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Chapter 2 — Polynomials

Class 10 · Mathematics

Overview

Introduction: This chapter introduces polynomials as expressions formed from variables and coefficients using only addition, subtraction and multiplication by non-negative integer powers of the variable. You learn standard terms (monomial, binomial, trinomial), degree of a polynomial, and special cases (zero and constant polynomials). Importance: Polynomials form a foundation for algebra and are used throughout mathematics (equations, functions, calculus) and applications (physics, engineering, computer science). Mastery of polynomials helps in factorisation, solving equations and understanding higher-level topics. Key themes: The chapter focuses on (1) recognising polynomials and determining their degree, (2) zeros (roots) of polynomials and their relation to factors, (3) the Division Algorithm for polynomials, (4) the Remainder Theorem and Factor Theorem, and (5) using these results to factorise polynomials and solve problems. What you will learn: You will be able to identify and classify polynomials, compute degrees, find zeros, apply the Division Algorithm to divide polynomials, use the Remainder Theorem to evaluate remainders quickly, use the Factor Theorem to test and…

Learning Objectives

  • Define a polynomial, coefficient, degree and leading coefficient with examples
  • Classify polynomials by degree (constant, linear, quadratic, cubic) and by number of terms (monomial, binomial, trinomial)
  • State the Remainder Theorem and use it to find the remainder when a polynomial is divided by (x − a)
  • Explain the Factor Theorem and use it to determine whether (x − a) is a factor of a polynomial
  • Apply polynomial long division and synthetic division to divide one polynomial by another and obtain quotient and remainder
  • Identify zeros (roots) of a polynomial and verify them by direct substitution
  • Relate the zeros of a quadratic polynomial to its coefficients and use the relation to find missing coefficients or zeros
  • Factorize quadratic and higher-degree polynomials into linear and irreducible factors using factor theorem and common factor techniques

Topics in this chapter

11 topics · tap a topic title to jump straight to it.

🔢1

Introduction to Polynomials

📐 MATHEMATICAL FORMULA / THEOREM

Introduction to Polynomials

Key Point: General form: p(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0, with a_n ≠ 0.

What is a polynomial?
A polynomial in one variable x is an expression of the form a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0, where each a_i is a constant (called a coefficient), n is a non‑negative integer, and a_n ≠ 0. The highest power n with a_n ≠ 0 is called the degree of the polynomial.

Parts of a polynomial
- Term: each a_i x^i is a term.
- Coefficient: the number a_i multiplying x^i.
- Constant term: a_0 (term with x^0).

Types by number of terms
- Monomial: one term (e.g. 5x^3).
- Binomial: two terms (e.g. x^2 + 3x).
- Trinomial: three terms (e.g. x^2 + x + 1).

Special polynomials by degree
- Constant polynomial: degree 0 (e.g. 7).
- Linear polynomial: degree 1 (e.g. 2x − 3).
- Quadratic polynomial: degree 2 (e.g. x^2 − 4x + 4).
- Cubic: degree 3, etc. The zero polynomial (0) has all coefficients 0; its degree is usually left undefined (or sometimes taken as −∞).

Valid vs. invalid expressions
Valid polynomials have nonnegative integer exponents and a finite number of terms. Examples that are NOT polynomials: x^{−1} (negative exponent), 1/x (variable in denominator), x^{1/2} (fractional exponent), and infinite series like 1 + x + x^2 + ⋯.

Key properties (short)
- A polynomial of degree n can have at most n distinct real zeros (roots).
- Remainder theorem: when a polynomial p(x) is divided by (x − a), the remainder is p(a).
- Factor theorem: (x − a) is a factor of p(x) iff p(a) = 0.

How polynomials appear in problems
Polynomials model many algebraic quantities: lengths, areas, volumes, cost or revenue models (linear or quadratic), and approximate curves in physics/engineering.

📌 Examples
  • Area of a square with side (x + 2): (x + 2)^2 = x^2 + 4x + 4 (a quadratic polynomial).
  • Perimeter of rectangle with sides x and (2x + 1): 2[x + (2x + 1)] = 6x + 2 (a linear polynomial).
  • Displacement under uniform acceleration: s(t) = ut + (1/2)at^2 (a quadratic polynomial in time t).
  • Volume of a cuboid with sides (x), (x + 1), (x + 2): x(x + 1)(x + 2) = x^3 + 3x^2 + 2x (a cubic polynomial).
  • Cost function: if fixed cost is 50 and cost per unit is 20, total cost for x units: C(x) = 20x + 50 (linear polynomial).
🧮 Formulas
  1. \[General form: p(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0\]
    \[with a_n ≠ 0.\]
  2. \[Degree of p(x): n (largest exponent with nonzero coefficient).\]
  3. \[Remainder theorem: remainder on dividing p(x) by (x − a) = p(a).\]
  4. \[Factor theorem: (x − a) is a factor of p(x) ⇔ p(a) = 0.\]
  5. \[Max number of real zeros = degree of polynomial (counted without multiplicity).\]
  6. \[Quadratic relations (for ax^2 + bx + c = 0): sum of roots = −b/a\]
    \[product of roots = c/a\]
    \[discriminant Δ = b^2 − 4ac (Δ > 0 → two distinct real roots\]
    \[Δ = 0 → one real repeated root\]
    \[Δ < 0 → complex roots).\]
🔢2

Degree of a Polynomial and Leading Coefficient

📐 MATHEMATICAL FORMULA / THEOREM

Degree of a Polynomial and Leading Coefficient

Key Point: If p(x)=a_n x^n+...+a_1 x + a_0 with a_n ≠ 0, then deg p(x) = n and leading coefficient = a_n

What is a polynomial? A polynomial in one variable x is an expression of the form a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0, where each a_i is a real number (coefficient) and n is a non‑negative integer.

Degree of a polynomial
The degree of a nonzero polynomial is the highest power (exponent) of x that has a nonzero coefficient. If p(x)=a_n x^n+...+a_0 with a_n ≠ 0, then deg p(x)=n.

Leading coefficient
The coefficient a_n of the highest-power term a_n x^n is called the leading coefficient of the polynomial.

Special case — zero polynomial
The zero polynomial 0 has every coefficient equal to 0. Its degree is undefined (often treated as −∞ for algebraic convenience) and it has no well-defined leading coefficient.

How to find degree and leading coefficient — steps

  • Combine like terms first (add coefficients of same power).
  • Identify the term with the highest exponent whose coefficient is not zero.
  • The exponent of that term is the degree; its coefficient is the leading coefficient.

