A polynomial is just a tidy sum of powers of x with number coefficients. Master how to add, multiply, factorise and find zeroes — and a whole branch of algebra opens up!
What is a polynomial?
An expression like 3x² − 5x + 2 with whole-number powers of x.
Degree
The highest power of the variable in the polynomial.
Zero of a polynomial
The value of x that makes p(x) = 0.
Factorisation
Writing a polynomial as a product of simpler factors.
1. Polynomials in one variable
A polynomial in one variable x is an algebraic expression of the form anxn + an−1xn−1 + ... + a1x + a0, where a0, a1, ..., an are constants (real numbers) and n is a non-negative integer. The constants are called coefficients. The key rule is that the powers of the variable must be whole numbers (0, 1, 2, 3, ...). So 2x² + 3x + 1 is a polynomial, but 2x½ + 3 or 5/x + 2 (which is 5x⁻¹) are NOT polynomials, because their powers are not whole numbers.
Each part separated by + or − is called a term. For example, in 4x³ − 2x + 7, the terms are 4x³, −2x and 7. The number multiplying a power of x is its coefficient: the coefficient of x³ is 4, the coefficient of x is −2, and 7 is the constant term. A polynomial with only one term is a monomial (like 5x), with two terms a binomial (like x + 2), and with three terms a trinomial (like x² + x + 1).
2. Degree of a polynomial
The degree of a polynomial is the highest power of the variable present in it. For 3x⁴ − x² + 7, the degree is 4. A non-zero constant like 5 has degree 0 (since 5 = 5x⁰). The zero polynomial (just 0) has no defined degree. Degree gives polynomials their special names:
Linear polynomial — degree 1 (e.g. 2x + 3). Quadratic polynomial — degree 2 (e.g. x² − 5x + 6). Cubic polynomial — degree 3 (e.g. x³ + 2x² − x + 1). A linear polynomial in one variable has at most one term of degree 1, a quadratic at most a square term, and so on.
3. Value and zeroes of a polynomial
If p(x) is a polynomial and you replace x by a number c, the result p(c) is the value of the polynomial at x = c. A zero (or root) of a polynomial p(x) is a number c such that p(c) = 0. For example, for p(x) = x − 2, putting x = 2 gives p(2) = 0, so 2 is a zero. To find the zero of a linear polynomial ax + b, set ax + b = 0, giving x = −b/a. A linear polynomial has exactly one zero, a quadratic has at most two zeroes, and in general a polynomial of degree n has at most n zeroes. Note: a non-zero constant polynomial has no zero, while every value of x is a zero of the zero polynomial.
4. Remainder Theorem
The Remainder Theorem states: if a polynomial p(x) of degree ≥ 1 is divided by the linear polynomial (x − a), then the remainder is p(a). This saves a lot of long division! Instead of dividing, you simply substitute x = a into p(x). For example, the remainder when x³ + 1 is divided by (x + 1) is found by putting x = −1: p(−1) = (−1)³ + 1 = 0. If you divide by (ax + b), substitute x = −b/a.
5. Factor Theorem
The Factor Theorem is a special case of the remainder theorem: (x − a) is a factor of p(x) if and only if p(a) = 0. In other words, if a number a is a zero of p(x), then (x − a) divides p(x) exactly with no remainder. This theorem is the engine behind factorising cubic and quadratic polynomials. To factorise a cubic, we hunt for a value a (usually a factor of the constant term) for which p(a) = 0, peel off (x − a), and continue factorising the leftover quadratic.
6. Factorisation of quadratics (splitting the middle term)
To factorise ax² + bx + c, we split the middle term: find two numbers whose product is a × c and whose sum is b. For x² + 5x + 6, we need two numbers with product 6 and sum 5, which are 2 and 3. So x² + 5x + 6 = x² + 2x + 3x + 6 = x(x + 2) + 3(x + 2) = (x + 2)(x + 3). This method always works for quadratics whose factors are rational.
7. Algebraic identities
Identities are equalities true for ALL values of the variables — they let us expand and factorise quickly. The Class 9 identities are: (x + y)² = x² + 2xy + y²; (x − y)² = x² − 2xy + y²; x² − y² = (x + y)(x − y); (x + a)(x + b) = x² + (a + b)x + ab. The three-variable identity (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx is very useful. The cube identities are (x + y)³ = x³ + y³ + 3xy(x + y) and (x − y)³ = x³ − y³ − 3xy(x − y). Finally, the powerful identity x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx); a beautiful result of this is that if x + y + z = 0, then x³ + y³ + z³ = 3xyz.
- (x + y)² = x² + 2xy + y²
- (x − y)² = x² − 2xy + y²
- x² − y² = (x + y)(x − y)
- (x + a)(x + b) = x² + (a + b)x + ab
- (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
- (x + y)³ = x³ + y³ + 3xy(x + y)
- (x − y)³ = x³ − y³ − 3xy(x − y)
- x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx)
- Remainder when p(x) ÷ (x − a) is p(a)
- (x − a) is a factor of p(x) ⇔ p(a) = 0
Factorise the cubic polynomial x³ − 3x² − 9x − 5 using the Factor Theorem.
- Try factors of the constant term (−5): test x = −1. p(−1) = (−1)³ − 3(−1)² − 9(−1) − 5 = −1 − 3 + 9 − 5 = 0. So (x + 1) is a factor.
- Divide p(x) by (x + 1) (or split terms): x³ − 3x² − 9x − 5 = (x + 1)(x² − 4x − 5).
- Factorise the quadratic x² − 4x − 5 by splitting the middle term: product = −5, sum = −4, so the numbers are −5 and +1.
- x² − 4x − 5 = x² − 5x + x − 5 = x(x − 5) + 1(x − 5) = (x − 5)(x + 1).
- Combine: x³ − 3x² − 9x − 5 = (x + 1)(x + 1)(x − 5) = (x + 1)²(x − 5).
Without actually calculating the cubes, evaluate (−12)³ + 7³ + 5³.
- Let x = −12, y = 7, z = 5. Check their sum: x + y + z = −12 + 7 + 5 = 0.
- Use the identity: if x + y + z = 0, then x³ + y³ + z³ = 3xyz.
- Substitute: 3xyz = 3 × (−12) × 7 × 5.
- Compute: 3 × (−12) = −36; −36 × 7 = −252; −252 × 5 = −1260.
For the Factor Theorem, remember "Zero in, factor out": if plugging a number in gives zero, then (x − that number) factors out. And for which numbers to try first — always test the factors of the constant term (like ±1, ±2, ±5).
Many students think any algebraic expression is a polynomial. Remember: powers of the variable must be whole numbers. So √x (= x1/2), 1/x (= x−1) and x3/2 are NOT terms of a polynomial. Also, never forget to check the sign carefully when substituting negative values into the Remainder/Factor Theorem.
Q1. Find the value of the polynomial p(x) = 5x² − 3x + 7 at x = −1.
Answer: Substitute x = −1: p(−1) = 5(−1)² − 3(−1) + 7 = 5(1) + 3 + 7 = 5 + 3 + 7 = 15.
Q2. Using the Remainder Theorem, find the remainder when x³ + 3x² + 3x + 1 is divided by (x + 2).
Answer: Here the divisor (x + 2) gives a = −2. By the Remainder Theorem, remainder = p(−2) = (−2)³ + 3(−2)² + 3(−2) + 1 = −8 + 3(4) − 6 + 1 = −8 + 12 − 6 + 1 = −1.
Q3. Factorise 6x² + 17x + 5 by splitting the middle term.
Answer: We need two numbers with product a×c = 6×5 = 30 and sum b = 17. These are 15 and 2. So 6x² + 17x + 5 = 6x² + 15x + 2x + 5 = 3x(2x + 5) + 1(2x + 5) = (2x + 5)(3x + 1).
Q4. Without multiplying directly, evaluate 105 × 95 using a suitable identity.
Answer: Write 105 = 100 + 5 and 95 = 100 − 5. Then 105 × 95 = (100 + 5)(100 − 5). Using the identity (x + y)(x − y) = x² − y² with x = 100, y = 5: = 100² − 5² = 10000 − 25 = 9975.
- ✅ A polynomial has only whole-number powers of the variable; degree = highest power.
- ✅ A zero of p(x) is a value c with p(c) = 0; a degree-n polynomial has at most n zeroes.
- ✅ Remainder Theorem: dividing p(x) by (x − a) leaves remainder p(a). Factor Theorem: (x − a) is a factor ⇔ p(a) = 0.
- ✅ Factorise quadratics by splitting the middle term; factorise cubics using the Factor Theorem.
- ✅ Memorise the identities — especially x³ + y³ + z³ − 3xyz, which gives x³ + y³ + z³ = 3xyz when x + y + z = 0.
Book a free demo class