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MathematicsCh-2 15 min comprehensive revision
NCERT Class 10 Mathematics — Chapter 2

Polynomials

Geometrical meaning of zeroes, parabolic curves, relations between zeroes and coefficients of quadratic polynomials, and symmetric algebraic identities.

Quick Key Takeaways:
Degree and Zeroes: A polynomial p(x)p(x) of degree nn has at most nn real zeroes, corresponding to the number of points where the graph y=p(x)y = p(x) intersects the X-axis.
Quadratic Zeroes Relations: For p(x)=ax2+bx+cp(x) = ax^2 + bx + c with zeroes α,β\alpha, \beta: α+β=−ba,αβ=ca\alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a}
Polynomial Formation: A quadratic polynomial with zeroes α,β\alpha, \beta is given by k[x2−(α+β)x+αβ]k[x^2 - (\alpha + \beta)x + \alpha\beta], where k≠0k \neq 0.
Symmetric Expressions: α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta and 1α+1β=α+βαβ\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta}.
Graph & Coordinate VisualizerExplore the geometric meaning of zeroes as x-intercepts of polynomial curves
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1

1. Geometrical Zeroes, Parabolic Curves & Degree Theorems

Axioms & Foundational Theory

Comprehensive mathematical foundations, definitions, formulas, and visual conceptual model for Polynomials.

• Geometrical Meaning of the Zeroes of a Polynomial
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Definition of a Zero: A real number kk is a zero of polynomial p(x)p(x) if p(k)=0p(k) = 0.
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Graphical Intersection with X-Axis: The real zeroes of y=p(x)y = p(x) are precisely the X-coordinates of the points where the graph intersects or touches the X-axis.
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Linear Polynomial ax+bax + b: Graph is a straight line intersecting the X-axis at exactly one point: (−ba,0)(-\frac{b}{a}, 0).
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Quadratic Polynomial ax2+bx+cax^2 + bx + c (a≠0a \neq 0): Graph is a parabola opening upwards if a>0a > 0, or downwards if a<0a < 0:
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Case 1 (D>0D > 0): Intersects X-axis at two distinct points   ⟹  \implies 2 distinct real zeroes.
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Case 2 (D=0D = 0): Touches X-axis at exactly one point   ⟹  \implies 2 equal real zeroes.
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Case 3 (D<0D < 0): Completely above or below X-axis   ⟹  \implies No real zeroes.
📊 Polynomials: Parabolic Zeroes & Geometric RootsVisual Model
2 Distinct ZeroesD > 0 (Intersects 2 times)αβ1 Real ZeroD = 0 (Touches x-axis)α = β0 Real ZeroesD < 0 (No intersection)No real roots

Visual schematic mapping the geometry of parabolas, X-axis intersection roots, and the algebraic zero-coefficient relations.

2

2. Relationship Between Zeroes and Coefficients & Symmetric Identities

Theorem Proofs & Derivations

Step-by-step rigorous geometric and algebraic theorem proofs and formula derivations for Polynomials.

• Derivation of Zeroes-Coefficient Relationship for Quadratic Polynomials
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Let α\alpha and β\beta be the zeroes of p(x)=ax2+bx+cp(x) = ax^2 + bx + c (a≠0a \neq 0).
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By Factor Theorem, (x−α)(x - \alpha) and (x−β)(x - \beta) are factors of p(x)p(x):
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ax2+bx+c=k(x−α)(x−β)=k[x2−(α+β)x+αβ]=kx2−k(α+β)x+kαβax^2 + bx + c = k(x - \alpha)(x - \beta) = k[x^2 - (\alpha + \beta)x + \alpha\beta] = kx^2 - k(\alpha+\beta)x + k\alpha\beta
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Equating coefficients of like powers of xx on both sides:
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x2x^2: a=ka = k
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xx: b=−k(α+β)=−a(α+β)  ⟹  α+β=−bab = -k(\alpha + \beta) = -a(\alpha + \beta) \implies \mathbf{\alpha + \beta = -\frac{b}{a}}
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Constant term: c=kαβ=aαβ  ⟹  αβ=cac = k\alpha\beta = a\alpha\beta \implies \mathbf{\alpha\beta = \frac{c}{a}}
• High-Yield Symmetric Polynomial Identities
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α2+β2=(α+β)2−2αβ=(−ba)2−2(ca)=b2−2aca2\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}
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(α−β)2=(α+β)2−4αβ=b2−4aca2  ⟹  ∣α−β∣=b2−4ac∣a∣(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = \frac{b^2 - 4ac}{a^2} \implies |\alpha - \beta| = \frac{\sqrt{b^2 - 4ac}}{|a|}
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α3+β3=(α+β)3−3αβ(α+β)=−b3a3+3bca2=3abc−b3a3\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = -\frac{b^3}{a^3} + \frac{3bc}{a^2} = \frac{3abc - b^3}{a^3}
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αβ+βα=α2+β2αβ=b2−2acac\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{b^2 - 2ac}{ac}
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3. Important Solved Board Examination Questions (3-Mark & 5-Mark)

