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Chapter 2: Polynomials

Overview

This page provides comprehensive Chapter 2: Polynomials – Board Exam Notes aligned with the latest CBSE 2025–26 syllabus. Covers zeros of polynomials (graphical & algebraic), relationship between zeros and coefficients of quadratic polynomials, solved board questions, and interactive quiz.

Zeros of Polynomials • Graphical Method • Zeros-Coefficients Relationship • Solved Board Questions

Exam Weightage & Blueprint

Total: 4-6 Marks

Polynomials falls under Unit II: Algebra (20 marks total). As per the latest syllabus, focus is on finding zeros graphically and algebraically, and verifying the relationship between zeros and coefficients of quadratic polynomials.

Question Type Marks Frequency Focus Topic
MCQ 1 High Graphs (No. of Zeroes)
Short Answer 2 or 3 High Relation b/w Zeroes & Coefficients
Case Study 4 Medium Parabolic Path Applications

Polynomial Basics

Degree: The highest power of $x$ in $p(x)$ is called the degree of the polynomial.
Type Degree General Form Max Zeroes
Linear 1 $ax + b$ 1
Quadratic 2 $ax^2 + bx + c$ 2

Geometrical Meaning of Zeroes: Graph of $y = ax^2 + bx + c$ (Parabola)

x y α β Case 1: 2 Zeroes Cuts x-axis at 2 points x y α = β Case 2: 1 Zero Touches x-axis at 1 point x y Case 3: 0 Zeroes No intersection with x-axis Number of real zeroes of $p(x)$ = Number of points where graph intersects the $x$-axis

Key Formulas & Relationships

1. Relationship (Quadratic)

For zeroes $\alpha$ and $\beta$ of $ax^2 + bx + c$:

$$ \text{Sum } (\alpha + \beta) = \frac{-b}{a} $$
$$ \text{Product } (\alpha \beta) = \frac{c}{a} $$

2. Forming a Polynomial

$$ p(x) = k [ x^2 - (\alpha + \beta)x + (\alpha \beta) ] $$

(where k is a non-zero constant)

3. High-Frequency Board Exam Shortcuts & Symmetric Identities

Condition on Zeros Mathematical Relation Board Shortcut Result
Reciprocal to each other $\alpha = \frac{1}{\beta} \implies \alpha\beta = 1$ $\frac{c}{a} = 1 \implies \mathbf{c = a}$
Equal in magnitude, opposite in sign $\alpha = -\beta \implies \alpha + \beta = 0$ $\frac{-b}{a} = 0 \implies \mathbf{b = 0}$
Sum of squares ($\alpha^2 + \beta^2$) $(\alpha+\beta)^2 - 2\alpha\beta$ $\left(\frac{-b}{a}\right)^2 - 2\left(\frac{c}{a}\right) = \frac{b^2 - 2ac}{a^2}$
Sum of reciprocals ($\frac{1}{\alpha} + \frac{1}{\beta}$) $\frac{\alpha+\beta}{\alpha\beta}$ $\frac{-b/a}{c/a} = \mathbf{-\frac{b}{c}}$
Ratio sum ($\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$) $\frac{\alpha^2+\beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta}$ $\frac{b^2 - 2ac}{ac}$
📌 Remember the Signs:
Sum of zeros = $-b/a$ (note the negative sign!)
Product of zeros = $c/a$ (constant term / leading coefficient)

Solved Examples (Board Marking Scheme)

Q1. Find zeroes of $x^2 - 2x - 8$ and verify relationship. (3 Marks)

Step 1: Factorization 1 Mark

$x^2 - 4x + 2x - 8 = x(x-4) + 2(x-4)$

$\Rightarrow (x+2)(x-4)$. Zeroes: $-2, 4$.

Step 2: Sum Verification 1 Mark

Sum $= -2 + 4 = 2$. Formula: $-(-2)/1 = 2$.

Step 3: Product Verification 1 Mark

Product $= -2 \times 4 = -8$. Formula: $-8/1 = -8$.

Exam Strategy & Mistake Bank

⚠️ Mistake Bank

Sign Error: Forgetting the negative in $-b/a$. If $b$ is already negative, result becomes positive!
X-axis only: In graph questions, count only X-axis intersections. Don't count Y-axis!

💡 Scoring Tips

Show Calculation: For 3M questions, explicitly write "Sum of Zeroes = ..." and "$-b/a = ...$" separately.
Identity Use: For $t^2 - 15$, use $a^2-b^2$ identity to get zeroes $\pm\sqrt{15}$.

Self-Assessment Mock Test (10 Marks)

Q1 (1M): The number of zeroes for a quadratic polynomial is exactly 2. (True/False?)


Q2 (2M): Find a quadratic polynomial whose zeroes are $1/4$ and $-1$.


Q3 (3M): Find zeroes of $4u^2 + 8u$ and verify relationship.


Q4 (4M): If $\alpha$ and $\beta$ are zeroes of $x^2 + 4x + 3$, find the value of $\alpha^2 + \beta^2$.

📈 Zeros from Graph (Graphical Method)

The number of zeros of a polynomial $p(x)$ = number of times the graph of $y = p(x)$ intersects the x-axis.

Linear ($ax + b$)

Graph is a straight line. Crosses x-axis at exactly 1 point.

→ 1 zero

Quadratic ($ax^2 + bx + c$)

Graph is a parabola. Can cross x-axis at 0, 1, or 2 points.

→ 0, 1, or 2 zeros

Key Cases for Quadratic Graph ($y = ax^2 + bx + c$):
  • Parabola Direction:
    • If $a > 0$: Parabola opens upwards ($\cup$) with a minimum point.
    • If $a < 0$: Parabola opens downwards ($\cap$) with a maximum point.
  • Number of Zeros (Intersection with x-axis):
    • 2 zeros: Parabola cuts x-axis at 2 distinct points (Discriminant $D > 0$)
    • 1 zero (coincident/repeated): Parabola touches x-axis at 1 point ($D = 0$)
    • 0 zeros: Parabola does not touch or intersect x-axis at all ($D < 0$)

where Discriminant $D = b^2 - 4ac$

⚠️ Common Mistake: Students count y-axis intersections as zeros. Only x-axis intersections are zeros! A zero is where $p(x) = 0$, i.e., $y = 0$.

📝 More Solved Board Questions

Q2. Find the zeros of the quadratic polynomial $4\sqrt{3}x^2 + 5x - 2\sqrt{3}$ and verify the relationship between the zeros and coefficients. 3 Marks (Standard Paper)

Sol. Product required $= 4\sqrt{3} \times (-2\sqrt{3}) = -24$. Sum required $= 5$. Factors are $+8$ and $-3$.

$4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3} = 0$

$\Rightarrow 4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) = 0$   (Note: $3 = \sqrt{3}\times\sqrt{3}$)

$\Rightarrow (4x - \sqrt{3})(\sqrt{3}x + 2) = 0 \implies x = \frac{\sqrt{3}}{4}, \; x = -\frac{2}{\sqrt{3}}$

Zeros: $\alpha = \frac{\sqrt{3}}{4}$, $\beta = -\frac{2}{\sqrt{3}}$

Sum of zeros: $\alpha + \beta = \frac{\sqrt{3}}{4} - \frac{2}{\sqrt{3}} = \frac{3 - 8}{4\sqrt{3}} = \frac{-5}{4\sqrt{3}} = \frac{-b}{a}$   Verified

Product of zeros: $\alpha\beta = \left(\frac{\sqrt{3}}{4}\right)\left(-\frac{2}{\sqrt{3}}\right) = -\frac{2}{4} = \frac{-2\sqrt{3}}{4\sqrt{3}} = \frac{c}{a}$   Verified

Q3. Find the zeros of $t^2 - 15$ and verify the relationship. 2 Marks

Sol. $t^2 - 15 = t^2 - (\sqrt{15})^2 = (t - \sqrt{15})(t + \sqrt{15}) = 0$

$\Rightarrow t = \sqrt{15}, \; -\sqrt{15}$. Here $a = 1, b = 0, c = -15$.