Key properties

  • If deg P = m and deg Q = n then deg(P+Q) ≤ max(m,n). (If leading terms cancel, the degree may be smaller.)
  • If deg P = m and deg Q = n then deg(P·Q) = m + n, provided the product is not the zero polynomial.
  • The leading coefficient of P·Q is the product of the leading coefficients of P and Q.
  • An nth-degree polynomial can have at most n real turning points and at most n real roots (counting multiplicity).

Why it matters
Degree and leading coefficient determine the polynomial's end behavior and overall shape: for large |x| the highest-degree term a_n x^n dominates. If n is even, both ends of the graph go in the same direction; if odd, ends go in opposite directions. The sign of a_n (leading coefficient) decides whether ends go up or down.

Example (brief)
For p(x)=4x^3 - x + 7: degree = 3, leading coefficient = 4. For q(x)=x^2 - 5x + 6: degree = 2, leading coefficient = 1.

Class 10 tip: Always combine like terms before deciding degree. If leading terms cancel when adding polynomials, re-evaluate the highest remaining power.

📌 Examples
  • p(x) = 4x^3 - x + 7 → degree = 3, leading coefficient = 4
  • q(x) = 0 (zero polynomial) → degree is undefined (often written −∞); no leading coefficient
  • r(x) = 5x^4 + 0·x^3 - 2x^2 + x → degree = 4, leading coefficient = 5
  • Add: P(x)=3x^3+2x^2 and Q(x)=-3x^3+4x → P+Q = 2x^2+4x → degree = 2 (leading terms cancelled)
  • Multiply: P(x)=2x^2+x and Q(x)=x^3-1 → deg(P)=2, deg(Q)=3 ⇒ deg(PQ)=5; leading coefficient = (2)(1)=2
🧮 Formulas
  1. \[If p(x)=a_n x^n+...+a_1 x + a_0 with a_n ≠ 0\]
    \[then deg p(x) = n and leading coefficient = a_n\]
  2. \[deg(P+Q) ≤ max(deg P\]
    \[deg Q) (equality may fail if leading terms cancel)\]
  3. \[deg(P·Q) = deg P + deg Q (unless product is zero polynomial)\]
  4. \[If leading coefficients are a_m and b_n\]
    \[leading coefficient of P·Q = a_m · b_n\]
  5. \[Degree of zero polynomial is undefined (commonly treated as −∞ in algebraic contexts)\]
🔢3

Types of Polynomials

📐 MATHEMATICAL FORMULA / THEOREM

Types of Polynomials

Key Point: General form: P(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0, with a_n ≠ 0.

What is a polynomial? A polynomial in one variable x is an expression of the form a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0, where a_n, ..., a_0 are real numbers and n is a non-negative integer. The highest power n with a_n ≠ 0 is called the degree of the polynomial and a_n is the leading coefficient.

Classification by number of terms

  • Monomial: A polynomial with one term (e.g. 5x^3, 7).
  • Binomial: Two terms (e.g. x^2 + 3x).
  • Trinomial: Three terms (e.g. x^2 + 5x + 6).
  • Polynomial (general): More than three terms (e.g. x^4 - 2x^3 + x - 5).

Classification by degree

  • Constant polynomial: Degree 0, e.g. 7.
  • Linear polynomial: Degree 1, e.g. 2x + 3. Graph is a straight line.
  • Quadratic polynomial: Degree 2, e.g. ax^2 + bx + c. Graph is a parabola.
  • Cubic polynomial: Degree 3, e.g. x^3 - 2x + 1. Graph has an S-like shape.
  • Quartic and higher: Degree 4, 5, ... with more complex shapes.

Special case — Zero polynomial: The polynomial 0 (all coefficients zero). Its degree is undefined or sometimes taken as minus infinity; treat it separately in classification.

Important properties

  • The degree of the sum or product of polynomials: deg(P + Q) ≤ max(deg P, deg Q), deg(P·Q) = deg P + deg Q (if neither is zero).
  • A polynomial of degree n has at most n distinct real zeros (roots).

These classifications help in solving equations, sketching graphs, and modelling real situations where relationships between quantities follow polynomial expressions.

📌 Examples
  • Linear (degree 1): Distance traveled at constant speed — distance = speed × time + initial distance (e.g. d = 60t + 5).
  • Quadratic (degree 2): Area of a square as a function of side length — A = s^2. Also projectile motion height vs time: h(t) = -4.9t^2 + vt + h0 (approx.).
  • Cubic (degree 3): Volume scaling with linear dimension — if length scales by x, some composite volume formulas become cubic in x; certain engineering stress–strain relations may be approximated by cubics.
  • Monomial: Cost of identical items — cost = price_per_item × number_of_items (e.g. 50x).
  • Binomial: Simple profit model with fixed and variable parts — P(x) = revenue(x) - cost(x) could be x^2 - 5x (if revenue ~ x^2 and cost ~ 5x).
🧮 Formulas
  1. \[General form: P(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0\]
    \[with a_n ≠ 0.\]
  2. \[Degree: deg(P) = n (the highest exponent with nonzero coefficient).\]
  3. \[Classification by degree: constant (deg 0)\]
    \[linear (deg 1)\]
    \[quadratic (deg 2)\]
    \[cubic (deg 3)\]
    \[etc.\]
  4. \[Classification by terms: monomial (1 term)\]
    \[binomial (2 terms)\]
    \[trinomial (3 terms).\]
  5. \[Degree of product: deg(P·Q) = deg P + deg Q (if P and Q are nonzero).\]
  6. \[Roots relation: A polynomial of degree n has at most n distinct real roots.\]
🌱4

Zeros (Roots) of a Polynomial

📐 MATHEMATICAL FORMULA / THEOREM

Zeros (Roots) of a Polynomial

Key Point: Zero condition: p(r) = 0 ⇔ (x − r) is a factor of p(x).

Definition: A zero (or root) of a polynomial p(x) is a number r such that p(r) = 0. Graphically, zeros are the x‑coordinates where the polynomial curve meets the x‑axis.

Factor Theorem: If r is a zero of p(x), then (x − r) is a factor of p(x). Conversely, if (x − r) is a factor, then r is a zero.

Remainder Theorem: When p(x) is divided by (x − c), the remainder is p(c). So p(c) = 0 exactly when (x − c) is a factor.

Multiplicity: If (x − r)k divides p(x) but (x − r)k+1 does not, r is a root of multiplicity k. Multiplicity affects the graph: odd multiplicity => curve crosses the x‑axis; even multiplicity => curve touches and bounces off the axis.

