avoid

Brain-Teaser Questions

1. The graph of y = p(x-2) is a parabola that intersects the x-axis at x=3 and x=7. What are the zeroes of the polynomial p(x)?

   💡 Answer: The zeroes of p(x) are 1 and 5. If x=3 is a zero of p(x-2), then p(3-2) = p(1) = 0. If x=7 is a zero of p(x-2), then p(7-2) = p(5) = 0. So the zeroes of p(x) are 1 and 5.

2. A quadratic polynomial p(x) = ax² + bx + c has two distinct zeroes. If the graph opens upwards (a > 0) and the y-intercept is negative (c < 0), can you prove it must have two distinct zeroes?

   💡 Answer: Yes. If a > 0, the parabola opens upwards. If c < 0, the y-intercept (0, c) is below the x-axis. Since an upward-opening parabola that passes through a point below the x-axis must eventually rise on both sides, it is guaranteed to cross the x-axis in two distinct places.

3. The graph of a polynomial y = p(x) intersects the x-axis at exactly three points: (-2, 0), (0, 0), and (2, 0). Can this be the graph of a quadratic polynomial?

   💡 Answer: No. A quadratic polynomial has a degree of 2. A polynomial of degree n can intersect the x-axis at a maximum of n points. Since this graph intersects the x-axis at 3 points, its degree must be at least 3. It cannot be a quadratic polynomial.

Mini Cheatsheet

Geometrical Meaning of the Zeroes of a Polynomial — Part 2

Geometrical Meaning of the Zeroes of a Polynomial — Part 2

Welcome back! In the previous lesson, we saw how the zeroes of a linear polynomial correspond to a single point where its graph—a straight line—intersects the x-axis. We also took our first look at quadratic polynomials and their beautiful U-shaped graphs called parabolas.

Today, we'll explore the other possibilities for quadratic polynomials. What if the graph doesn't cut the x-axis twice? What if it just touches it, or doesn't touch it at all? We'll then leap into the world of cubic polynomials and uncover a universal rule that connects a polynomial's degree to the maximum number of zeroes it can have. This visual connection between algebra (equations) and geometry (graphs) is a cornerstone of mathematics!

{{FORMULA: expr=y = p(x) | symbols=p(x):a polynomial in variable x, y:the value of the polynomial for a given x, Zeroes:the x-coordinates where the graph intersects the x-axis}}

The Three Cases for a Quadratic Polynomial

When we graph a quadratic polynomial y = ax² + bx + c, its parabola can interact with the x-axis in three distinct ways. This interaction directly tells us the number of real zeroes the polynomial has.

Remember, a zero of a polynomial p(x) is any value of x for which p(x) = 0. Geometrically, this is where the graph y = p(x) meets the line y = 0, which is simply the x-axis!

Let's examine the remaining two cases not covered previously.

Case (ii): The Graph Touches the x-axis at Exactly One Point

Imagine a parabola that swoops down (or up) and just barely kisses the x-axis at a single point before turning back. This single point of contact represents one real zero.

Because the parabola touches the axis and reverses direction at that point, this zero is sometimes called a repeated root or a zero with a multiplicity of two. For our purposes, it means there is only one unique value of x that makes the polynomial equal to zero.

{{VISUAL: diagram: Two parabolas for Case (ii). One opens upwards (a>0) and touches the x-axis at its vertex. The other opens downwards (a<0) and also touches the x-axis at its vertex. Both are labeled to show "Exactly one zero".}}

Case (iii): The Graph Does Not Intersect the x-axis at All

What if the parabola is entirely above or entirely below the x-axis? In this scenario, the graph of y = p(x) never crosses the line y = 0.

This means there is no real value of x for which p(x) is zero. Therefore, the quadratic polynomial has no real zeroes. The equation ax² + bx + c = 0 has no solution in the set of real numbers.

{{VISUAL: diagram: Two parabolas for Case (iii). One opens upwards and is entirely above the x-axis. The other opens downwards and is entirely below the x-axis. Both are labeled to show "No real zeroes".}}

{{KEY: type=concept | title=Zeroes of a Quadratic Polynomial | text=A quadratic polynomial can have at most two zeroes. Geometrically, this means its parabolic graph can intersect the x-axis at most two times. The three possibilities are: two distinct zeroes (two intersection points), one zero (one touching point), or no real zeroes (no intersection).}}

What About Cubic Polynomials?

Now, let's increase the complexity. A cubic polynomial has the general form ax³ + bx² + cx + d, where the highest power (degree) is 3.

What does its graph look like? Unlike the simple 'U' shape of a parabola, the graph of a cubic polynomial typically has an 'S' shape. Let's consider the polynomial p(x) = x³ – 4x.

If we plot points for y = x³ – 4x, we get a curve that wiggles its way across the x-axis.

The table shows that y is 0 when x is -2, 0, and 2. These are the three zeroes of the polynomial. The graph will intersect the x-axis at these three points.

{{VISUAL: diagram: A graph of the cubic polynomial y = x³ - 4x. The curve starts from the bottom-left, goes up through (-2, 0), reaches a local maximum, comes down through (0, 0), reaches a local minimum, and goes up again through (2, 0) into the top-right. The three x-intercepts are clearly marked.}}

However, a cubic polynomial doesn't always have three zeroes.

• The graph of y = x³ only intersects the x-axis at one point, x = 0.
• The graph of y = x³ – x² = x²(x – 1) touches the x-axis at x = 0 and crosses it at x = 1, giving it two distinct zeroes.

The key takeaway is that a cubic polynomial can have at most three zeroes.

The General Rule

This leads us to a powerful general conclusion that connects a polynomial's degree to its number of zeroes.

Remark: In general, given a polynomial p(x) of degree n, the graph of y = p(x) intersects the x-axis at at most n points. Therefore, a polynomial p(x) of degree n has at most n zeroes.

This simple rule is fundamental!

• Degree 1 (Linear) → At most 1 zero.
• Degree 2 (Quadratic) → At most 2 zeroes.
• Degree 3 (Cubic) → At most 3 zeroes.
• Degree n → At most n zeroes.

Solved Examples

Let's apply these visual concepts to interpret some graphs.

Example 1: Identifying Zeroes from a Parabola (Easy)

Given: The graph of a quadratic polynomial p(x) that opens upwards and its lowest point (vertex) is at (3, 2).

To Find: The number of zeroes of p(x).

Solution:

1. The graph is a parabola that opens upwards.
2. The lowest point on the graph is the vertex, which is given as (3, 2).
3. Since the y-coordinate of the lowest point is 2 (which is greater than 0), the entire parabola lies above the x-axis.
4. A graph that never touches or crosses the x-axis represents a polynomial with no real zeroes.

Final Answer: <The polynomial has 0 zeroes.>

Example 2: Finding Zeroes of a Perfect Square Trinomial (Medium)

Given: The quadratic polynomial p(x) = x² – 6x + 9.

To Find: The number of zeroes by analysing its graph's interaction with the x-axis.

Solution:

1. First, let's try to factorize the polynomial. We can recognize this as a perfect square trinomial.

   p(x) = x² – 2(x)(3) + 3²
   

2. Using the identity a² - 2ab + b² = (a - b)², we get:

   p(x) = (x - 3)²
   

3. The zeroes of p(x) are the values of x for which p(x) = 0.

   (x - 3)² = 0
   

4. This equation is only true when x - 3 = 0, which means x = 3.

5. There is only one value of x that makes the polynomial zero. Geometrically, this means the graph of y = x² – 6x + 9 is a parabola that touches the x-axis at exactly one point.

Final Answer: <The polynomial has 1 zero (at x = 3).>

Example 3: Analysing a Cubic Polynomial Graph (Hard)

Given: The graph of a cubic polynomial y = p(x) which passes through the points (-4, 0), touches the x-axis at (1, 0), and passes through (3, 8).

To Find: The number of zeroes of p(x).

Solution:

1. The zeroes of a polynomial are the x-coordinates of the points where its graph intersects or touches the x-axis.
2. We are given that the graph passes through (-4, 0). This is an intersection point on the x-axis. So, x = -4 is one zero.
3. We are told the graph touches the x-axis at (1, 0). A touching point also corresponds to a zero. So, x = 1 is another zero.
4. The point (3, 8) is just another point on the curve and is not on the x-axis, so it doesn't represent a zero.
5. The graph intersects the x-axis at x = -4 and touches it at x = 1. These are two distinct points on the x-axis.

Final Answer: <The polynomial has 2 distinct zeroes.>

Example 4: Deducing Polynomial Degree from its Graph (Tricky)

Given: A graph of a polynomial p(x) that starts in the top-left, crosses the x-axis at x = -2, goes down, turns back up to cross the x-axis at x = 1, goes up, and turns back down to cross the x-axis at x = 3, continuing downwards to the bottom-right.

