Class 10 Maths Chapter 2: Polynomials Notes & Formulas

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

CBSE 2026–27 Syllabus Status: Polynomials

Unit II: Algebra • 4–5 Marks

This chapter is strictly structured according to the CBSE Rationalised Mathematics Syllabus. Focus your board preparation only on prescribed topics and avoid studying deleted sections:

In 2026–27 Board Syllabus:

Deleted / Not in Board Exam:

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)

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 of zeroes } (\alpha + \beta) = -\frac{b}{a} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} $
$ \text{Product of zeroes } (\alpha\beta) = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} $

2. Forming a Polynomial

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

(where $k$ is any non-zero real 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 $\left(\alpha^2 + \beta^2\right)$	$(\alpha+\beta)^2 - 2\alpha\beta$	$\left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) = \frac{b^2 - 2ac}{a^2}$
Sum of reciprocals $\left(\frac{1}{\alpha} + \frac{1}{\beta}\right)$	$\frac{\alpha+\beta}{\alpha\beta}$	$\frac{-b/a}{c/a} = -\frac{b}{c}$
Ratio sum $\left(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\right)$	$\frac{\alpha^2+\beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta}$	$\frac{b^2 - 2ac}{ac}$

📌 Remember the Signs:
Sum of zeros = $-\frac{b}{a}$ (note the negative sign!)
Product of zeros = $\frac{c}{a}$ (constant term / leading coefficient)

⚡ High-Yield Conceptual Rules (Board MCQs & A-R):

Definition of Polynomial: Exponents of variables must be strictly non-negative integers. Expressions with $x^{1/2}, \frac{1}{x}, \sqrt{x}$ are NOT polynomials.
Vertex of Parabola ($y = ax^2 + bx + c$): The turning point has coordinates $\left(-\frac{b}{2a}, -\frac{D}{4a}\right)$.
• If $a > 0$: Opens upwards, minimum value at vertex.
• If $a < 0$: Opens downwards, maximum value at vertex.
Graph vs Discriminant ($D = b^2 - 4ac$):
• $D > 0$: Cuts $x$-axis at 2 distinct points (2 distinct real zeros).
• $D = 0$: Touches $x$-axis at exactly 1 point (2 equal real zeros).
• $D < 0$: Does not touch or cut $x$-axis (0 real zeros).
Signs of Zeros & Coefficients:
• Both zeros positive $\implies -b/a > 0$ and $c/a > 0$.
• Both zeros negative $\implies -b/a < 0$ and $c/a > 0$.
• Zeros of opposite signs $\implies c/a < 0$ ($c$ and $a$ have opposite signs).
• One zero reciprocal of other $\implies c = a$.
• One zero negative of other $\implies b = 0$.

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: $\alpha\beta = (\frac{\sqrt{3}}{4})(-\frac{2}{\sqrt{3}}) = -\frac{2}{4} = \frac{-2\sqrt{3}}{4\sqrt{3}} = \frac{c}{a}$ 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 $a=2, b=-5, c=3$

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

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

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

🎯 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: $\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$

CBSE Section E • 4 Marks

Case Study: Highway Overpass Parabolic Arch

Scenario: Highway engineers design a suspension overpass whose supportive steel cable hangs in the shape of a downward parabola given by the quadratic model $p(x) = -x^2 + 2x + 8$, where $x$ is the horizontal distance (in decameters) from the left embankment and $p(x)$ is the height above road level.

Q1: Find the zeroes of the polynomial $p(x) = -x^2 + 2x + 8$ to determine the ground anchor points. 1 Mark

Q2: What is the total horizontal span between the two ground anchors? 1 Mark

Q3: Find the maximum height of the parabolic arch above road level. (OR) If another bridge follows $q(x) = x^2 - 4x + 3$, find a quadratic polynomial whose zeroes are $(2α+1)$ and $(2β+1)$ where $α, β$ are zeroes of $q(x)$. 2 Marks

📋 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: $k[x^2 - (sum)x + (product)]$
✅ $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$
✅ $1/\alpha + 1/\beta = (\alpha+\beta)/(\alpha\beta) = -b/c$
✅ $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta}$
✅ Always verify: Sum and Product match $-b/a$ and $c/a$
✅ Non-negative integer power rule: variable in denominator or under square root $\implies$ not a polynomial
✅ Parabola vertex $\left(-\frac{b}{2a}, -\frac{D}{4a}\right)$: minimum if $a > 0$, maximum if $a < 0$
✅ $D > 0 \implies 2$ x-intercepts; $D = 0 \implies 1$ touch point; $D < 0 \implies 0$ x-intercepts

💡 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:

Frequently Asked Questions: Class 10 Maths Chapter 2

What is the marks weightage of Polynomials in CBSE Class 10?

Polynomials is part of Unit 2: Algebra (total 20 marks) and typically contributes 3 to 4 marks through 1-mark MCQs and 2-mark or 3-mark questions on zeroes and coefficients.

Is polynomial division algorithm included in the syllabus?

No, the Division Algorithm for Polynomials has been deleted from the rationalized CBSE syllabus. The active focus is finding zeroes geometrically and algebraically, and verifying the relationship between zeroes and coefficients.

What are the formulas for sum and product of zeroes in a quadratic polynomial?

For p(x) = ax^2 + bx + c, the sum of zeroes is alpha + beta = -b/a and the product of zeroes is alpha * beta = c/a. A quadratic polynomial with zeroes alpha and beta can be formed as k[x^2 - (alpha + beta)x + alpha*beta].

What does the number of zeroes of a polynomial represent geometrically?

The number of real zeroes corresponds to the exact number of points where the graph y = p(x) intersects the x-axis.

Chapter 2 Study Modules & Practice Resources

Complete Notes & Formulas • NCERT Solutions • Previous Year Questions • Chapter Tests & Quizzes • Practice Worksheets

Chapter 2: Polynomials

Overview