Overview
Chapter: Polynomials (Class 9, NCERT Mathematics). This chapter introduces polynomials as algebraic expressions in one variable, explains their classification (monomial, binomial, trinomial) and degree, and presents the Division Algorithm for polynomials. It develops two fundamental tools: the Remainder Theorem (remainder on dividing by x − a is p(a)) and the Factor Theorem (x − a is a factor iff p(a) = 0), and shows how to use them to find zeroes and factorise polynomials (especially quadratic and simple cubic examples). Importance: Polynomials form the foundation for algebraic manipulation, solving equations, and later topics (quadratic equations, coordinate geometry, calculus). What the student will learn: precise definitions, how to perform polynomial division, how to compute and interpret remainders and factors, how to find and use zeros of polynomials, how zeros relate to coefficients (for quadratic), and how to factorise polynomials to solve problems and simplify expressions.
Learning Objectives
- Define polynomial, monomial, binomial and trinomial with examples.
- Identify degree and coefficients of a given polynomial.
- Classify polynomials as constant, linear, quadratic or of higher degree.
- Find zeros (roots) of linear and quadratic polynomials.
- Apply the Remainder Theorem to compute the remainder when a polynomial is divided by (x − a).
- Use the Factor Theorem to determine whether (x − a) is a factor of a polynomial.
- Perform polynomial division by a linear polynomial to obtain quotient and remainder.
- Factorise quadratic polynomials by using their zeros and factorisation methods.
Topics in this chapter
11 topics · tap a topic title to jump straight to it.
Introduction to Polynomials
Introduction to Polynomials
Key Point: General form: p(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0, where a_n ≠ 0 and n ∈ {0,1,2,...}.
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 the coefficients a_i are real numbers (or from some field) and n is a non‑negative integer. Each a_i x^i is called a term. The highest power n with a_n ≠ 0 is the degree of the polynomial.
Parts of a polynomial: coefficient (a_i), variable (x), exponent (i), constant term (a_0).
Classification by degree: degree 0 — constant polynomial (e.g., 5); degree 1 — linear (e.g., 3x − 2); degree 2 — quadratic (e.g., x^2 + x + 1); degree 3 — cubic, etc. A polynomial with one term is a monomial, two terms is a binomial, three terms is a trinomial.
Zero polynomial: the polynomial 0 (all coefficients zero) is called the zero polynomial; its degree is not defined (or sometimes said to be −∞).
Standard form and like terms: Write terms in descending powers of x to get the standard form. Combine like terms (same power of x) by adding coefficients.
Evaluation and zeros: To evaluate p(x) at x = a, substitute a for x and compute p(a). If p(a) = 0, then a is a root (zero) of the polynomial.
Important idea (Remainder Theorem): When a polynomial p(x) is divided by (x − a), the remainder is p(a). This links evaluation to division.
- p(x) = 4x^3 − x^2 + 2x + 7: degree = 3 (cubic), coefficients 4, −1, 2, 7.
- q(x) = 0: zero polynomial (no defined degree).
- r(x) = 6: constant polynomial of degree 0.
- s(x) = x + 5: linear polynomial. Graph is a straight line.
- Area of a rectangle with length (x + 3) and width 4: A(x) = 4(x + 3) = 4x + 12 (a linear polynomial modelling area).
- Volume of a cuboid with dimensions x, (x + 1), and 2: V(x) = 2x(x + 1) = 2x^2 + 2x (a quadratic polynomial modelling volume).
- \[General form: p(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0\]\[where a_n ≠ 0 and n ∈ {0,1,2,...}.\]
- \[Degree: deg(p) = n (highest exponent with nonzero coefficient).\]
- \[Classification by degree: constant (0)\]\[linear (1)\]\[quadratic (2)\]\[cubic (3)\]\[etc.\]
- \[Number of terms classification: monomial (1 term)\]\[binomial (2 terms)\]\[trinomial (3 terms).\]
- \[Evaluation: p(a) = a_n a^n + a_{n-1} a^{n-1} + ... + a_1 a + a_0.\]
- \[Remainder Theorem: remainder when p(x) is divided by (x − a) is p(a).\]
Types of Polynomials
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
In Class 9 Mathematics, 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_i are real numbers and a_n ≠ 0. The highest power n with nonzero coefficient is the degree of the polynomial.
Polynomials are classified in two common ways:
- By degree (highest power of x):
- Constant polynomial: degree 0 (e.g., 5)
- Linear polynomial: degree 1 (e.g., 2x + 3)
- Quadratic polynomial: degree 2 (e.g., x^2 + 4x + 4)
- Cubic polynomial: degree 3 (e.g., x^3 − 2x^2 + x)
- Quartic: degree 4; and so on (degree n gives an n-th degree polynomial).
- Zero polynomial (0): special case — its degree is not defined in CBSE convention.
- By number of nonzero terms:
- Monomial: one term (e.g., 7x^3)
- Binomial: two terms (e.g., x + 5)
- Trinomial: three terms (e.g., x^2 + 2x + 1)
- Polynomial (general): more than three terms (e.g., x^4 − x^3 + 3x − 2)
Other useful concepts:
- Leading coefficient: the coefficient a_n of the highest-degree term. If a_n = 1 the polynomial is called monic.
- Standard form: arrange terms in descending powers of x: a_n x^n + ... + a_1 x + a_0.
- Operations and degree rules: degree(p + q) ≤ max(degree(p), degree(q)); degree(p · q) = degree(p) + degree(q) provided neither is the zero polynomial.