To Find: The number of zeroes and the minimum possible degree of the polynomial.

Solution:

1. Finding the zeroes: The number of zeroes is the number of times the graph intersects the x-axis.

2. The graph crosses the x-axis at three distinct points: x = -2, x = 1, and x = 3.

   Number of zeroes = 3
   

3. Finding the minimum degree: We know that a polynomial of degree n has at most n zeroes.

4. Since this polynomial has 3 zeroes, its degree must be at least 3.

5. A linear polynomial (degree 1) has at most 1 zero. A quadratic polynomial (degree 2) has at most 2 zeroes.

6. A cubic polynomial (degree 3) can have up to 3 zeroes. This fits our observation. Therefore, the minimum possible degree for a polynomial with 3 zeroes is 3.

Final Answer: <The polynomial has 3 zeroes, and its minimum possible degree is 3.>

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. A polynomial graph is shaped like the letter 'W'. What is the minimum possible degree of this polynomial?

   💡 Answer: The graph has 3 turns (down-up, up-down, down-up). A polynomial of degree n has at most n-1 turns. So, n-1 must be at least 3, which means n must be at least 4. The minimum possible degree is 4.

2. If a polynomial p(x) has a degree of 5, what is the minimum number of real zeroes it must have?

   💡 Answer:

   1. A polynomial with an odd degree (like 5) will have its graph point in opposite directions at the very ends (one end goes to +∞, the other to -∞). To get from one to the other, the graph must cross the x-axis at least once. Therefore, it must have at least one real zero.

3. Can the graph of a quadratic polynomial and the graph of a linear polynomial intersect at exactly 3 points? Why or why not?

   💡 Answer: No. To find the intersection points, we set the polynomials equal: ax² + bx + c = mx + n. Rearranging gives ax² + (b-m)x + (c-n) = 0. This is a quadratic equation, which can have at most 2 solutions. Therefore, their graphs can intersect at most twice.

Mini Cheatsheet

Solved NCERT Exercises

EXERCISE 2.1

Q1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

{{VISUAL: diagram: The six graphs from NCERT Exercise 2.1, Figure 2.10, clearly labeled (i) to (vi).}}

Solution:

The number of zeroes of a polynomial p(x) is the number of points where its graph y = p(x) intersects the x-axis.

(i)

• Observation: The graph is a straight line parallel to the x-axis and does not intersect it at any point.
• Conclusion: There are no points where the y-value is 0.

Final Answer: <The number of zeroes is 0.>

(ii)

• Observation: The graph intersects the x-axis at exactly one point.
• Conclusion: There is one value of x for which p(x) = 0.

Final Answer: <The number of zeroes is 1.>

(iii)

• Observation: The graph intersects the x-axis at three distinct points.
• Conclusion: There are three values of x for which p(x) = 0.

Final Answer: <The number of zeroes is 3.>

(iv)

• Observation: The graph is a parabola that intersects the x-axis at two distinct points.
• Conclusion: There are two values of x for which p(x) = 0.

Final Answer: <The number of zeroes is 2.>

(v)

• Observation: The graph intersects the x-axis at four distinct points.
• Conclusion: There are four values of x for which p(x) = 0.

Final Answer: <The number of zeroes is 4.>

(vi)

• Observation: The graph intersects the x-axis at one point and touches the x-axis at two other points. In total, it meets the x-axis at 3 distinct points.
• Conclusion: There are three values of x for which p(x) = 0.

Final Answer: <The number of zeroes is 3.>

Relationship between Zeroes and Coefficients of a Polynomial — Part 1

Page 4 of 5: Relationship between Zeroes and Coefficients of a Polynomial — Part 1

Concept Introduction

Imagine you're a quality inspector at a manufacturing plant that produces custom mirrors. Each mirror has a parabolic curved edge described by a quadratic equation. If the mirror breaks at exactly two points (the zeroes), can you predict those break points just by looking at the equation coefficients?

The answer is yes! This is precisely what the relationship between zeroes and coefficients reveals. Instead of factorizing or solving the equation every time, you can instantly calculate the sum and product of zeroes using simple ratios of coefficients.