Fundamental Theorem of Algebra: A polynomial of degree n (with complex coefficients) has exactly n roots in the complex numbers, counted with multiplicity. For real polynomials some roots may be complex (non‑real) and appear in conjugate pairs.

How to find zeros (common methods):

  • Factorization: express p(x) as product of linear/quadratic factors.
  • Quadratic formula: for ax2+bx+c=0, use x = [−b ± sqrt(b2 − 4ac)]/(2a).
  • Synthetic division or polynomial long division to reduce degree once a root is known.
  • Graphing or numerical methods for difficult cases.

Relation between coefficients and zeros (Vieta's formulas): For p(x)=anxn + an−1xn−1 + ... + a1x + a0, with roots r1, r2, …, rn (counted with multiplicity):

  • Sum of roots = r1 + r2 + ... + rn = −an−1/an.
  • Product of roots = r1·r2·...·rn = (−1)n a0/an.

Discriminant (quadratic): D = b2 − 4ac determines nature of roots: D > 0 two distinct real roots, D = 0 one repeated real root, D < 0 two complex conjugate roots.

Interpretation and checking: After finding candidate roots, substitute into p(x) to verify p(r) = 0. If a root is known, divide p(x) by (x − r) to reduce degree and find remaining roots.

📌 Examples
  • Linear: p(x) = 2x − 6. Solve 2x − 6 = 0 ⇒ x = 3. So zero = 3 and factor = (x − 3).
  • Quadratic (factorable): p(x) = x^2 − 5x + 6 = (x − 2)(x − 3). Zeros: x = 2, 3.
  • Quadratic (formula): p(x) = 2x^2 − 4x + 1. D = (−4)^2 − 4·2·1 = 16 − 8 = 8. Roots: x = [4 ± √8]/(4) = (4 ± 2√2)/4 = 1 ± (√2)/2.
  • Cubic (using factor theorem): p(x) = x^3 − 6x^2 + 11x − 6. Try x = 1 ⇒ p(1)=0, so (x − 1) factor. Divide to get (x − 1)(x^2 − 5x + 6) = (x − 1)(x − 2)(x − 3). Zeros: 1, 2, 3.
  • Real‑life (projectile): Height h(t) = −5t^2 + 20t (height in metres). Solve −5t^2 + 20t = 0 ⇒ −5t(t − 4) = 0 ⇒ t = 0 s (launch), t = 4 s (hits ground).
🧮 Formulas
  1. \[Zero condition: p(r) = 0 ⇔ (x − r) is a factor of p(x).\]
  2. \[Remainder Theorem: remainder of p(x) ÷ (x − c) = p(c).\]
  3. \[Quadratic formula: For ax^2 + bx + c = 0\]
    \[x = [−b ± √(b^2 − 4ac)] / (2a).\]
  4. \[Discriminant: D = b^2 − 4ac (D &gt\]
    \[0 two real roots\]
    \[D = 0 one repeated root\]
    \[D &lt\]
    \[0 complex roots).\]
  5. \[Vieta (general): For degree n\]
    \[sum of roots = −a_{n−1}/a_n\]
    \[product = (−1)^n a_0 / a_n.\]
⚖️5

Operations on Polynomials

📐 MATHEMATICAL FORMULA / THEOREM

Operations on Polynomials

Key Point: Addition/subtraction: combine like terms. Example: (a_n x^n + ... + a_0) ± (b_n x^n + ... + b_0) = (a_n ± b_n) x^n + ... + (a_0 ± b_0).

What is a polynomial? A polynomial in a variable x is an expression of the form a_n xn + a_{n-1} xn-1 + ... + a_1 x + a_0 where coefficients a_i are real numbers and n is a non-negative integer. The term with the highest power n gives the degree of the polynomial. (The zero polynomial has no well-defined degree.)

Basic operations:

  • Addition / Subtraction: Add or subtract corresponding coefficients of like powers (terms with same power of x). The degree of the sum is at most the maximum of the degrees of the addends; cancellation can make it smaller.
  • Multiplication: Use the distributive law: multiply every term of one polynomial by every term of the other and collect like terms. The degree of the product equals the sum of the degrees of the factors (unless cancellation reduces it).
  • Division: Polynomial long division (or synthetic division for divisors of form x - a) produces a quotient and a remainder. Division algorithm:
    dividend = divisor × quotient + remainder, where degree(remainder) < degree(divisor).
  • Remainder theorem: If a polynomial f(x) is divided by (x − a), the remainder is f(a). Useful for quick evaluation and checking factors.
  • Factor theorem: (x − a) is a factor of f(x) ⇔ f(a) = 0. This helps factor polynomials to find roots.

Techniques and shortcuts: Use algebraic identities (binomial square, difference of squares, sum/difference of cubes) to multiply or factor polynomials quickly. For dividing by linear factors, synthetic division saves work.

Worked idea (short): To add 3x2 + 5x − 2 and 2x2 − 4x + 7, line up like terms: (3x2+2x2) + (5x−4x) + (−2+7) = 5x2 + x + 5.

Why it matters (applications): Polynomials model areas, volumes, cost/revenue functions, trajectories (quadratics), and are building blocks for algebra, calculus and many applied problems.

📌 Examples
  • Addition: Add P(x)=3x^2+5x−2 and Q(x)=2x^2−4x+7. Result: P(x)+Q(x)=5x^2+x+5.
  • Multiplication (binomials): Multiply (x+2)(x−3). Using FOIL: x^2−3x+2x−6 = x^2−x−6.
  • Use identity: Expand (x+1)^3. (x+1)^3 = x^3+3x^2+3x+1 (binomial cube).
  • Division & remainder: Divide f(x)=2x^3+3x^2−5x+1 by (x−2). Synthetic division (or long division) gives quotient 2x^2+7x+9 and remainder 19, so f(x) = (x−2)(2x^2+7x+9)+19. By remainder theorem f(2)=19.
  • Factor theorem: If f(x)=x^3−6x^2+11x−6, test x=1: f(1)=0 so (x−1) is a factor. Dividing gives (x−1)(x−2)(x−3).
🧮 Formulas
  1. \[Addition/subtraction: combine like terms\]
    \[Example: (a_n x^n + ... + a_0) ± (b_n x^n + ... + b_0) = (a_n ± b_n) x^n + ... + (a_0 ± b_0).\]
  2. \[Multiplication degree rule: deg(P·Q) = deg(P) + deg(Q) (in general\]
    \[if leading coefficients don't cancel).\]
  3. \[Division algorithm: Dividend = Divisor × Quotient + Remainder\]
    \[with deg(Remainder) < deg(Divisor).\]
  4. \[Remainder theorem: Remainder when f(x) is divided by (x−a) = f(a).\]
  5. \[Factor theorem: (x−a) is a factor of f(x) ⇔ f(a) = 0.\]
  6. \[Binomial identities: (a+b)^2 = a^2 + 2ab + b^2\]
    \[(a−b)^2 = a^2 − 2ab + b^2\]
6

Division Algorithm for Polynomials

📐 MATHEMATICAL FORMULA / THEOREM

Division Algorithm for Polynomials

Key Point: Division Algorithm: p(x) = g(x)·q(x) + r(x), where g(x) ≠ 0 and either r(x) = 0 or deg r(x) < deg g(x).

Statement. For any two polynomials p(x) (dividend) and g(x) (divisor) with g(x) ≠ 0, there exist unique polynomials q(x) (quotient) and r(x) (remainder) such that

p(x) = g(x)·q(x) + r(x),

where either r(x) = 0 or deg r(x) < deg g(x). This is called the Division Algorithm for Polynomials.

Why it holds (idea). By repeated subtraction of suitable multiples of g(x) from p(x) (just like integer long division) we lower the degree step by step. When the remaining polynomial has degree less than deg g(x), we stop — that remainder is r(x) and all subtractions collected form q(x). Uniqueness follows because if two different pairs (q,r) and (q',r') satisfied the relation, subtracting gives (q−q')·g(x) = r'−r; degree considerations force q=q' and r=r'.

How to perform the division (long division — procedure).

  1. Arrange both p(x) and g(x) in descending powers of x, inserting zero-coefficient terms if necessary.
  2. Divide the leading term of p(x) by the leading term of g(x) to get the first term of q(x).
  3. Multiply g(x) by that term and subtract from p(x); the result is a new smaller-degree polynomial.
  4. Repeat steps 2–3 until the remaining polynomial has degree less than deg g(x). That remainder is r(x).

Special useful results.

  • Remainder Theorem: If you divide p(x) by (x − a), the remainder is p(a). So p(x) = (x − a)·q(x) + p(a).
  • Factor Theorem: (x − a) is a factor of p(x) iff p(a) = 0.

Practical notes. For linear divisors x − a, synthetic division (a tabular method) is faster than polynomial long division. For higher-degree divisors use long division or compare coefficients after multiplication.

Checks and uniqueness. After division, verify by computing g(x)·q(x) + r(x) and confirming it equals p(x). The quotient and remainder are unique for the given p and g.

📌 Examples
  • Example 1 (linear divisor, using Remainder Theorem): Divide p(x) = x^3 - 2x^2 + 4x - 3 by (x - 1). Remainder = p(1) = 1 - 2 + 4 - 3 = 0, so (x - 1) is a factor. Using long/synthetic division gives quotient q(x) = x^2 - x + 3 and remainder r(x) = 0.
  • Example 2 (long division): Divide p(x) = 2x^3 + 3x^2 - x + 5 by g(x) = x^2 + 1. Long division yields quotient q(x) = 2x + 3 and remainder r(x) = -3x + 2, so 2x^3 + 3x^2 - x + 5 = (x^2 + 1)(2x + 3) + (-3x + 2). Note deg r (1) &lt; deg g (2).
  • Example 3 (synthetic division quick use): Divide p(x) = 3x^4 - x^3 + 0x^2 + 4x - 2 by (x + 2) i.e. a = -2. Synthetic division with -2 gives quotient coefficients and remainder p(-2). Compute p(-2) directly if only remainder is needed.
🧮 Formulas
  1. \[Division Algorithm: p(x) = g(x)·q(x) + r(x)\]
    \[where g(x) ≠ 0 and either r(x) = 0 or deg r(x) &lt\]
    \[deg g(x).\]
  2. \[Remainder Theorem: Remainder when p(x) is divided by (x − a) is p(a).\]
  3. \[Factor Theorem: (x − a) is a factor of p(x) ⇔ p(a) = 0.\]
  4. \[Degree condition: deg r(x) &lt\]
    \[deg g(x).\]
  5. \[Uniqueness: For fixed p and g (g ≠ 0) the quotient q and remainder r are unique.\]
🔢7

Remainder Theorem

📐 MATHEMATICAL FORMULA / THEOREM

Remainder Theorem

Key Point: Remainder Theorem: If f(x) is divided by (x − a), remainder R = f(a).

Statement. If a polynomial f(x) is divided by (x − a), then the remainder is f(a).

Short proof. By polynomial division, f(x) = (x − a)·q(x) + r(x), where the degree of r(x) < degree of (x − a) = 1. Hence r(x) is a constant r. Put x = a to get f(a) = r. Thus remainder = f(a).

Consequences and extensions.

  • Factor Theorem: (x − a) is a factor of f(x) if and only if f(a) = 0.
  • If the divisor is (ax − b) (a ≠ 0), you can rewrite it as a(x − b/a). The remainder when f(x) is divided by (ax − b) equals f(b/a).
  • For a divisor g(x) of degree n, the remainder is a polynomial r(x) of degree < n. To find r(x) (of degree < n) write f(x) = g(x)·Q(x) + r(x) and determine the coefficients of r(x) by equating coefficients or by substituting n distinct values of x (for example, the n roots of g(x) if available).

How this helps: It gives a quick way to find the remainder without doing full long division, helps test if a linear factor divides a polynomial (factor theorem), and is a standard tool for factoring and root-finding of polynomials in Class 10 problems.

📌 Examples
  • Example 1: Let f(x) = x^3 − 4x^2 + x + 6. Remainder on division by (x − 2) is f(2) = 8 − 16 + 2 + 6 = 0. So (x − 2) is a factor.
  • Example 2: For the same f(x), remainder on division by (x − 1) is f(1) = 1 − 4 + 1 + 6 = 4. So remainder = 4.
  • Example 3: If divisor is (2x − 3), set x = 3/2. For f(x) = x^3 − 4x^2 + x + 6, remainder = f(3/2) = 15/8 = 1.875.
  • Example 4 (divisor of higher degree): Divide f(x) = x^3 + x + 2 by g(x) = x^2 + 1. Do polynomial division or write remainder as ax + b. Division gives quotient x + 0 and remainder 2 (so remainder is constant 2), i.e. x^3 + x + 2 = (x^2 + 1)(x) + 2.
🧮 Formulas
  1. \[Remainder Theorem: If f(x) is divided by (x − a)\]
    \[remainder R = f(a).\]
  2. \[Factor Theorem: (x − a) is a factor of f(x) ⇔ f(a) = 0.\]
  3. \[General linear divisor: If divisor is (ax − b)\]
    \[remainder = f(b/a).\]
  4. \[Degree rule: If divisor has degree n\]
    \[remainder has degree &lt\]
    \[n.\]
  5. \[Polynomial identity: f(x) = (x − a)·q(x) + f(a) for some polynomial q(x).\]
🔢8

Factor Theorem and Factorization

📐 MATHEMATICAL FORMULA / THEOREM

Factor Theorem and Factorization

Key Point: Remainder Theorem: Remainder when p(x) is divided by (x − a) is p(a).

What is a factor of a polynomial? A polynomial q(x) is a factor of a polynomial p(x) if there exists a polynomial r(x) such that p(x) = q(x)·r(x). In particular, a linear polynomial (x − a) is a factor of p(x) if p(x) can be written as (x − a)·g(x).

Remainder Theorem (brief): If a polynomial p(x) is divided by (x − a), the remainder is p(a).

Factor Theorem (statement): For a polynomial p(x), (x − a) is a factor of p(x) if and only if p(a) = 0. Equivalently, a is a zero/root of p(x) exactly when (x − a) divides p(x).

Why it works (short proof idea): By division algorithm, p(x) = (x − a)·q(x) + r, where r is a constant. Putting x = a gives p(a) = r. So p(a) = 0 ⇔ r = 0 ⇔ p(x) is exactly divisible by (x − a).

Use of Factor Theorem in factorization: To factor a polynomial with integer coefficients, we

  1. Guess a possible rational root a (by Rational Root Test: possible roots are factors of constant term divided by factors of leading coefficient).
  2. Evaluate p(a). If p(a) = 0, then (x − a) is a factor.
  3. Divide p(x) by (x − a) (long division or synthetic division) to get a lower-degree polynomial q(x).
  4. Factor q(x) further (repeat the process) until p(x) is expressed as product of irreducible factors (linear or irreducible quadratics).

Types of factorization often used in Class 10: factoring quadratics (ax^2 + bx + c) into linear factors using zeroes or the quadratic formula; factoring cubic polynomials using the factor theorem and synthetic division; factorization by grouping for polynomials with four terms.

Multiplicity of roots: If p(a) = p'(a) = ... = p^{(k-1)}(a) = 0 but p^{(k)}(a) ≠ 0, then (x − a)^k is a factor and a is a root of multiplicity k. Graphically, multiplicity affects whether the curve crosses (odd multiplicity) or touches and turns (even multiplicity) at x = a.

Practical tips: Start trying small integer divisors of the constant term (±1, ±2, …). Use synthetic division for quick division. Check for common factors first. For quadratics, use sum-product relations (sum of roots = −b/a, product = c/a) to help factor.

📌 Examples
  • Example 1 — Factor x^2 − 5x + 6: Find roots by solving x^2 − 5x + 6 = 0. Factors of 6 that sum to 5 are 2 and 3. So x^2 − 5x + 6 = (x − 2)(x − 3). Verify: p(2)=0 and p(3)=0, so (x−2) and (x−3) are factors.
  • Example 2 — Factor x^3 − 6x^2 + 11x − 6: Try possible integer roots ±1, ±2, ±3, ±6. p(1)=0, so (x − 1) is a factor. Divide by (x − 1) to get x^2 − 5x + 6, which factors as (x − 2)(x − 3). Full factorization: (x − 1)(x − 2)(x − 3).
  • Example 3 — Factor 2x^3 + x^2 − 5x − 2: Possible rational roots are ±1, ±2, ±1/2. Evaluate p(1)= (2+1−5−2)=−4; p(2)= (16+4−10−2)=8; p(−1)= (−2+1+5−2)=2; p(1/2)= (2·1/8 + 1/4 − 5/2 −2)=0 → so x=1/2 is a root. Hence (x − 1/2) or equivalently (2x −1) is a factor. Divide to get (2x −1)(x^2 + x + 2) and check whether quadratic factors further (here it does not over reals).
  • Example 4 — Factor by grouping: x^3 + 3x^2 + 2x + 6. Group as (x^3 + 3x^2) + (2x + 6) = x^2(x + 3) + 2(x + 3) = (x + 3)(x^2 + 2).
🧮 Formulas
  1. \[Remainder Theorem: Remainder when p(x) is divided by (x − a) is p(a).\]
  2. \[Factor Theorem: (x − a) is a factor of p(x) ⇔ p(a) = 0.\]
  3. \[Division identity: p(x) = (x − a)·q(x) + r (where r is constant)\]
    \[and r = p(a).\]
  4. \[Quadratic relations: For ax^2 + bx + c = 0 with roots α, β: α + β = −b/a, α·β = c/a.\]
  5. \[Rational Root Test (useful rule): Any rational root of a polynomial with integer coefficients is of the form ±(factor of constant term)/(factor of leading coefficient).\]
  6. \[Multiplicity: If (x − a)^k divides p(x) but (x − a)^{k+1} does not\]
    \[root a has multiplicity k\]
    \[Even multiplicity → curve touches x-axis\]
    \[odd multiplicity → curve crosses.\]
🔢9

Relationship between Zeros and Coefficients

📐 MATHEMATICAL FORMULA / THEOREM

Relationship between Zeros and Coefficients

Key Point: General polynomial: If p(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_0 = a_n Π_{i=1}^n (x - r_i), then r1 + r2 + ... + rn = -a_{n-1}/a_n

Definition: For a polynomial p(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 (a_n ≠ 0), a zero (or root) is a number r such that p(r)=0. If the zeros are r1, r2, ..., rn (counted with multiplicity), the polynomial can be written as a_n(x - r1)(x - r2)...(x - rn).

Vieta's relations (core idea): Expanding a_n(x - r1)(x - r2)...(x - rn) and comparing coefficients with a_n x^n + a_{n-1} x^{n-1} + ... + a_0 gives symmetric relations between the zeros and the coefficients. In particular:

  • Sum of the zeros: r1 + r2 + ... + rn = -a_{n-1}/a_n.
  • Sum of products of zeros taken two at a time: Σ_{i
  • More generally, the k-th elementary symmetric sum of the zeros equals (-1)^k a_{n-k}/a_n.
  • Product of all zeros: r1 r2 ... rn = (-1)^n a_0 / a_n.

Quadratic special case: For ax^2 + bx + c (a ≠ 0) with zeros α and β,

  • α + β = -b/a
  • αβ = c/a

These give quick checks and help reconstruct a quadratic if zeros are known: a(x - α)(x - β) = ax^2 - a(α + β)x + aαβ.

Consequences and useful facts:

  • If coefficients are real, non-real complex zeros occur in conjugate pairs.
  • For a quadratic ax^2 + bx + c, the axis of symmetry x-coordinate is (α + β)/2 = -b/(2a); the vertex y-value can be expressed using coefficients or discriminant.
  • The sign of the product of real zeros equals the sign of c/a for a quadratic; zero constant term a_0 = 0 implies x = 0 is a root.
  • Vieta's relations let you find one root when others are known, or check correctness of factorization.

Why it works (brief derivation): Write p(x)=a_nΠ(x - r_i). Expanding, the coefficient of x^{n-1} is -a_n Σ r_i, so a_{n-1} = -a_n Σ r_i. Similarly, the constant term is a_n (-1)^n Π r_i, so a_0 = a_n (-1)^n Π r_i. This yields the formula pattern above.

📌 Examples
  • Example 1 (Quadratic): p(x) = 2x^2 - 5x + 3. Factor or find roots: Discriminant Δ = (-5)^2 - 4·2·3 = 1, roots = (5 ± 1) / (2·2) = 3/2 and 1. Check: sum = 3/2 + 1 = 5/2 = -b/a = -(-5)/2; product = (3/2)·1 = 3/2 = c/a = 3/2.
  • Example 2 (Cubic): p(x) = x^3 - 6x^2 + 11x - 6. Known factorization: (x-1)(x-2)(x-3). Zeros 1,2,3. Sum = 6 = -(-6)/1, sum of pairwise products = 1·2 + 1·3 + 2·3 = 11 = a_{1}/a_3, product = 1·2·3 = 6 = (-1)^3 a_0/a_3 = 6.
  • Example 3 (Real-life — break-even/profit): If profit P(x) = -2x^2 + 40x - 150 (x = number of units), the zeros of P give break-even production levels. Using α + β = -b/a = -40/(-2) = 20 and αβ = c/a = -150/(-2) = 75, so the two break-even units multiply to 75 and add to 20; solving gives x = 5 and x = 15.
  • Example 4 (Projectile motion): Vertical displacement s(t) = -5t^2 + 20t + 15 (meters). Zeros are times when projectile hits ground. Sum of times = -b/a = -20/(-5) = 4 s, so if first nonnegative root is t1 and second t2, they add to 4 s. (Compute exact roots for times.)
🧮 Formulas
  1. \[General polynomial: If p(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_0 = a_n Π_{i=1}^n (x - r_i)\]
    \[then r1 + r2 + ... + rn = -a_{n-1}/a_n\]
  2. \[Sum of products of zeros taken two at a time: Σ_{i<j} r_i r_j = a_{n-2}/a_n\]
  3. \[k-th elementary symmetric sum: Σ_{1≤i1<...<ik≤n} r_{i1}...r_{ik} = (-1)^k a_{n-k}/a_n\]
  4. \[Product of all zeros: r1 r2 ... rn = (-1)^n a_0 / a_n\]
  5. \[Quadratic special case: For ax^2 + bx + c, α + β = -b/a and αβ = c/a\]
  6. \[Axis of symmetry (quadratic): x_vertex = (α + β)/2 = -b/(2a)\]
🔢10

Formation of Polynomials from Given Zeros

📐 MATHEMATICAL FORMULA / THEOREM

Formation of Polynomials from Given Zeros

Key Point: General: If zeros are α1, α2, ..., αn, polynomial P(x) = k ∏_{i=1}^n (x − αi), where k ≠ 0 (choose k = 1 for a monic polynomial).

Idea: If α, β, γ, ... are the zeros (roots) of a polynomial, then the polynomial (up to a nonzero constant factor k) can be written as the product of linear factors (x−α)(x−β)(x−γ).... The degree of the polynomial equals the number of zeros counting multiplicity.

Steps to form a polynomial from given zeros:

  • Write the factor for each zero: (x − zero). If a zero has multiplicity m, include that factor m times.
  • Multiply the factors to get the polynomial. Optionally choose a constant k (often 1) so the polynomial has the desired leading coefficient or integer coefficients.
  • If zeros are fractions, clear denominators by multiplying by an appropriate k to get integer coefficients.
  • If a complex number a + bi (b ≠ 0) is given as a zero and you want a polynomial with real coefficients, include its conjugate a − bi as a zero too (complex roots occur in conjugate pairs).

Relations (Vieta's formulas) for coefficients: For a monic polynomial with zeros α, β (quadratic): x^2 − (α+β)x + αβ. For cubic with zeros α, β, γ: x^3 − (α+β+γ)x^2 + (αβ+βγ+γα)x − αβγ.

Common pitfalls: Remember multiplicity (even multiplicity => the graph touches the x-axis; odd multiplicity => it crosses). When zeros are rational fractions, multiply by the LCM of denominators to avoid fractional coefficients.

📌 Examples
  • Example 1 (two simple integer zeros): Zeros 2 and −3. Polynomial (take k = 1): (x − 2)(x + 3) = x^2 + x − 6.
  • Example 2 (fractional zeros): Zeros 1/2 and −3/4. Form factors: (x − 1/2)(x + 3/4) = x^2 + (1/4)x − 3/8. Multiply by 8 (LCM of denominators) to clear fractions: 8x^2 + 2x − 3 (k = 8).
  • Example 3 (repeated root): Zeros 2 (multiplicity 2) and −1. Polynomial: (x − 2)^2(x + 1) = (x^2 − 4x + 4)(x + 1) = x^3 − 3x^2 + 4.
  • Example 4 (complex conjugate pair): Zeros 2 + i and 2 − i (for real coefficients). Polynomial: (x − (2+i))(x − (2−i)) = (x − 2 − i)(x − 2 + i) = (x − 2)^2 + 1 = x^2 − 4x + 5.
🧮 Formulas
  1. \[General: If zeros are α1, α2, ..., αn\]
    \[polynomial P(x) = k ∏_{i=1}^n (x − αi)\]
    \[where k ≠ 0 (choose k = 1 for a monic polynomial).\]
  2. \[Quadratic (monic) with zeros α, β: x^2 − (α + β)x + αβ.\]
  3. \[Cubic (monic) with zeros α, β, γ: x^3 − (α + β + γ)x^2 + (αβ + βγ + γα)x − αβγ.\]
  4. \[If zeros include complex a ± bi and polynomial has real coefficients\]
    \[include both conjugates so imaginary parts cancel.\]
🔢11

Applications and Problem Solving

📐 MATHEMATICAL FORMULA / THEOREM

Applications and Problem Solving

Key Point: Division algorithm: For polynomials P(x) and non-zero D(x), there exist unique Q(x) and R(x) such that P(x)=D(x)Q(x)+R(x), where deg R < deg D.

Overview
In Class 10 Polynomials, 'Applications and Problem Solving' means modeling real situations by polynomials, using algebraic tools (factorisation, division algorithm, remainder and factor theorems) to solve the model and interpreting results in context. Typical problems ask for unknown lengths, times, counts, costs, or to check whether a given expression is a factor or to find remainders quickly.

Common steps to solve application problems

  • Understand the situation and choose a variable (say x) for the unknown quantity.
  • Translate relationships into an algebraic expression or polynomial.
  • Simplify to get a polynomial equation (often quadratic or cubic).
  • Use factorisation, the remainder theorem, the factor theorem, or division algorithm to solve or reduce the polynomial.
  • Check roots for extraneous or non-physical values (negative lengths, etc.) and interpret the solution in context.

Key techniques and how they help

  • Factorisation: converts a polynomial equation into simpler linear factors so you can find roots quickly.
  • Remainder theorem: evaluate P(a) to get remainder when P(x) is divided by (x − a) — faster than long division.
  • Factor theorem: if P(a)=0 then (x − a) is a factor; used to test and factor polynomials and find unknown coefficients.
  • Division algorithm: P(x)=D(x)Q(x)+R(x) is used when dividing polynomials; R(x) has smaller degree than D(x).

Real-life contexts: area/perimeter problems (lengths expressed as linear expressions lead to quadratic area), profit/cost/revenue modeled by polynomials, mixture problems, sequence sums, and design/layout partitioning (dividing land, tiling), and checking feasibility constraints (non-negative roots).

Tips

  • Always write units and check whether roots make sense physically.
  • If you obtain a polynomial with degree > 2, try rational root test or factor by grouping.
  • Sketching the graph of the polynomial helps visualize number and multiplicity of real roots.
📌 Examples
  • 1) Remainder theorem (quick evaluation). Problem: Find remainder when P(x)=2x^3−3x^2+4x−5 is divided by x−2. Solution: By remainder theorem remainder = P(2)=2(8)−3(4)+4(2)−5=16−12+8−5=7. So remainder = 7.
  • 2) Factor theorem (find factor and root). Problem: For P(x)=x^3−6x^2+11x−6 show x−1 is a factor and find other factors. Solution: P(1)=1−6+11−6=0 so x−1 is a factor. Divide P(x) by x−1 or factor by inspection: P(x)=(x−1)(x−2)(x−3). Roots: x=1,2,3.
  • 3) Area problem (model → quadratic → factor). Problem: A rectangle has length (x+5) m and width (x−2) m. If area = 84 m², find x. Solution: (x+5)(x−2)=84 → x^2+3x−10−84=0 → x^2+3x−94=0. Solve by quadratic formula: x=[−3±√(9+376)]/2=[−3±√385]/2. Only positive root used (approx x≈8.31). Then actual dimensions ≈ 13.31 m and 6.31 m. (If problem expects integer result, numbers would be chosen to factor.)
  • 4) Practical partitioning (using factorisation). Problem: A square plot of side (x+3) m is divided into a smaller square of side (x−1) m and a rectangular strip whose area is 20 m². Write equation and solve. Solution: Total area (x+3)^2 = (x−1)^2 + 20. Expand: x^2+6x+9 = x^2−2x+1 + 20 → 8x + 9 = −2x + 21 → 10x = 12 → x = 1.2 m. Interpret dimensions accordingly.
  • 5) Using division algorithm to find unknown coefficient. Problem: P(x)=x^3+ax^2+bx+6 leaves remainder 0 when divided by x−1 and remainder 8 when divided by x−2. Find a and b. Solution: P(1)=1+a+b+6=0 → a+b=−7. P(2)=8+4a+2b+6=8 → 4a+2b+14=8 → 4a+2b=−6 → divide by 2: 2a+b=−3. Solve with a+b=−7: subtract gives a=4, then b=−11.
🧮 Formulas
  1. \[Division algorithm: For polynomials P(x) and non-zero D(x)\]
    \[there exist unique Q(x) and R(x) such that P(x)=D(x)Q(x)+R(x)\]
    \[where deg R < deg D.\]
  2. \[Remainder theorem: Remainder when P(x) is divided by (x−a) is P(a).\]
  3. \[Factor theorem: (x−a) is a factor of P(x) iff P(a)=0.\]
  4. \[Common factorisations: x^2−y^2=(x−y)(x+y)\]
    \[x^2+2xy+y^2=(x+y)^2\]
    \[x^2−2xy+y^2=(x−y)^2.\]
  5. \[Quadratic relations (useful when quadratics arise): For ax^2+bx+c=0\]
    \[sum of roots = −b/a\]
    \[product of roots = c/a.\]

Key Concepts

Polynomial
An expression formed by finite sum of terms of the form a_n x^n where n is a non-negative integer and coefficients a_n are real numbers.
Monomial
A polynomial with only one term (single term) of the form ax^n.
Binomial
A polynomial with exactly two terms.
Trinomial
A polynomial with exactly three terms.
Degree of a polynomial
The highest power (exponent) of x that appears with a non-zero coefficient in the polynomial.
Leading term
The term of a polynomial which contains the highest power of x.
Leading coefficient
The coefficient of the leading term (coefficient of the highest power of x).
Coefficient
The numerical factor multiplied by a power of x in a term of a polynomial.
Constant term
The term in a polynomial that does not contain the variable x (degree 0 term).
Standard form
A polynomial written with terms in descending powers of x (highest power first).
Zero (root) of a polynomial
A value a for which f(a) = 0; i.e., it makes the polynomial equal to zero.
Zero polynomial
The polynomial 0 (all coefficients are zero). Its degree is not defined (often left undefined or treated as -∞).
Factor of a polynomial
A polynomial g(x) is a factor of f(x) if f(x) can be expressed as g(x) · h(x) for some polynomial h(x).
Linear polynomial
A polynomial of degree 1, of the form ax + b with a ≠ 0.
Quadratic polynomial
A polynomial of degree 2, of the form ax^2 + bx + c with a ≠ 0.
Remainder Theorem
When a polynomial f(x) is divided by (x - a), the remainder is f(a).
Factor Theorem
If f(a) = 0 for a polynomial f(x), then (x - a) is a factor of f(x); conversely, if (x - a) is a factor then f(a)=0.
Multiplicity of a root
The number of times a particular root a is repeated as a factor (x - a)^k in the factorization; k is the multiplicity.
Division algorithm (polynomial division)
For polynomials f(x) and g(x) (g ≠ 0) there exist unique polynomials q(x) (quotient) and r(x) (remainder) such that f(x) = g(x)q(x) + r(x), where degree r < degree g.
Irreducible polynomial
A non-constant polynomial that cannot be expressed as a product of two non-constant polynomials with real coefficients.

Practice Questions

  1. Define the degree and leading coefficient of a polynomial, and give them for p(x) = 4x³ − x + 7. / किसी बहुपद की घात और अग्रणी गुणांक को परिभाषित करें, और p(x) = 4x³ − x + 7 के लिए बताएं।
    Show answer

    The degree is the highest power of x with a nonzero coefficient, and the leading coefficient is the coefficient of that highest-power term. For p(x) = 4x³ − x + 7, the degree is 3 and the leading coefficient is 4. / घात x की वह उच्चतम घात है जिसका गुणांक शून्येतर हो, और अग्रणी गुणांक उस उच्चतम घात वाले पद का गुणांक है। p(x) = 4x³ − x + 7 के लिए घात 3 और अग्रणी गुणांक 4 है।

  2. State the Remainder Theorem and use it to find the remainder when p(x) = 2x³ − 3x² + 4x − 5 is divided by (x − 2). / शेषफल प्रमेय बताएं और इसका उपयोग करके p(x) = 2x³ − 3x² + 4x − 5 को (x − 2) से विभाजित करने पर शेषफल ज्ञात करें।
    Show answer

    The Remainder Theorem states that the remainder when p(x) is divided by (x − a) is p(a). So remainder = p(2) = 2(8) − 3(4) + 4(2) − 5 = 16 − 12 + 8 − 5 = 7. / शेषफल प्रमेय कहती है कि p(x) को (x − a) से विभाजित करने पर शेषफल p(a) होता है। अतः शेषफल = p(2) = 2(8) − 3(4) + 4(2) − 5 = 16 − 12 + 8 − 5 = 7।

  3. Using the Factor Theorem, show that (x − 1) is a factor of p(x) = x³ − 6x² + 11x − 6 and fully factorize it. / गुणनखंड प्रमेय का उपयोग करके दिखाएं कि (x − 1), p(x) = x³ − 6x² + 11x − 6 का एक गुणनखंड है और इसका पूर्ण गुणनखंडन करें।
    Show answer

    p(1) = 1 − 6 + 11 − 6 = 0, so by the Factor Theorem (x − 1) is a factor. Dividing gives x² − 5x + 6 = (x − 2)(x − 3), so p(x) = (x − 1)(x − 2)(x − 3). / p(1) = 1 − 6 + 11 − 6 = 0, अतः गुणनखंड प्रमेय से (x − 1) एक गुणनखंड है। विभाजित करने पर x² − 5x + 6 = (x − 2)(x − 3) मिलता है, अतः p(x) = (x − 1)(x − 2)(x − 3)।

  4. For the quadratic 2x² − 5x + 3, find the sum and product of its zeros using the coefficient relations and verify. / द्विघात 2x² − 5x + 3 के लिए, गुणांक संबंधों का उपयोग करके इसके शून्यकों का योग और गुणनफल ज्ञात करें और सत्यापित करें।
    Show answer

    Sum of zeros = −b/a = −(−5)/2 = 5/2 and product = c/a = 3/2. The zeros are 3/2 and 1; check: 3/2 + 1 = 5/2 and (3/2)(1) = 3/2, which match. / शून्यकों का योग = −b/a = −(−5)/2 = 5/2 और गुणनफल = c/a = 3/2। शून्यक 3/2 और 1 हैं; जाँच: 3/2 + 1 = 5/2 और (3/2)(1) = 3/2, जो मेल खाते हैं।

  5. Form a quadratic polynomial whose zeros are 2 and −3. / एक द्विघात बहुपद बनाएं जिसके शून्यक 2 और −3 हैं।
    Show answer

    Taking k = 1, the polynomial is (x − 2)(x − (−3)) = (x − 2)(x + 3) = x² + x − 6. / k = 1 लेने पर, बहुपद (x − 2)(x − (−3)) = (x − 2)(x + 3) = x² + x − 6 है।

  6. Apply the Division Algorithm to divide p(x) = 2x³ + 3x² − x + 5 by g(x) = x² + 1, and state the degree condition on the remainder. / विभाजन एल्गोरिथम लागू करके p(x) = 2x³ + 3x² − x + 5 को g(x) = x² + 1 से विभाजित करें, और शेषफल पर घात शर्त बताएं।
    Show answer

    Long division gives quotient q(x) = 2x + 3 and remainder r(x) = −3x + 2, so 2x³ + 3x² − x + 5 = (x² + 1)(2x + 3) + (−3x + 2). The remainder satisfies deg r(x) = 1 < deg g(x) = 2. / दीर्घ विभाजन से भागफल q(x) = 2x + 3 और शेषफल r(x) = −3x + 2 मिलता है, अतः 2x³ + 3x² − x + 5 = (x² + 1)(2x + 3) + (−3x + 2)। शेषफल deg r(x) = 1 < deg g(x) = 2 को संतुष्ट करता है।

  7. Explain how the multiplicity of a root affects the graph of a polynomial at that root. / किसी मूल की बहुलता उस मूल पर बहुपद के ग्राफ को कैसे प्रभावित करती है, समझाएं।
    Show answer

    If (x − r) has odd multiplicity, the curve crosses the x-axis at x = r; if it has even multiplicity, the curve touches the x-axis and turns back (bounces off) without crossing. / यदि (x − r) की बहुलता विषम है, तो वक्र x = r पर x-अक्ष को काटता है; यदि यह सम है, तो वक्र x-अक्ष को छूता है और बिना काटे वापस मुड़ जाता है।

  8. If p(x) = x³ + ax² + bx + 6 leaves remainder 0 when divided by (x − 1) and remainder 8 when divided by (x − 2), find a and b. / यदि p(x) = x³ + ax² + bx + 6 को (x − 1) से विभाजित करने पर शेषफल 0 और (x − 2) से विभाजित करने पर शेषफल 8 रहता है, तो a और b ज्ञात करें।
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    p(1) = 1 + a + b + 6 = 0 gives a + b = −7; p(2) = 8 + 4a + 2b + 6 = 8 gives 4a + 2b = −6, i.e. 2a + b = −3. Subtracting: a = 4, then b = −11. / p(1) = 1 + a + b + 6 = 0 से a + b = −7; p(2) = 8 + 4a + 2b + 6 = 8 से 4a + 2b = −6, अर्थात् 2a + b = −3। घटाने पर: a = 4, फिर b = −11।

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