CHAPTER I - THEORY OF EQUATIONS DEFINITION 1 A function defined by n n-1 f ( x ) = p 0 x + p 1 x + ... + p n-1 x + p n where p 0 ≠ 0, n is a positive integer or zero and p i ( i = 0,1,2...,n) are fixed complex numbers, is called a polynomial function of degree n in the indeterminate x . A complex number a is called a zero of a polynomial f ( x ) if f (a ) = 0. Theorem 1 – Fundamental Theorem of Algebra Every polynomial function of degree greater than or equal to 1 has atleast one zero. Definition 2 n n-1 Let f ( x ) = p 0 x + p 1 x + ... + p n-1 x + p n where p 0 ≠ 0, n is a positive integer . Then f ( x ) = 0 is a polynomial equation of nth degree. Definition 3 A complex number a is called a root ( solution) of a polynomial equation f (x) = 0 if f ( a) = 0. THEOREM 2 – DIVISION ALGORITHM FOR POLYNOMIAL FUNCTIONS If f(x) and g( x ) are two polynomial functions with degree of g(x) is greater than or equal to 1 , then there are unique polynomials q ( x) and r (x ) , called respectively quotient and reminder , such that f (x ) = g (x) q(x) + r (x) with the degree of r (x) less than the degree of g (x). Theorem 3- Reminder Theorem If f (x) is a polynomial , then f(a ) is the remainder when f(x) is divided by x – a . Theorem 4 Every polynomial equation of degree n ≥ 1 has exactly n roots. Theorem 5 In a polynomial equation with real coefficients , imaginary roots occur in conjugate pairs. Corollary 1 If f (a) = 0 , the polynomial has a factor x – a. Corollary 2 – Factor Theorem A number a is a zero of the polynomial f(x) if and only if x – a is a factor of f (x) . 1. Two roots of the equation x 4 – 6 x 3 + 18 x 2 – 30x + 25 = 0 are of the form a + i b and b + i a . find all roots of the equation. 2.Two roots of the equation x 4 - 6 x 3 + a x 2 + b x + 25 = 0 are of the form p + i q and q + i p . Find the values of a and b and also the roots of equation. Theorem -6 In an equation with rational coefficients quadratic surd roots occur in pairs. That is , if a + √ b is a root of f (x) =0 , then a - √ b is also a root of the equation. 1. Solve the equation x 5 – x 4 + 8 x 2 – 9 x – 15 = 0, one root being -√ 3 and another 1 + 2√ -1 . 2.Find all the roots of the quartic equation x 4 +x 3 – x 2 – 2x -2 = 0, given that one of its roots is √ 2 . 3. Find the equation with rational coefficients, one of whose roots is √5 +√ 2. .