Topper Step Solutions

Standard CBSE board exam numericals and riders with complete step-by-step mathematical working and justifications.

• 3-Mark Standard Board Question: If α\alpha and β\beta are the zeroes of the quadratic polynomial f(x)=x2−5x+kf(x) = x^2 - 5x + k such that α−β=1\alpha - \beta = 1, find the value of kk.
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Step 1 (Coefficients): In f(x)=x2−5x+kf(x) = x^2 - 5x + k, a=1,b=−5,c=ka = 1, b = -5, c = k.
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Step 2 (Sum and Product):
α+β=−−51=5— (1)\alpha + \beta = -\frac{-5}{1} = 5 \quad \text{--- (1)}
αβ=k1=k— (2)\alpha\beta = \frac{k}{1} = k \quad \text{--- (2)}
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Step 3 (Using Identity): Given α−β=1\alpha - \beta = 1.
(α−β)2=(α+β)2−4αβ(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta
(1)2=(5)2−4(k)(1)^2 = (5)^2 - 4(k)
1=25−4k  ⟹  4k=24  ⟹  k=61 = 25 - 4k \implies 4k = 24 \implies k = 6
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Final Boxed Answer: k=6\mathbf{k = 6}
• 5-Mark Heavyweight Board Problem / Rider: Find the zeroes of the quadratic polynomial p(x)=6x2−3−7xp(x) = 6x^2 - 3 - 7x and verify the relationship between the zeroes and the coefficients.
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Step 1 (Standard Form): Rewrite in descending powers of xx: p(x)=6x2−7x−3p(x) = 6x^2 - 7x - 3.
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Step 2 (Factorization by Splitting Middle Term):
Product =6×(−3)=−18= 6 \times (-3) = -18; Sum =−7= -7. Factors: −9-9 and +2+2.
6x2−9x+2x−3=3x(2x−3)+1(2x−3)=(2x−3)(3x+1)6x^2 - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (2x - 3)(3x + 1)
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Step 3 (Zeroes):
2x−3=0  ⟹  α=322x - 3 = 0 \implies \alpha = \frac{3}{2}
3x+1=0  ⟹  β=−133x + 1 = 0 \implies \beta = -\frac{1}{3}
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Step 4 (Verification of Sum):
- LHS: α+β=32+(−13)=9−26=76\text{LHS: } \alpha + \beta = \frac{3}{2} + \left(-\frac{1}{3}\right) = \frac{9 - 2}{6} = \frac{7}{6}
- RHS: −ba=−−76=76  ⟹  LHS=RHS\text{RHS: } -\frac{b}{a} = -\frac{-7}{6} = \frac{7}{6} \implies \text{LHS} = \text{RHS}.
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Step 5 (Verification of Product):
- LHS: αβ=(32)(−13)=−36=−12\text{LHS: } \alpha\beta = \left(\frac{3}{2}\right)\left(-\frac{1}{3}\right) = -\frac{3}{6} = -\frac{1}{2}
- RHS: ca=−36=−12  ⟹  LHS=RHS\text{RHS: } \frac{c}{a} = \frac{-3}{6} = -\frac{1}{2} \implies \text{LHS} = \text{RHS}.
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Conclusion: Relationships are verified.
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4. CBSE Case-Study Modeling: Highway Overpass Parabolic Arch Modeling

Case Study Mastery (4 Marks)

Real-world mathematical modeling scenario with structured multi-part questions and full step solutions.