Sum of zeros: $\sqrt{15} + (-\sqrt{15}) = 0 = \frac{-0}{1} = \frac{-b}{a}$   Verified

Product of zeros: $(\sqrt{15})(-\sqrt{15}) = -15 = \frac{-15}{1} = \frac{c}{a}$   Verified

Q4. If one zero of $(k-1)x^2 + kx + 1$ is the negative of the other, find $k$. 1 Mark (MCQ)

Sol. Let zeros be $\alpha$ and $-\alpha$.

Sum of zeros $= \alpha + (-\alpha) = 0 \implies \frac{-b}{a} = 0 \implies \frac{-k}{k-1} = 0 \implies \mathbf{k = 0}$.

Q5. If one zero of $3x^2 + 8x + k$ is the reciprocal of the other, find $k$. 1 Mark (MCQ)

Sol. Let zeros be $\alpha$ and $\frac{1}{\alpha}$.

Product of zeros $= \alpha \cdot \frac{1}{\alpha} = 1 \implies \frac{c}{a} = 1 \implies \frac{k}{3} = 1 \implies \mathbf{k = 3}$.

Q6. If $\alpha$ and $\beta$ are zeros of $2x^2 - 5x + 3$, find the value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$. 3 Marks

Sol. Here $a=2, b=-5, c=3$

$\alpha + \beta = \frac{-(-5)}{2} = \frac{5}{2}$,   $\alpha\beta = \frac{3}{2}$

$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta}$

$= \frac{\left(\frac{5}{2}\right)^2 - 2\left(\frac{3}{2}\right)}{\frac{3}{2}} = \frac{\frac{25}{4} - 3}{\frac{3}{2}} = \frac{\frac{13}{4}}{\frac{3}{2}} = \frac{13}{4} \times \frac{2}{3} = \mathbf{\frac{13}{6}}$.

🎯 Board Pattern Shortcuts to Memorize:
  • Reciprocal zeros $\implies c = a$
  • Zeros opposite in sign $\implies b = 0$
  • Sum of squares: $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$
  • Sum of cubes: $\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$

📋 Board Revision Checklist

  • ✅ Zeros of $p(x)$ = x-axis intersections of graph of $y = p(x)$ (ignore y-axis)
  • ✅ Parabola opens upwards if $a > 0$, downwards if $a < 0$
  • ✅ Max zeros of polynomial of degree $n$ = $n$
  • ✅ Quadratic: Sum of zeros $= -b/a$, Product $= c/a$
  • ✅ Reciprocal zeros ($\alpha = 1/\beta$) → $c = a$
  • ✅ Zeros equal in magnitude, opposite in sign ($\alpha = -\beta$) → $b = 0$
  • ✅ Forming polynomial: $k[x^2 - (\text{sum})x + (\text{product})]$
  • ✅ $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$
  • ✅ $1/\alpha + 1/\beta = (\alpha+\beta)/(\alpha\beta) = -b/c$
  • ✅ $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta}$
  • ✅ Always verify: Sum and Product match $-b/a$ and $c/a$
💡 Exam Tip:
In 3-mark questions, always write verification separately — “Sum of zeros = ... = $-b/a$ = ... ✅ Verified”. This earns full marks even if factorisation has a minor slip.

Concept Mastery Quiz 🎯

Test your readiness for the board exam.

1. The graph of a quadratic polynomial is a:

2. If zeros of $x^2 + 7x + 10$ are $\alpha, \beta$, then $\alpha + \beta$ = ?

3. A quadratic polynomial whose graph does not cross the x-axis has:

4. If product of zeros of $3x^2 + kx - 9$ is $-3$, then $k$ = ?

5. The number of zeros a cubic polynomial can have is: