Form of any Equation. by any power of x[1], and used to denote the remaining Now the given equation is, by hypothesis, satisfied by the The terms not multiplied by 2[1], or a power of z[1], must, therefore, cancel each other; and the first member of the given equation becomes Px[1], which is divisible by [1], or its equal x x'. 162. Corollary. If the equation xn + axn-1 + &c. = 0 is divided by x -x', the first term of the quotient is n-1; and if the coefficients of x2-2, 2-3, &c. in the quotient are denoted by a', b', &c., the quotient is xn-1 + α' χη-2 + b xn-3 + &c.; and the equation is (x - x') (xn-1 + a' xn-2 + bxn-3 + &c.) = 0; which is satisfied either by the value of x, x = x', or by the roots of the equation xn-1 + a' xn-2 + bxn-3 + &c. = 0. If now x" is one of the roots of this last equation, we may have in the same way xn-1+a'xn-2+&c. =(x-x") (xn-2+a" xn-3+&c.)=0, Number of the Roots of an Equation. and the given equation becomes (x-x') (x - x) (xn-2+a" xn-3+&c.) = 0; which is satisfied by the value of x", x = x"; so that z" is a root of the given equation. By proceeding in this way to find the roots x'', x", &c., the given equation may be reduced to the form (х - х') (х-х") (x - x) (x-x) &c. = 0, in which the number of factors х-х, х x", &c. is the same with the degree n of the given equation; and, therefore, the number of roots of an equation is denoted by the degree of the equation; that is, an equation of the third degree has three roots, one of the fourth degree has four roots, &c. 163. Scholium. Some of the roots x', x", x', &c. are often equal to each other, and in this case the number of unequal roots is less than the degree of the equation. Thus the number of unequal roots of the equation of the 9th degree, (x - 7) (x + 4)3 (x - 1)5 = 0, is but three, namely, 7, 4, and 1, and yet it is to be regarded as having 9 roots, one equal to 7, three equal to - 4, and five equal to 1. would appear to have but one root, that is, n Imaginary Roots. but it must, by the preceding reasoning, have n roots, or rather, the nth root of a must have n different values. must be divisible by x - 1, and we have x3 1 = (x - 1)(x2 + x + 1) = 0. Now the roots of the equation are x2 + x + 1 = 0 x=(-1+√ - 3), and = (-1--3). Hence the required roots are x = 1, = + (−1+√-3), and = (-1--3). 3. Find the four roots of the equation must be divisible by x - 1, and we have x5 - 1 = (x - 1) (x4 + x3 + x2 + x + 1) = 0. Now the roots of the equation x4 + x3 + x2 + x + 1 = 0 can be found by the following peculiar process. Divide by x2, and we have x2+x+1++ 1 1 0. x2 which, being substituted in the preceding equation, gives y2 + y-1 = 0; the roots of which are y=(-1+√5), and = (-1-5). But the values of x deduced from the equation are Solution of Equations of a peculiar Form. x= [y+(y2-4)], and = [y- (y2 -4)], in which y, being substituted, gives x = [-1-√5±√(−10 + 2√5)], and = [-1-√5±√(-10-25)]. 6. Find the six roots of the equation x6 = 1. Ans. x = 1, = − 1, = 1 (−1±√-3), We might proceed in the same way to higher equations, such as the 8th, 9th, 12th, &c.; but, since much more simple solutions are given by the aid of trigonometry, this subject will be postponed to a more advanced part of the course. 165. Corollary. Before proceeding farther, we may remark that the method of solution used in the last example of the preceding article may be applied to any equation of an even degree, in which the successive coefficients of the different powers of x are the same, whether the equation is arranged according to the ascending or according to the descending powers of x, as is the case in the following equation. Ax2 + B x2n-1 + C x 2 n −2+ &c. EXAMPLES. 1. Solve the equation Ax + Bx3 + Cx2 + Bx + A = 0. Solution. Divide by x2, and we have |