• Practical Application Context: Highway Overpass Parabolic Arch Modeling
A highway overpass arch is constructed in the shape of a downward parabola given by the quadratic polynomial h(x)=−x2+2x+8h(x) = -x^2 + 2x + 8, where h(x)h(x) represents the height of the arch in meters and xx represents the horizontal distance from the left support.
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Q1: Find the horizontal span (width) between the arch foundations. →\rightarrow Set h(x)=0  ⟹  −x2+2x+8=0  ⟹  x2−2x−8=0  ⟹  (x−4)(x+2)=0  ⟹  x=4,x=−2h(x) = 0 \implies -x^2 + 2x + 8 = 0 \implies x^2 - 2x - 8 = 0 \implies (x - 4)(x + 2) = 0 \implies x = 4, x = -2. Horizontal distance span =4−(−2)=6 meters= 4 - (-2) = \mathbf{6\text{ meters}}.
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Q2: Find the maximum height of the parabolic arch. →\rightarrow Axis of symmetry x=−b2a=−22(−1)=1x = -\frac{b}{2a} = -\frac{2}{2(-1)} = 1. Maximum height h(1)=−(1)2+2(1)+8=−1+2+8=9 metersh(1) = -(1)^2 + 2(1) + 8 = -1 + 2 + 8 = \mathbf{9\text{ meters}}.
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Q3: What are the sum and product of the zeroes of this polynomial? →\rightarrow Sum =−ba=−2−1=2= -\frac{b}{a} = -\frac{2}{-1} = \mathbf{2}. Product =ca=8−1=−8= \frac{c}{a} = \frac{8}{-1} = \mathbf{-8}.
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5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

Important Solved Board Questions

Examiner step-marking allocations, common algebraic/arithmetic deduction traps, and presentation guidelines.

• Step-by-Step Marking Rubric & Presentation Guidelines
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1 Mark: Reorder polynomial into standard form ax2+bx+cax^2 + bx + c.
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1 Mark: Correct splitting of middle term and factor extraction.
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1 Mark: Calculation of zeroes α,β\alpha, \beta.
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1 Mark: Explicit verification steps for both sum (α+β=−b/a\alpha+\beta = -b/a) and product (αβ=c/a\alpha\beta = c/a).
• Common Calculation Traps & Verification Checklist
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Trap 1: Omitting the constant kk when forming a polynomial k[x2−(α+β)x+αβ]k[x^2 - (\alpha+\beta)x + \alpha\beta].
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Trap 2: Forgetting the negative sign in α+β=−b/a\alpha + \beta = -b/a.
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Trap 3: Misidentifying the number of zeroes from a graph by counting intersections with the Y-axis instead of the X-axis.
Authentic Board Question (3 Marks)Topic: Polynomials Algebraic Formulations & Exact Methods
Solve the algebraic equation, find unknown coefficients, and apply standard theorem methods for Polynomials.

Official CBSE Step-by-Step Marking Breakdown:

Step 1: Standard Form Equation & Algebraic Setup: Convert given conditions into standard algebraic form (ax2+bx+c=0ax^2+bx+c=0, an=a+(n−1)da_n = a+(n-1)d, or system of equations).
1 Mark
Step 2: Step-by-Step Factorisation / Reduction Method: Apply formal algebraic method (splitting the middle term, quadratic formula, elimination, or AP summation).
1 Mark
Step 3: Final Solution & Domain Verification: State the final values of x,n,dx, n, d, verifying against real-life boundary conditions (e.g. n>0n > 0, positive speeds).
1 Mark
Model Student Answer (Target: Full 3/3 Marks):
To score full 3 marks on Polynomials in CBSE Mathematics:

1. Algebraic Setup: Express the given problem into standard algebraic form.
2. Solving Technique: Carry out step-by-step algebraic manipulation showing all factorization or formula substitution lines.
3. Boxed Result: State and box the final root values, discarding any non-viable negative or fractional answers where context dictates.
Examiner Mark Deduction Traps:
•Always check boundary conditions (e.g., number of terms nn in an AP must be a positive natural number).
•Show all factorization steps clearly—do not jump directly from equation to roots.

High-Frequency Conceptual Doubts & FAQs

Curated answers to the most common questions asked by Class 10 students.
The zeroes of a polynomial y=p(x)y = p(x) correspond precisely to the xx-coordinates of the points where the graph of the polynomial intersects the xx-axis. A polynomial of degree nn can intersect the xx-axis at most nn times, and therefore has at most nn real zeroes.

Related YouTube Videos & Masterclasses

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