This classification helps in solving, graphing and modeling problems. For example, many physical relationships are modeled by linear or quadratic polynomials (straight-line motion, area formulas, projectile motion approximations, revenue/cost functions, etc.).
- Constant: 7 (degree 0)
- Linear: 3x − 5 (degree 1) — graph is a straight line; models simple cost = fixed + variable × units
- Quadratic: x^2 + 4x + 4 (degree 2) — graph is a parabola; models area or projectile height
- Cubic: x^3 − 2x^2 + x (degree 3) — S-shaped curve, appears in volume or certain growth models
- Monomial: 5x^3 (one term)
- Binomial: x + 6 (two terms)
- \[General form: p(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0\]\[with a_n ≠ 0\]
- \[Degree: deg(p) = n (highest power with nonzero coefficient)\]\[For 0 polynomial degree is not defined (CBSE convention).\]
- \[Leading coefficient: coef. of x^{deg(p)} is a_n\]\[If a_n = 1\]\[p(x) is monic.\]
- \[Standard form: write terms in descending powers: a_n x^n + ... + a_1 x + a_0\]
- \[Addition: deg(p + q) ≤ max(deg(p)\]\[deg(q))\]\[equality may fail if leading terms cancel.\]
- \[Multiplication: deg(p · q) = deg(p) + deg(q) (if neither polynomial is the zero polynomial).\]
Degree and Coefficients
Degree and Coefficients
Key Point: If f(x) = a_n x^n + ... + a_1 x + a_0 with a_n ≠ 0, then degree(f) = 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 constant (real number) and n is a non‑negative integer. Each a_i x^i is called a term (or monomial), and a_i is its coefficient.
Coefficient: The coefficient of a term is the constant multiplying the variable power. Example: in 5x^3 − 2x + 4, the coefficient of x^3 is 5, of x is −2, and of x^0 (the constant term) is 4. If a power of x is missing, its coefficient is 0 (e.g., in x^3 + 2, the coefficient of x^2 is 0).
Degree of a term (monomial): For a single term a x^k (a ≠ 0) the degree is k (the exponent of x).
Degree of a polynomial (single variable): After combining like terms, the degree of a nonzero polynomial is the highest exponent of x that appears with a nonzero coefficient. For f(x) = a_n x^n + ... + a_0 with a_n ≠ 0, degree(f) = n, and a_n is called the leading coefficient. A nonzero constant has degree 0. The zero polynomial (0) has no well‑defined degree (often said to be "undefined" or −∞).
Multivariable polynomials: For a monomial x^a y^b z^c, the degree is a + b + c (sum of exponents). The degree of a multivariable polynomial is the maximum degree among its monomials.
Key properties:
- To determine degree, first simplify/combine like terms.
- For two nonzero polynomials f and g: degree(f + g) ≤ max(degree(f), degree(g)). Equality may fail if leading terms cancel.
- For nonzero polynomials: degree(f · g) = degree(f) + degree(g). The leading coefficient of the product is the product of leading coefficients.
- Leading term (a_n x^n) determines end behavior: for large |x| the polynomial behaves like its leading term.
Practical note: Always write the polynomial in descending powers (highest degree first) to identify the degree and leading coefficient quickly.
- Example 1: f(x) = 3x^4 - 5x^2 + 2x - 7. Degree = 4 (highest power). Leading coefficient = 3. Coefficients: of x^4 is 3, x^3 is 0, x^2 is -5, x is 2, constant is -7.
- Example 2: g(x) = 6x^3 + x^3 - 4x + 2. First combine like terms: g(x) = 7x^3 - 4x + 2. Degree = 3, leading coefficient = 7.
- Example 3 (constant and zero): h(x) = 5. Degree = 0 (leading coefficient 5). For the zero polynomial 0, degree is undefined (or sometimes said to be -∞).
- Example 4 (multivariable): p(x,y) = 4x^2y + 3xy^3 - 7y. Degrees of monomials: 4x^2y has degree 3 (2+1), 3xy^3 has degree 4 (1+3), -7y has degree 1. Degree(p) = 4 (from 3xy^3).
- Example 5 (cancellation): f(x) = x^3 + 2x^2 - x^3 + x. Simplify: f(x) = 2x^2 + x. Degree = 2 (x^3 terms cancel).
- \[If f(x) = a_n x^n + ... + a_1 x + a_0 with a_n ≠ 0\]\[then degree(f) = n and leading coefficient = a_n.\]
- \[Coefficient of x^k in a polynomial is the constant multiplying x^k after simplifying (0 if missing).\]
- \[Degree of constant (nonzero) polynomial = 0\]\[degree(0) is undefined (or -∞).\]
- \[For nonzero polynomials f and g: degree(f + g) ≤ max(degree(f)\]\[degree(g)).\]
- \[For nonzero polynomials f and g: degree(f · g) = degree(f) + degree(g).\]
- \[For monomial in several variables x^a y^b ...\]\[degree = a + b + ...\]\[degree of polynomial = maximum of its monomial degrees.\]
Zeros (Roots) of a Polynomial
Zeros (Roots) of a Polynomial
Key Point: Remainder Theorem: remainder when f(x) is divided by (x − a) is f(a).
Definition: A zero (or root) of a polynomial f(x) is a number 'a' such that f(a) = 0. Graphically, zeros are the x-coordinates where the graph of y = f(x) meets the x-axis.
Why zeros matter: Zeros help us factor polynomials, solve equations, find intercepts in graphs and model real situations (like times when profit is zero or when a projectile hits the ground).
Key ideas and methods:
- Remainder Theorem: If a polynomial f(x) is divided by (x − a), the remainder is f(a). So f(a) = 0 implies (x − a) divides f(x).
- Factor Theorem: (x − a) is a factor of f(x) if and only if f(a) = 0. This is used to find factors and reduce degree step by step.
- Multiplicity: If (x − a)^k is a factor of f(x), then a is a root of multiplicity k. If k is odd the graph crosses the x-axis at a; if k is even it touches and turns around at a.
- Number of roots: A non-zero polynomial of degree n has at most n real roots. (Over the complex numbers it has exactly n roots counting multiplicity.)
How to find zeros (practical steps):
- Try simple integer candidates using the Remainder Theorem (factor pairs of the constant term are good starting points).
- When you find a zero a, divide the polynomial by (x − a) (long division or synthetic division) to lower the degree.
- Factor the resulting polynomial further (quadratic formula, factorisation, or repeated application of the factor theorem) to find all zeros.
Connections with coefficients (quadratic): For ax^2 + bx + c with roots r1 and r2: r1 + r2 = −b/a and r1·r2 = c/a. These relations help check answers or form equations from given roots.
- Example 1 (Linear): f(x) = 2x − 6. Solve f(x) = 0 ⇒ 2x − 6 = 0 ⇒ x = 3. So 3 is the zero and (x − 3) is a factor.
- Example 2 (Quadratic, factorable): f(x) = x^2 − 5x + 6. Factor: (x − 2)(x − 3). Zeros: x = 2 and x = 3.
- Example 3 (Quadratic, no real zeros): f(x) = x^2 + 4x + 5. Discriminant Δ = 16 − 20 = −4 < 0, so no real zeros (two complex roots). Graph does not cross the x-axis.
- Example 4 (Using Remainder/Factor Theorem): f(x) = x^3 − 2x^2 + 3x − 6. Test x = 2: f(2) = 8 − 8 + 6 − 6 = 0, so (x − 2) is a factor. Divide to get quotient x^2 + 0x + 3, so f(x) = (x − 2)(x^2 + 3). Zeros: x = 2 and x = ±√(−3) (two complex roots).
- Example 5 (Multiplicity): f(x) = (x − 1)^2(x + 2). Zeros: x = 1 (multiplicity 2, graph touches at x=1) and x = −2 (multiplicity 1, graph crosses at x=−2).
- \[Remainder Theorem: remainder when f(x) is divided by (x − a) is f(a).\]
- \[Factor Theorem: (x − a) is a factor of f(x) ⇔ f(a) = 0.\]
- \[Quadratic relations: For ax^2 + bx + c with roots r1\]\[r2: r1 + r2 = −b/a and r1·r2 = c/a.\]
- \[Cubic relations (useful check): For ax^3 + bx^2 + cx + d with roots r1\]\[r2\]\[r3: r1 + r2 + r3 = −b/a\]\[r1r2 + r2r3 + r3r1 = c/a\]\[r1r2r3 = −d/a.\]
- \[Multiplicity: If (x − a)^k divides f(x) then a is a root of multiplicity k (even multiplicity ⇒ touch-and-turn\]\[odd ⇒ cross).\]
Polynomials Operations (brief)
Polynomials Operations (brief)
Key Point: Addition/Subtraction: combine like terms: a_n x^n + b_n x^n = (a_n + b_n) x^n
What is a polynomial: A polynomial in x is an expression of the form a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 where coefficients a_i are real numbers and n is a nonnegative integer. The highest power with nonzero coefficient is the degree.
Basic operations
- Addition and subtraction: Combine like terms (terms with the same power of x) by adding or subtracting their coefficients. Degree of the result is at most the maximum degree of the operands.
- Multiplication: Multiply every term of the first polynomial by every term of the second and then combine like terms. Exponents add when you multiply powers of x: x^m * x^n = x^{m+n}. Degree of product = sum of degrees.
- Division (Division algorithm): For polynomials f(x) and g(x) with g(x) not zero, there exist unique q(x) and r(x) such that f(x) = g(x)q(x) + r(x) where either r(x)=0 or deg r < deg g. You can perform polynomial long division or synthetic division when dividing by linear terms of the form x - a.
- Remainder theorem: When a polynomial f(x) is divided by (x - a), the remainder is f(a). If f(a)=0, then (x - a) is a factor of f(x) (factor theorem).
Quick rules and consequences
- Degrees: deg(sum) ≤ max(deg(f),deg(g)); deg(product) = deg(f)+deg(g) except when leading coefficients cancel.
- Zero polynomial has undefined degree (sometimes called -infinity) and all coefficients zero.
These operations let you simplify polynomial expressions, factor polynomials, solve polynomial equations and understand how polynomial expressions model quantities in real life.
- Addition: (3x^2 + 2x - 5) + (x^2 - 4x + 7) = (3x^2 + x^2) + (2x - 4x) + (-5 + 7) = 4x^2 - 2x + 2.
- Subtraction: (5x^3 - 2x + 1) - (x^3 + 3x - 4) = (5x^3 - x^3) + (-2x - 3x) + (1 + 4) = 4x^3 - 5x + 5.
- Multiplication: (x + 2)(x^2 - x + 3). Distribute: x*(x^2 - x + 3) + 2*(x^2 - x + 3) = x^3 - x^2 + 3x + 2x^2 - 2x + 6 = x^3 + x^2 + x + 6.
- Division (example + remainder theorem): Divide f(x)=2x^3 + 3x^2 - x + 5 by g(x)=x - 1. Using synthetic division with a = 1: coefficients 2, 3, -1, 5 produce quotient coefficients 2, 5, 4 and remainder 9. So 2x^3 + 3x^2 - x + 5 = (x - 1)(2x^2 + 5x + 4) + 9. By the remainder theorem f(1)=9, matching the remainder.
- \[Addition/Subtraction: combine like terms: a_n x^n + b_n x^n = (a_n + b_n) x^n\]
- \[Multiplication of powers: x^m * x^n = x^{m+n}\]
- \[Multiplying terms: (a x^m)(b x^n) = ab x^{m+n}\]
- \[Distributive law: p(x)q(x) = sum over all termwise products\]
- \[Square and cube identities: (a + b)^2 = a^2 + 2ab + b^2\]\[(a - b)^2 = a^2 - 2ab + b^2\]\[(a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3\]
- \[Division algorithm: For f(x) and g(x) ≠ 0 there exist unique q(x)\]\[r(x) with f(x) = g(x) q(x) + r(x) and either r(x)=0 or deg r < deg g\]
Division Algorithm for Polynomials
Division Algorithm for Polynomials
Key Point: Division identity: 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) and g(x) (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).
What it means (intuitively): When you divide one polynomial by another you get a quotient polynomial and a remainder polynomial just like with integers. The remainder’s degree is always smaller than the divisor’s degree.
Division procedure (long division / repeated subtraction):
- Arrange both polynomials in descending powers of x (fill missing terms with 0 coefficients).
- Divide the leading term of the dividend by the leading term of the divisor to get the first term of the quotient.
- Multiply the divisor by that term and subtract the result from the dividend.
- Bring down the next term(s) and repeat until the degree of the remaining polynomial (the current remainder) is less than the degree of the divisor.
- The final leftover is r(x); the collected terms are q(x).
Proof idea of existence and uniqueness (sketch): Existence is shown by performing the long division algorithm—each step reduces the degree of the dividend or stops when remainder degree is smaller than divisor. Uniqueness: if p = gq1 + r1 = gq2 + r2 then g(q1 − q2) = (r2 − r1). The left side is divisible by g, but the degree of (r2 − r1) is less than deg g unless r1 = r2, forcing q1 = q2 and r1 = r2.
Important corollaries:
- Remainder Theorem: The remainder on dividing p(x) by (x − a) is p(a).
- Factor Theorem: (x − a) is a factor of p(x) iff p(a) = 0.
Tips: For division by linear factors (x − a), use synthetic division (fast). For higher-degree divisors, use polynomial long division. Keep coefficients organized (use zeros for missing powers).
- Example 1 (division by linear polynomial using synthetic division): Divide p(x)=2x^3 + 3x^2 − 5x + 6 by g(x)=x − 2. Synthetic steps with root 2: coefficients 2, 3, −5, 6. Bring down 2; multiply 2·2=4 add to 3 →7; multiply 7·2=14 add to −5 →9; multiply 9·2=18 add to 6 →24. So quotient q(x)=2x^2 + 7x + 9 and remainder r(x)=24. Check: p(x)= (x−2)(2x^2+7x+9)+24. Also p(2)=24 (remainder theorem).
- Example 2 (division by quadratic using long division): Divide p(x)=x^3 + 2x^2 − 5x + 3 by g(x)=x^2 + 1. Long division: leading term x^3/x^2 = x; multiply g by x → x^3 + x; subtract: (x^3+2x^2−5x+3) − (x^3+x) = 2x^2 − 6x + 3. Next term 2x^2/x^2 = 2; multiply g by 2 → 2x^2 + 2; subtract: (2x^2 − 6x + 3) − (2x^2 + 2) = −6x + 1. Remainder r(x)=−6x + 1 (degree 1 < 2). Quotient q(x)=x + 2 so p(x)=(x^2+1)(x+2)+(−6x+1).
- \[Division identity: p(x) = g(x)·q(x) + r(x)\]\[where g(x) ≠ 0 and either r(x)=0 or deg r(x) < deg g(x).\]
- \[Degree inequality: deg p(x) = max(deg[g(x)·q(x)]\]\[deg r(x)) and deg r(x) <\]\[deg g(x).\]
- \[Remainder theorem: remainder of p(x) on division by (x − a) is r = p(a).\]
- \[Factor theorem: (x − a) is a factor of p(x) ⇔ p(a) = 0.\]
Remainder Theorem
Remainder Theorem
Key Point: Division algorithm: For polynomials f(x) and non-zero g(x), there exist q(x) and r(x) such that f(x) = g(x)·q(x) + r(x) where either r(x) = 0 or deg r(x) < deg g(x).
Definition: The Remainder Theorem gives a quick way to find the remainder when a polynomial f(x) is divided by a linear polynomial of the form (x − a).
Statement: If a polynomial f(x) is divided by (x − a), then the remainder is f(a).
Why it works (short proof): By the division algorithm for polynomials, f(x) can be written as
f(x) = (x − a)·q(x) + r,
where q(x) is the quotient and r is the remainder. Because the divisor (x − a) has degree 1, r is a constant. Substituting x = a gives f(a) = r. Hence the remainder equals f(a).
Corollary (Factor Theorem): (x − a) is a factor of f(x) if and only if f(a) = 0. That is, f(a) = 0 ⇔ (x − a) divides f(x) exactly.
Generalizations:
- If the divisor is of the form (ax − b), then the remainder is f(b/a) (substitute x = b/a).
- For a divisor g(x) of degree m, division gives f(x) = g(x)·q(x) + r(x) where deg r(x) < m.
How to use the theorem (steps):
- Identify the linear divisor (x − a). If divisor is (ax − b), rewrite as x − (b/a) and use a = b/a.
- Compute f(a) by substituting x = a in the polynomial.
- That value f(a) is the remainder.
Practical notes for Class 9 students: The theorem avoids long polynomial division for linear divisors and is often used to check whether a given linear expression is a factor of a polynomial (use the Factor Theorem).
- Example 1: Find the remainder when f(x) = x^3 − 2x^2 + 3x − 5 is divided by (x − 2). Using the Remainder Theorem, remainder = f(2) = 2^3 − 2·2^2 + 3·2 − 5 = 8 − 8 + 6 − 5 = 1.
- Example 2: Remainder when f(x) = x^4 + 1 is divided by (x + 1). Here (x + 1) = (x − (−1)), so a = −1. Remainder = f(−1) = (−1)^4 + 1 = 1 + 1 = 2.
- Example 3 (Factor test): f(x) = x^2 − 4x + 3. Test if (x − 1) is a factor. Compute f(1) = 1 − 4 + 3 = 0, so (x − 1) is a factor. Also f(3) = 9 − 12 + 3 = 0, so (x − 3) is a factor. Hence f(x) = (x − 1)(x − 3).
- Example 4 (non-monic linear divisor): Find remainder when f(x) = 2x^2 + 3x + 1 is divided by (4x − 2). Write 4x − 2 = 4(x − 1/2). So a = 1/2. Remainder = f(1/2) = 2(1/4) + 3(1/2) + 1 = 1/2 + 3/2 + 1 = 3.
- Example 5 (using division-degree fact): If a divisor g(x) has degree 2, the remainder will be of degree < 2 (a linear or constant polynomial). The Remainder Theorem applies directly only to linear divisors; for higher-degree divisors find r(x) by polynomial division or comparison.
- \[Division algorithm: For polynomials f(x) and non-zero g(x)\]\[there exist q(x) and r(x) such that f(x) = g(x)·q(x) + r(x) where either r(x) = 0 or deg r(x) <\]\[deg g(x).\]
- \[Remainder Theorem (linear divisor): If f(x) is divided by (x − a)\]\[remainder = f(a).\]
- \[Factor Theorem: (x − a) is a factor of f(x) ⇔ f(a) = 0.\]
- \[General linear divisor: If divisor is (ax − b)\]\[remainder = f(b/a).\]
- \[Degree rule: If divisor g(x) has degree m\]\[then deg(remainder) <\]\[m.\]
Factor Theorem
Factor Theorem
Key Point: Remainder form: p(x) = (x − a)·q(x) + r, where r is constant (r = p(a)).
Definition (Factor Theorem): For a polynomial p(x), (x − a) is a factor of p(x) if and only if p(a) = 0.
Relation to Remainder Theorem: When p(x) is divided by (x − a), the remainder is p(a). So p(a) = 0 implies remainder 0, hence (x − a) divides p(x).
Proof sketch: By polynomial division, p(x) = (x − a)q(x) + r where q(x) is a polynomial and r is a constant. Put x = a to get p(a) = r. Thus p(a) = 0 ⇔ r = 0 ⇔ (x − a) is a factor.
How to use it (step-by-step):
- Given p(x), list possible rational roots (for integer-coefficient polynomials these are divisors of the constant term divided by divisors of the leading coefficient).
- Evaluate p(a) for each candidate a. If p(a) = 0, then (x − a) is a factor.
- Divide p(x) by (x − a) (long division or synthetic division) to obtain a lower-degree polynomial q(x).
- Repeat the process on q(x) until fully factored.
Notes for Class 9: Often you will try integer candidates that divide the constant term (±1, ±2, ±3, ...). For quadratics, finding one root lets you factor it into two linear factors.
Example illustration (quick):
p(x) = x^2 − 5x + 6. Try x = 2: p(2)=4−10+6=0 ⇒ (x−2) is a factor. Divide by (x−2) ⇒ quotient (x−3) ⇒ p(x)=(x−2)(x−3).
- Example 1 (Quadratic): p(x) = x^2 − 5x + 6. Test x = 2: p(2)=0 so (x − 2) is a factor. Divide to get p(x) = (x − 2)(x − 3). Roots: x = 2, 3.
- Example 2 (Cubic with three integer roots): p(x) = x^3 − 6x^2 + 11x − 6. Test x = 1: p(1)=0 ⇒ (x − 1) factor. Divide to get x^2 − 5x + 6 = (x − 2)(x − 3). So p(x) = (x − 1)(x − 2)(x − 3).
- Example 3 (Use synthetic division): p(x) = x^3 + 2x^2 − 5x − 6. Try x = 2: p(2)=8+8−10−6=0 ⇒ (x − 2) factor. Synthetic divide to obtain quotient x^2 + 4x + 3 = (x + 1)(x + 3). So p(x) = (x − 2)(x + 1)(x + 3).
- Real-life example 1 (Break-even / economics): If profit P(q) for q units is modeled by a polynomial, roots of P(q) (found using factor theorem) are break-even production levels where profit is zero.
- Real-life example 2 (Projectile / physics approximation): If height h(t) of an object is approximated by a polynomial in time t, roots h(t)=0 give times when the object hits ground. Factor theorem helps find such times if polynomial factors conveniently.
- \[Remainder form: p(x) = (x − a)·q(x) + r\]\[where r is constant (r = p(a)).\]
- \[Remainder Theorem: remainder when p(x) is divided by (x − a) is p(a).\]
- \[Factor Theorem: p(a) = 0 ⇔ (x − a) is a factor of p(x)\]\[Equivalently p(x) = (x − a)·q(x) when p(a)=0.\]
- \[Rational root candidates (useful hint): possible rational roots = ±(factors of constant term)/(factors of leading coefficient)\]\[For integer coefficient polynomials testing integer divisors of constant term often suffices.\]
Factorization of Polynomials
Factorization of Polynomials
Key Point: a^2 − b^2 = (a − b)(a + b)
What is factorization? Factorization of a polynomial means writing it as a product of two or more polynomials of lower degree. For example, x^2 + 5x + 6 = (x + 2)(x + 3).
Why it matters: Factoring helps find zeros (roots) of polynomials, simplifies algebraic expressions, and solves equations and real-life problems (areas, arrangements, simplification).
Key idea — zeros and factors: If p(a) = 0 for a polynomial p(x), then (x − a) is a factor of p(x). Conversely, if (x − a) is a factor, then a is a zero of p(x).
Common methods used in Class 9:
- Taking out the common factor: Factor out the greatest common divisor of all terms. Example: 4x^3 − 8x^2 + 12x = 4x(x^2 − 2x + 3).
- Factorization of quadratic trinomials (ax^2 + bx + c): Use the split-the-middle-term method: find two numbers whose product = a·c and sum = b, then split the middle term and factor by grouping. Example: 6x^2 + 11x + 3: numbers 9 and 2 because 9·2 = 18 (=6·3) and 9 + 2 = 11. So 6x^2 + 9x + 2x + 3 = 3x(2x + 3) + 1(2x + 3) = (2x + 3)(3x + 1).
- Difference of squares: a^2 − b^2 = (a − b)(a + b). Example: x^2 − 9 = (x − 3)(x + 3).
- Perfect square trinomials: a^2 + 2ab + b^2 = (a + b)^2 and a^2 − 2ab + b^2 = (a − b)^2. Example: x^2 + 6x + 9 = (x + 3)^2.
- Factorization by grouping: Used when a polynomial has four terms. Group terms in pairs and factor each pair. Example: x^3 + 3x^2 + x + 3 = x^2(x + 3) + 1(x + 3) = (x + 3)(x^2 + 1).
Practical steps to factor a quadratic ax^2 + bx + c:
- Compute product a·c.
- Find two integers m and n such that m·n = a·c and m + n = b.
- Write bx as mx + nx and factor by grouping.
Tip about discriminant (useful idea): For ax^2 + bx + c, the quantity D = b^2 − 4ac tells whether real linear factors exist: if D > 0 two distinct real linear factors; D = 0 one repeated factor; D < 0 no real linear factors (irreducible over reals).
- Factor x^2 + 5x + 6: Find two numbers multiply to 6 and add to 5 → 2 and 3. So x^2 + 5x + 6 = (x + 2)(x + 3).
- Factor 6x^2 + 11x + 3: a·c = 18, find 9 and 2. Split: 6x^2 + 9x + 2x + 3 = 3x(2x + 3) + 1(2x + 3) = (2x + 3)(3x + 1).
- Factor x^2 − 9 using difference of squares: x^2 − 3^2 = (x − 3)(x + 3).
- Common factor: 12x^3 − 18x^2 = 6x^2(2x − 3).
- Grouping: x^3 + 3x^2 + x + 3 = (x^2 + 1)(x + 3).
- \[a^2 − b^2 = (a − b)(a + b)\]
- \[a^2 + 2ab + b^2 = (a + b)^2\]
- \[a^2 − 2ab + b^2 = (a − b)^2\]
- \[Quadratic factor relation: For ax^2 + bx + c with roots α, β → (x − α)(x − β) = ax^2 + bx + c (up to scaling)\]\[Sum of roots = −b/a\]\[Product of roots = c/a\]
- \[Discriminant D = b^2 − 4ac (D >\]\[0 two real distinct linear factors\]\[D = 0 repeated factor\]\[D <\]\[0 no real linear factors)\]
Relationship between Zeros and Coefficients (Quadratic focus)
Relationship between Zeros and Coefficients (Quadratic focus)
Key Point: For ax² + bx + c (a ≠ 0) with zeros α and β: α + β = −b/a
Consider a quadratic polynomial f(x) = ax² + bx + c with a ≠ 0. Let its zeros (roots) be α and β. If we factor the quadratic using its zeros we get f(x) = a(x − α)(x − β).
Expanding: a(x − α)(x − β) = ax² − a(α + β)x + aαβ. Comparing coefficients with ax² + bx + c gives the fundamental relations:
α + β = −b/a, and αβ = c/a.
These relations are powerful: they let you form a quadratic from given zeros (a(x − α)(x − β)), check roots quickly, and connect geometric features of the parabola with algebraic coefficients. For example the axis of symmetry (x-coordinate of vertex) is x = (α + β)/2 = −b/(2a). The discriminant D = b² − 4ac determines the nature of roots: D > 0 → two distinct real roots, D = 0 → one (repeated) real root, D < 0 → two complex conjugate roots.
Another useful derived relation for the distance between roots is |α − β| = sqrt((α + β)² − 4αβ) = sqrt(b² − 4ac)/|a|. This ties the spread of the zeros to the discriminant.
- Form a quadratic polynomial whose zeros are 2 and 3. Solution: f(x) = k(x−2)(x−3) = k(x² − 5x + 6). For a monic polynomial (k = 1): x² − 5x + 6. Here α + β = 5 and αβ = 6, matching −b/a = 5 and c/a = 6.
- Given f(x) = 2x² − 5x + 3. Find sum and product of zeros and the zeros. Sum = −b/a = −(−5)/2 = 5/2. Product = c/a = 3/2. Roots (by formula) are x = (5 ± 1)/4 → x = 3/2 and x = 1. Check: 3/2 + 1 = 5/2, (3/2)(1) = 3/2.
- Real-life: projectile motion simplified vertically: h(t) = −t² + 4t gives zeros t = 0 s and t = 4 s (times when height is zero). Sum of zeros = 4 = −b/a and product = 0 = c/a. These zeros are the times of launch and landing (break-even times).
- \[For ax² + bx + c (a ≠ 0) with zeros α and β: α + β = −b/a\]
- \[αβ = c/a\]
- \[Quadratic from zeros: k(x − α)(x − β) = k(x² − (α + β)x + αβ)\]
- \[Axis of symmetry / x-coordinate of vertex: x = (α + β)/2 = −b/(2a)\]
- \[Discriminant: D = b² − 4ac (D >\]\[0 two real roots\]\[D = 0 repeated root\]\[D <\]\[0 complex roots)\]
- \[Distance between roots: |α − β| = sqrt((α + β)² − 4αβ) = sqrt(b² − 4ac)/|a|\]
Applications and Problems
Applications and Problems
Key Point: Division algorithm: For polynomials f(x) and g(x) (g(x) ≠ 0), f(x) = g(x)·q(x) + r(x), where deg r(x) < deg g(x).
In Class 9 Polynomials, the section 'Applications and Problems' shows how polynomial expressions model real situations and how algebraic tools (factorisation, division algorithm, remainder theorem and factor theorem) are used to solve practical questions. Typical tasks are: form a polynomial from a word description, find zeroes, check or find factors, determine remainders on division, compare coefficients to find unknowns, and use polynomial expressions to represent areas, revenues or other quantities.
Key methods:
- Model: Express the quantity (length, area, profit, etc.) as a polynomial in a variable x.
- Simplify or expand the polynomial to standard form.
- Use factorisation or the factor theorem to find roots/zeros or simplify expressions.
- Apply the division algorithm/ remainder theorem to find remainders when divided by linear factors.
- Compare coefficients or substitute specific x-values to determine unknown constants.
Practical problem types include geometric problems (areas/perimeters expressed in x), algebraic verification problems (proving identities), constructing polynomials with given zeros, and using remainders/factors to test divisibility. Strategies: always try substitution (for remainder/factor checks), factor where possible, and interpret roots in context (e.g., a root may represent a dimension giving zero area — often not physically admissible but useful algebraically).
- Example 1 (Area modelling): A rectangle has length (x + 5) m and width (x - 3) m. Write the area as a polynomial, factor it and find area when x = 4. Solution: Area = (x+5)(x-3) = x^2 + 2x - 15. Factorised: (x+5)(x-3). For x = 4, area = 4^2 + 2·4 -15 = 16 + 8 -15 = 9 m^2.
- Example 2 (Remainder theorem): Find the remainder when f(x) = 2x^3 + 3x^2 - 5x + 7 is divided by (x - 2). Solution: By remainder theorem, remainder = f(2) = 2·8 + 3·4 - 5·2 + 7 = 16 + 12 -10 +7 = 25.
- Example 3 (Factor theorem): Prove x = 1 is a zero of f(x) = x^3 - 6x^2 + 11x - 6 and factorise f(x). Solution: f(1)=1-6+11-6=0, so (x-1) is a factor. Divide or perform synthetic division to get f(x) = (x-1)(x^2 -5x +6) = (x-1)(x-2)(x-3).
- Example 4 (Construct polynomial from zeros): Form a monic polynomial of least degree with zeros 2 and -3. Solution: Polynomial = (x-2)(x+3) = x^2 + x -6.
- Example 5 (Word problem with cost/revenue): Revenue R(x) = x(50 - x) (selling x items at price (50-x) each). Profit P(x) = R(x) - C(x) where cost C(x)=20x + 100. Write P(x) as a polynomial and find for x=10. Solution: R(x)=50x - x^2, so P(x)= -x^2 + 50x - (20x +100)= -x^2 +30x -100. For x=10, P(10)= -100 +300 -100 =100.
- Example 6 (Finding unknown coefficient): If polynomial x^3 + ax^2 + bx + 6 is divisible by (x-1) and (x-2), find a and b. Solution: If divisible by (x-1) and (x-2) then f(1)=0 and f(2)=0. f(1)=1+a+b+6=0 => a+b = -7. f(2)=8+4a+2b+6=0 => 4a+2b = -14 => dividing by 2: 2a+b = -7. Subtracting: (2a+b)-(a+b)= -7 - (-7) => a=0. Then b = -7.
- \[Division algorithm: For polynomials f(x) and g(x) (g(x) ≠ 0)\]\[f(x) = g(x)·q(x) + r(x)\]\[where deg r(x) < deg g(x).\]
- \[Remainder theorem: Remainder when f(x) is divided by (x - a) is f(a).\]
- \[Factor theorem: (x - a) is a factor of f(x) iff f(a) = 0.\]
- \[For quadratic ax^2 + bx + c: sum of zeros = -b/a\]\[product of zeros = c/a (useful when constructing polynomials from zeros).\]
Key Concepts
- Polynomial
- An algebraic expression made of one or more terms with non-negative integer powers of a variable and coefficients.
- Monomial
- A polynomial with only one term.
- Binomial
- A polynomial with exactly two terms.
- Trinomial
- A polynomial with exactly three terms.
- Degree of a polynomial
- The highest exponent of the variable in the polynomial (for the zero polynomial degree is not defined).
- Leading coefficient
- The coefficient of the term with the highest degree.
- Coefficient
- A constant factor multiplying a variable power in a term.
- Constant term
- The term in a polynomial that does not contain the variable.
- Zero polynomial
- The polynomial 0 whose all coefficients are zero; its degree is not defined.
- Like terms
- Terms that have the same variable factors raised to the same powers.
- Unlike terms
- Terms that differ in variable part or powers and cannot be combined directly.
- Factor (of a polynomial)
- A polynomial that divides the given polynomial exactly (without remainder).
- Factorization
- Expressing a polynomial as a product of its factors.
- Zero (root) of a polynomial
- A value of the variable that makes the polynomial equal to zero.
- Remainder Theorem
- When a polynomial f(x) is divided by (x - a), the remainder is f(a).
- Factor Theorem
- x - a is a factor of f(x) if and only if f(a) = 0.
- Division Algorithm (for polynomials)
- For polynomials f(x) and g(x) (g ≠ 0) there exist unique q(x) and r(x) such that f = qg + r and deg r < deg g.
- Linear polynomial
- A polynomial of degree 1.
- Quadratic polynomial
- A polynomial of degree 2.
- Cubic polynomial
- A polynomial of degree 3.
Practice Questions
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What is the degree of the polynomial 5x³ − 2x + 7? (a) 1 (b) 2 (c) 3 (d) 7 / बहुपद 5x³ − 2x + 7 की घात क्या है? (a) 1 (b) 2 (c) 3 (d) 7
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(c) 3 / (c) 3 — बहुपद में चर x की सबसे बड़ी घात 3 है, इसलिए इसकी डिग्री 3 है। / The highest power of x with a non-zero coefficient is 3, so the degree is 3.
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Using the Remainder Theorem, find the remainder when p(x) = x³ − 2x² + 3x − 5 is divided by (x − 2). (a) 1 (b) 3 (c) 5 (d) 9 / शेषफल प्रमेय से p(x) = x³ − 2x² + 3x − 5 को (x − 2) से भाग देने पर शेषफल ज्ञात करें। (a) 1 (b) 3 (c) 5 (d) 9
Show answer
(a) 1 / (a) 1 — शेषफल = p(2) = 8 − 8 + 6 − 5 = 1. / By the Remainder Theorem, remainder = p(2) = 8 − 8 + 6 − 5 = 1.
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Which of the following is a zero of p(x) = x² − 5x + 6? (a) 1 (b) 2 (c) 4 (d) 5 / p(x) = x² − 5x + 6 का शून्यांक कौन-सा है? (a) 1 (b) 2 (c) 4 (d) 5
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(b) 2 / (b) 2 — p(2) = 4 − 10 + 6 = 0; अत: x = 2 एक शून्यांक है। इसी प्रकार x = 3 भी शून्यांक है, लेकिन विकल्पों में 2 उपलब्ध है। / p(2) = 4 − 10 + 6 = 0, confirming x = 2 is a zero of the polynomial.
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Factorise the polynomial x² − 5x + 6. / बहुपद x² − 5x + 6 का गुणनखंडन करें।
Show answer
x² − 5x + 6 = (x − 2)(x − 3) / x² − 5x + 6 = (x − 2)(x − 3) — दो ऐसी संख्याएँ चाहिए जिनका गुणनफल 6 और योग −5 हो: −2 और −3. / Find two numbers with product 6 and sum −5: they are −2 and −3, so the factorisation is (x − 2)(x − 3).
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For the quadratic polynomial ax² + bx + c with zeros α and β, the product of the zeros equals ______. / द्विघात बहुपद ax² + bx + c के शून्यांकों α और β का गुणनफल ______ होता है।
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c/a / c/a — द्विघात बहुपद के शून्यांकों का गुणनफल = c/a होता है। / For ax² + bx + c with zeros α and β, their product αβ = c/a (a standard Vieta's relation).
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True or False: The polynomial x² + 4 has real zeros. / सत्य या असत्य: बहुपद x² + 4 के वास्तविक शून्यांक होते हैं।
Show answer
False / असत्य — विविक्तकर D = 0² − 4(1)(4) = −16 < 0; अत: इसके कोई वास्तविक शून्यांक नहीं हैं। / The discriminant D = 0 − 16 = −16 < 0, so x² + 4 has no real zeros.
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State the Factor Theorem and use it to check whether (x − 1) is a factor of p(x) = x³ − 6x² + 11x − 6. / गुणनखंड प्रमेय लिखें और जाँचें कि (x − 1), p(x) = x³ − 6x² + 11x − 6 का गुणनखंड है या नहीं।
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Factor Theorem: (x − a) is a factor of p(x) iff p(a) = 0. p(1) = 1 − 6 + 11 − 6 = 0. So (x − 1) is a factor. / गुणनखंड प्रमेय: (x − a), p(x) का गुणनखंड है यदि और केवल यदि p(a) = 0. p(1) = 0, अत: (x − 1) गुणनखंड है।
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Perform the division: Divide p(x) = x² + 7x + 10 by (x + 2) and state the quotient and remainder. / भाग करें: p(x) = x² + 7x + 10 को (x + 2) से विभाजित कर भागफल और शेषफल बताएँ।
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Quotient = (x + 5), Remainder = 0. So x² + 7x + 10 = (x + 2)(x + 5). / भागफल = (x + 5), शेषफल = 0. अत: x² + 7x + 10 = (x + 2)(x + 5). Since p(−2) = 4 − 14 + 10 = 0, (x + 2) is a factor and the Division Algorithm gives remainder 0.
Related Laws & Principles
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