This powerful shortcut not only saves time in exams but also helps engineers, computer scientists, and data analysts quickly analyze quadratic models without tedious calculations. In this section, we'll uncover this beautiful relationship for quadratic polynomials and verify it through multiple examples.

{{FORMULA: expr=α + β = -b/a, αβ = c/a | symbols=α:first zero, β:second zero, a:coefficient of x², b:coefficient of x, c:constant term}}

Definitions & Formulas

Derivation: How Zeroes Connect to Coefficients

Let's derive this relationship logically using factor form.

1. Start with a general quadratic polynomial:

p(x) = ax² + bx + c, where a ≠ 0

2. If α and β are the zeroes, then by Factor Theorem:

The polynomial can be written as:

p(x) = a(x - α)(x - β)

Here, a is the leading coefficient that remains constant.

3. Expand the factored form:

p(x) = a[x² - (α + β)x + αβ]

p(x) = ax² - a(α + β)x + aαβ

4. Compare with the standard form ax² + bx + c:

Matching coefficients of corresponding terms:

• Coefficient of x²: a = a ✓
• Coefficient of x: b = -a(α + β)
• Constant term: c = aαβ

5. Solve for sum of zeroes:

From the coefficient of x:

b = -a(α + β)

α + β = -b/a

6. Solve for product of zeroes:

From the constant term:

c = aαβ

αβ = c/a

{{KEY: type=theorem | title=The Golden Rule for Quadratic Zeroes | text=For any quadratic polynomial ax² + bx + c, Sum of zeroes = -(coefficient of x)/(coefficient of x²) and Product of zeroes = (constant term)/(coefficient of x²). These ratios work REGARDLESS of factorization difficulty!}}

Solved Examples

Example 1: Basic Verification (Easy)

Given: Quadratic polynomial x² + 7x + 10

To Find: Zeroes and verify the relationship between zeroes and coefficients

Solution:

1. Factorize the polynomial by splitting the middle term:

x² + 7x + 10 = x² + 5x + 2x + 10

= x(x + 5) + 2(x + 5)

= (x + 2)(x + 5)

2. Set each factor to zero:

x + 2 = 0  →  x = -2

x + 5 = 0  →  x = -5

So the zeroes are α = -2 and β = -5.

3. Calculate sum of zeroes directly:

α + β = (-2) + (-5) = -7

4. Verify using the formula (a = 1, b = 7, c = 10):

Sum = -b/a = -7/1 = -7  ✓

5. Calculate product of zeroes directly:

αβ = (-2) × (-5) = 10

6. Verify using the formula:

Product = c/a = 10/1 = 10  ✓

Final Answer: Zeroes are -2 and -5; relationships verified successfully

Example 2: Coefficient Greater Than 1 (Medium)

Given: Quadratic polynomial 2x² - 8x + 6

To Find: Zeroes and verify sum/product relationships

Solution:

1. Factor out the common factor first:

2x² - 8x + 6 = 2(x² - 4x + 3)

2. Factorize the expression inside:

= 2(x² - 3x - x + 3)

= 2[x(x - 3) - 1(x - 3)]

= 2(x - 1)(x - 3)

3. Find the zeroes:

x - 1 = 0  →  α = 1

x - 3 = 0  →  β = 3

4. Verify sum of zeroes (a = 2, b = -8, c = 6):

α + β = 1 + 3 = 4

-b/a = -(-8)/2 = 8/2 = 4  ✓

5. Verify product of zeroes:

αβ = 1 × 3 = 3

c/a = 6/2 = 3  ✓

Final Answer: Zeroes are 1 and 3; both relationships hold true

Example 3: Irrational Zeroes (Hard)

Given: Quadratic polynomial x² - 3

To Find: Zeroes and verify the coefficient relationships

Solution:

1. Recognize this as a difference of squares (a² - b² = (a - b)(a + b)):

x² - 3 = x² - (√3)²

= (x - √3)(x + √3)

2. Find the zeroes:

x - √3 = 0  →  α = √3

x + √3 = 0  →  β = -√3

3. Note the coefficients: a = 1, b = 0, c = -3

4. Verify sum of zeroes:

α + β = √3 + (-√3) = 0

-b/a = -0/1 = 0  ✓

5. Verify product of zeroes:

αβ = (√3) × (-√3) = -3

c/a = -3/1 = -3  ✓

Final Answer: Zeroes are √3 and -√3; formulas work even for irrational roots

Example 4: Reverse Problem (Tricky)

Given: Sum of zeroes = -3, Product of zeroes = 2

To Find: A quadratic polynomial with these properties

Solution:

1. Use the relationship formulas in reverse:

We know:

α + β = -b/a = -3

αβ = c/a = 2

2. Choose a = 1 for simplicity (the simplest polynomial):

-b/1 = -3  →  b = 3

c/1 = 2  →  c = 2

3. Construct the polynomial:

p(x) = 1·x² + 3·x + 2

p(x) = x² + 3x + 2

4. Verify by factorization:

x² + 3x + 2 = (x + 1)(x + 2)

Zeroes are -1 and -2.

5. Check: sum = -1 + (-2) = -3 ✓, product = (-1)×(-2) = 2 ✓

6. Note: ANY polynomial of form k(x² + 3x + 2) works, where k ≠ 0

Examples: 2x² + 6x + 4, 5x² + 15x + 10, etc.

Final Answer: x² + 3x + 2 (or any non-zero multiple of it)

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

Q1: If one zero of the polynomial 3x² + kx + 12 is 2, find the value of k and the other zero.

💡 Answer: Use product formula: αβ = c/a → 2 × β = 12/3 → β = 2. Both zeroes are 2 (repeated root). Now use sum formula: α + β = -k/3 → 2 + 2 = -k/3 → 4 = -k/3 → k = -12.

Q2: Can a quadratic polynomial have sum of zeroes = 5 and product of zeroes = -5? If yes, construct it.

💡 Answer: Yes! Using x² - (sum)x + (product) → x² - 5x + (-5) → x² - 5x - 5. Any non-zero multiple works: 2x² - 10x - 10, 3x² - 15x - 15, etc.

Q3: If α and β are zeroes of 2x² + 5x + k and α² + β² = 9, find k.

💡 Answer: Use identity: α² + β² = (α + β)² - 2αβ We have α + β = -5/2 and αβ = k/2 → 9 = (-5/2)² - 2(k/2) → 9 = 25/4 - k → k = 25/4 - 9 = 25/4 - 36/4 = -11/4.

Mini Cheatsheet

Relationship between Zeroes and Coefficients of a Polynomial — Part 2 & Summary

{{FORMULA: expr=α + β = -b/a | symbols=α,β:zeroes of the polynomial, a:coefficient of x², b:coefficient of x}}

Relationship between Zeroes and Coefficients

Welcome back! In the previous sections, we learned how to find the zeroes of a polynomial and what they represent geometrically. Now, we'll uncover a fascinating and powerful connection: a direct mathematical relationship between the zeroes of a polynomial and its coefficients. This relationship is a cornerstone of algebra, allowing us to deduce properties of the zeroes without actually finding them, and vice-versa.

Imagine an architect designing a parabolic arch bridge. The points where the arch touches the ground are the 'zeroes' of the quadratic equation that models the arch. The coefficients of this equation are determined by the height (a), position (b), and starting elevation (c) of the arch. The relationship we're about to study allows the architect to know the span of the bridge (distance between the zeroes) just by looking at the coefficients, making design calculations faster and more intuitive.

Definitions & Formulas

For any quadratic polynomial in the standard form p(x) = ax² + bx + c, where a ≠ 0, let its zeroes be the Greek letters alpha (α) and beta (β). The relationship between these zeroes and the coefficients a, b, and c is defined as follows:

Derivation of the Relationship

How do we know these formulas are correct? The logic comes from the fundamental relationship between the zeroes and factors of a polynomial.

1. If α and β are the zeroes of the polynomial p(x) = ax² + bx + c, then (x - α) and (x - β) must be its factors.

2. Therefore, we can express the polynomial as a product of its factors, multiplied by some constant k.

   ax² + bx + c = k(x - α)(x - β)
   

3. Now, let's expand the right-hand side of the equation.

   k(x - α)(x - β) = k(x² - βx - αx + αβ)
   

4. Grouping the terms with x gives us:

   = k[x² - (α + β)x + αβ]
   

5. Distributing the constant k across the terms, we get:

   = kx² - k(α + β)x + kαβ
   

6. Now we have two forms of the same polynomial. We can compare the coefficients of the powers of x on both sides of ax² + bx + c = kx² - k(α + β)x + kαβ.

   • Comparing x² terms: a = k
   • Comparing x terms: b = -k(α + β)
   • Comparing constant terms: c = kαβ

7. From these comparisons, we can isolate the sum and product of the zeroes.

   • For the sum (α + β): Since b = -k(α + β) and a = k, we can substitute a for k: b = -a(α + β). Rearranging this gives the formula for the sum of zeroes:

     α + β = -b/a
     

   • For the product (αβ): Since c = kαβ and a = k, we can again substitute a for k: c = a(αβ). Rearranging this gives the formula for the product of zeroes:

     αβ = c/a
     

{{KEY: type=formula | title=The Core Relationships | text=For a quadratic polynomial ax² + bx + c, the sum of zeroes α + β is -b/a (negative of the x-coefficient over the x²-coefficient) and the product of zeroes αβ is c/a (the constant term over the x²-coefficient).}}

Solved Examples

Let's apply these formulas to some problems from the NCERT textbook. We will find the zeroes and then verify the relationship.

Example 1: Simple Factoring (Easy)

NCERT Example 2 Given: The quadratic polynomial p(x) = x² + 7x + 10.

To Find: The zeroes of the polynomial and verify the relationship between the zeroes and the coefficients.

Solution:

1. First, we find the zeroes by factorizing the polynomial using the middle-term splitting method. We need two numbers that add up to 7 and multiply to 10. These numbers are 2 and 5.

   x² + 7x + 10 = x² + 2x + 5x + 10
   

2. Factor by grouping.

   = x(x + 2) + 5(x + 2) = (x + 2)(x + 5)
   

3. To find the zeroes, we set p(x) = 0. This means either x + 2 = 0 or x + 5 = 0.

   x = -2  or  x = -5
   

   So, the zeroes are α = -2 and β = -5.

4. Now, let's verify the relationship. First, identify the coefficients from x² + 7x + 10. Here, a = 1, b = 7, and c = 10.

5. Verification of the Sum of Zeroes: Sum from our found zeroes: α + β = (-2) + (-5) = -7. Sum from the formula: -b/a = -(7)/1 = -7. The values match.

6. Verification of the Product of Zeroes: Product from our found zeroes: αβ = (-2) × (-5) = 10. Product from the formula: c/a = 10/1 = 10. The values match.

Final Answer:

The zeroes are -2 and -5. The relationship between the zeroes and coefficients is verified.

Example 2: Using an Identity (Medium)

NCERT Example 3 Given: The quadratic polynomial p(x) = x² - 3.

To Find: The zeroes of the polynomial and verify the relationship between the zeroes and the coefficients.

Solution:

1. To find the zeroes, we can rewrite the polynomial using the algebraic identity a² - b² = (a - b)(a + b). Here, 3 can be written as (√3)².

   x² - 3 = x² - (√3)² = (x - √3)(x + √3)
   

2. Set p(x) = 0 to find the zeroes. This means x - √3 = 0 or x + √3 = 0.

   x = √3  or  x = -√3
   

   So, the zeroes are α = √3 and β = -√3.

3. Now, let's verify. First, identify the coefficients. The polynomial x² - 3 can be written in standard form as x² + 0x - 3. Here, a = 1, b = 0, and c = -3.

4. Verification of the Sum of Zeroes: Sum from found zeroes: α + β = (√3) + (-√3) = 0. Sum from the formula: -b/a = -(0)/1 = 0. The values match.

5. Verification of the Product of Zeroes: Product from found zeroes: αβ = (√3) × (-√3) = -3. Product from the formula: c/a = -3/1 = -3. The values match.

Final Answer:

The zeroes are √3 and -√3. The relationship is verified.

Example 3: Reversing the Problem (Hard)

NCERT Example 4 Given: The sum of zeroes is -3 and the product of zeroes is 2.

To Find: The quadratic polynomial.

Solution:

1. Let the quadratic polynomial be ax² + bx + c, and its zeroes be α and β.

2. We are given the sum of the zeroes.

   α + β = -3
   

3. We know the formula for the sum of zeroes is -b/a.

   -b/a = -3
   

4. We are also given the product of the zeroes.

   αβ = 2
   

5. We know the formula for the product of zeroes is c/a.

   c/a = 2
   

6. To find the simplest polynomial, we can assume a = 1. If a = 1, then from -b/a = -3, we get -b/1 = -3, which means b = 3. If a = 1, then from c/a = 2, we get c/1 = 2, which means c = 2.

7. Substitute these values of a, b, and c back into the standard form ax² + bx + c.

   (1)x² + (3)x + (2)
   

Final Answer:

A quadratic polynomial that fits the conditions is x² + 3x + 2.
Note: Any polynomial of the form k(x² + 3x + 2) for a real number k would also be correct.

Example 4: Rearranging Terms (Tricky)

From NCERT Exercise 2.2 Given: The quadratic polynomial p(x) = 6x² - 3 - 7x.

To Find: The zeroes of the polynomial and verify the relationship between the zeroes and the coefficients.

Solution:

1. Crucial first step: Rearrange the polynomial into standard form ax² + bx + c.

   p(x) = 6x² - 7x - 3
   

   Now, we can correctly identify a = 6, b = -7, and c = -3.

2. Find the zeroes by splitting the middle term. We need two numbers that add to -7 and multiply to 6 × (-3) = -18. These numbers are -9 and 2.

   6x² - 9x + 2x - 3
   

3. Factor by grouping.

   = 3x(2x - 3) + 1(2x - 3) = (3x + 1)(2x - 3)
   

4. Set p(x) = 0. This means 3x + 1 = 0 or 2x - 3 = 0.

   x = -1/3  or  x = 3/2
   

   The zeroes are α = -1/3 and β = 3/2.

5. Verification of the Sum of Zeroes: Sum from found zeroes: α + β = (-1/3) + (3/2) = (-2 + 9)/6 = 7/6. Sum from the formula: -b/a = -(-7)/6 = 7/6. The values match.

6. Verification of the Product of Zeroes: Product from found zeroes: αβ = (-1/3) × (3/2) = -3/6 = -1/2. Product from the formula: c/a = -3/6 = -1/2. The values match.

Final Answer:

The zeroes are -1/3 and 3/2. The relationship is verified.

Tips & Tricks

Here are a few shortcuts to speed up your work and build intuition.

Common Mistakes

Many students make small errors that can lead to incorrect answers. Here's what to watch out for.

Brain-Teaser Questions

Ready to test your understanding at a higher level?

1. If the zeroes of the polynomial p(x) = ax² + bx + c are reciprocal of each other, what is the relationship between a and c?

   💡 Answer: If zeroes are α and 1/α, their product is α × (1/α) = 1. We also know the product is c/a. So, c/a = 1, which means c = a.

2. If α and β are the zeroes of x² - 6x + 8, find a quadratic polynomial whose zeroes are -α and -β.

   💡 Answer: The new sum is (-α) + (-β) = -(α + β). The original sum α + β = -(-6)/1 = 6. So, the new sum is -6. The new product is (-α) × (-β) = αβ. The original product αβ = 8/1 = 8. So, the new product is 8. The new polynomial is x² - (New Sum)x + (New Product) → x² - (-6)x + 8 → x² + 6x + 8.

3. The sum of the zeroes of the polynomial kx² + 2x + 3k is equal to their product. What is the value of k?

   💡 Answer: Here, a=k, b=2, c=3k. Sum = -b/a = -2/k. Product = c/a = 3k/k = 3. We are given Sum = Product, so -2/k = 3. Solving for k, we get 3k = -2, so k = -2/3.

Chapter Summary & Mini Cheatsheet

This chapter introduced you to the world of polynomials. We explored their degrees, their geometric meaning, and the powerful link between their zeroes and coefficients.

Key Learnings from the Chapter:

• Degrees: Polynomials of degree 1, 2, and 3 are called linear, quadratic, and cubic polynomials, respectively.
• Standard Form: A quadratic polynomial is of the form ax² + bx + c, where a, b, c are real numbers and a ≠ 0.
• Zeroes: The zeroes of a polynomial p(x) are the x-coordinates where the graph of y = p(x) intersects the x-axis. A quadratic polynomial can have at most 2 zeroes.
• Relationship (Quadratic): For a quadratic polynomial ax² + bx + c with zeroes α and β:
  • Sum of zeroes: α + β = -b/a
  • Product of zeroes: αβ = c/a
• Relationship (Cubic): For a cubic polynomial ax³ + bx² + cx + d with zeroes α, β, and γ:
  • α + β + γ = -b/a
  • αβ + βγ + γα = c/a
  • αβγ = -d/a (Note: Cubic polynomials are generally not part of the board examination syllabus for the 2024-25 session, but the relationship is good to know.)

Here is a cheatsheet to screenshot for your last-minute revision.

Mini Cheatsheet