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 MarksPolynomials 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
| 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)
Number of real zeroes of $p(x)$ = Number of points where graph intersects the $x$-axisKey Formulas & Relationships
1. Relationship (Quadratic)
For zeroes $\alpha$ and $\beta$ of $ax^2 + bx + c$:
$$ \text{Product } (\alpha \beta) = \frac{c}{a} $$
2. Forming a Polynomial
(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 ($\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$) | $\frac{\alpha^2+\beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta}$ | $\frac{b^2 - 2ac}{ac}$ |
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)
$x^2 - 4x + 2x - 8 = x(x-4) + 2(x-4)$
$\Rightarrow (x+2)(x-4)$. Zeroes: $-2, 4$.
Sum $= -2 + 4 = 2$. Formula: $-(-2)/1 = 2$.
Product $= -2 \times 4 = -8$. Formula: $-8/1 = -8$.
Exam Strategy & Mistake Bank
⚠️ Mistake Bank
💡 Scoring Tips
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
- 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 $D = b^2 - 4ac$
📝 More Solved Board Questions
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
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
Sol. Let zeros be $\alpha$ and $-\alpha$.
Sum of zeros $= \alpha + (-\alpha) = 0 \implies \frac{-b}{a} = 0 \implies \frac{-k}{k-1} = 0 \implies \mathbf{k = 0}$.
Sol. Let zeros be $\alpha$ and $\frac{1}{\alpha}$.
Product of zeros $= \alpha \cdot \frac{1}{\alpha} = 1 \implies \frac{c}{a} = 1 \implies \frac{k}{3} = 1 \implies \mathbf{k = 3}$.
Sol. Here $a=2, b=-5, c=3$
$\alpha + \beta = \frac{-(-5)}{2} = \frac{5}{2}$, $\alpha\beta = \frac{3}{2}$
$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta}$
$= \frac{\left(\frac{5}{2}\right)^2 - 2\left(\frac{3}{2}\right)}{\frac{3}{2}} = \frac{\frac{25}{4} - 3}{\frac{3}{2}} = \frac{\frac{13}{4}}{\frac{3}{2}} = \frac{13}{4} \times \frac{2}{3} = \mathbf{\frac{13}{6}}$.
- 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 $c/a$
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: