A polynomial in one variable x is an expression of the form an · xn + an-1 · xn-1 + ... + a1 · x + a0, where the exponents are non-negative integers and the coefficients are real numbers.
Degree of a polynomial: the highest power of the variable. A polynomial of degree 1 is linear, degree 2 is quadratic, degree 3 is cubic.
Zeroes of a polynomial: A value c is a zero of p(x) if p(c) = 0.
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) if and only if p(a) = 0.
- Algebraic Identities:
- (a + b)2 = a2 + 2ab + b2
- (a - b)2 = a2 - 2ab + b2
- (a + b)(a - b) = a2 - b2
- (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
- (a + b)3 = a3 + 3a2 · b + 3a · b2 + b3
- (a - b)3 = a3 - 3a2 · b + 3a · b2 - b3
- a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
Find the degree of p(x) = 5x3 - 4x2 + 7x - 1.
The highest exponent is 3, so the degree is 3 (cubic polynomial).
Using Remainder Theorem, find the remainder when p(x) = x3 - 2x2 + 5 is divided by (x - 1).
p(1) = 1 - 2 + 5 = 4. The remainder is 4.
Is (x - 2) a factor of x3 - 3x2 + 4?
p(2) = 8 - 12 + 4 = 0. Since p(2) = 0, (x - 2) is a factor.
Expand (2x + 3y)2.
= (2x)2 + 2 · (2x) · (3y) + (3y)2 = 4x2 + 12xy + 9y2.
Factorise x2 + 5x + 6.
Find two numbers that multiply to 6 and add to 5: they are 2 and 3.
So x2 + 5x + 6 = (x + 2)(x + 3).
Evaluate 1032 using identities.
1032 = (100 + 3)2 = 10000 + 600 + 9 = 10609.
Factorise 8x3 + 27y3.
Using a3 + b3 = (a + b)(a2 - ab + b2) with a = 2x, b = 3y:
= (2x + 3y)(4x2 - 6xy + 9y2).
Common mistakes
Students often forget the middle term when expanding (a + b)2, writing it as a2 + b2. Always include 2ab. Also, the degree of the zero polynomial is undefined, not zero.
Summary
Polynomials are classified by degree. The Remainder and Factor Theorems are powerful tools for dividing polynomials and finding factors. Algebraic identities allow fast computation and factorisation without long multiplication.