Solution: x2+3x – 4) x4+ 4x3– 31x2– 94x + 120 (x2+x–30 x4+ 3x3– 4x2– – + x3– 27x2– 94x + 120 x3+ 3x2– 4x – – + – 30x2– 90x + 120 – 30x2– 90x + 120 + + – 0 ∴the quotient is x2+ x – 30 and the remainder is ‘0’ Relations between Roots and Coefficients: An nthorder equation has n roots. Corresponding to every root, there is a factor. If αis a root of f(x) = 0, then x - αis a factor of f(x). Sometimes (x – α)2may also be a factor. In such a case, αis said to be a double root. Similarly equations can have triple roots, quadruple roots and roots of multiplicity m. If m is the greatest value of k, for which (x – α)kis a factor of f(a), then αis said to be a root of multiplicity m. If all the roots are counted by taking their multiplicity into account, the number of roots is equal to n, the degree of the equation. If α1, α2,….., αn (not necessarily distinct) are the roots of f(x) = 0, then f(x) = an(x – α1) (x – α2) ….(x – αn) = an[xn– S1xn–1 + S2xn–2…..+(–1)nSn] where S1= the sum of the roots S2= the sum of the products of the roots taken 2 at a time S3= the sum of the product of the roots taken 3 at a time and so on. Sn= the 'sum' of the product of the roots taken n(or all) at a time. Thus, Snis a single term. Sn= α1α2…..αnLet us write down the polynomial f(x) in two forms: The standard form f(x) = anxn+ an–1xn–1 + an–2 xn–2…+a1x + a0In terms of the roots of the corresponding equation. f(x) = an[xn– S1xn–1 + S2xn–2+ ….+(–1)x–1Sn–1x + (–1)nSn] These polynomials are identically equal, i.e, equal for all values of x. Therefore the corresponding coefficients are equal. The sum of the roots S1= –an –1/ an The sum of the products of the roots, taken two at a time, S2= an–2/an The sum of the products of the roots, taken three at a time, S3= –an–3/anand so on. The 'sum' of the 'products' of the roots taken m (m ≤n) at a time Sm= Σα1 α2 α3…..αm = (–1)mnmnaa-∴Sn= α1 α2 α3…..αn = (–1)nn0aaFor example, consider the polynomial equation (x – 1)(x – 2)(x – 3) = x3– 6x2+ 11x – 6 = 0 (We can see immediately that the roots are 1, 2, 3) The sum of roots = (1 + 2 + 3) = –(–6)/1 The sum of the products of the roots, taken two at a time S2= 1(2) + 1(3) + 2(3) = 11 = 11/1 We can drop the word 'sum' and 'products' for the last relation, because there is only one term (only one product). The product = 1(2)(3) = 6 = –(–6)/1. Roots of Equations and Descartes’ Rule: If the coefficients are all real and the complex number z1, is a root of f(x) = 0, then the conjugate of z1, viz, 1zis also a root of f(x) = 0. Thus, for equations with real, coefficients, complex roots occur in pairs. A consequence of this is that any equation of an odd degree must have at least one real root.