If now, we suppose xa, the second member will reduce to zero, consequently, the first will reduce to zero, and hence a will be a root of the equation (Art. 245). It is evident, from the nature of division, that the quotient Q' will be of the form +R'x + U' = 0. xm-iP'xm—2 247. It follows from what has preceded, that in order to discover whether any polynomial is exactly divisible by the binoa, it is sufficient to see if the substitution of a for æ will reduce the polynomial to zero. mial x Conversely, if any polynomial is exactly divisible by x-a, then we know, that if the polynomial be placed equal to zero, a will be a root of the resulting equation. The property which we have demonstrated above, enables us to diminish the degree of an equation by 1 when we know one of its roots, by a simple division; and if two or more roots are known, the degree of the equation may be still further diminished by successive divisions. are 3 and 5: what does the equation become when freed of them? Ans. x24x + 1 = ( 3. A root of the equation, x36x211x - 6 = 0, is 1 what is the reduced equation? Ans. x25x + 6 = 0. 4. Two roots of the equation, 4x4 14x3 5x2+31x + 6 = 0, are 2 and 3: find the reduced equation. Ans. 4x2 + 6x + 1 = 0. Second Property. 248. Every equation involving but one unknown quantity, has is many roots as there are units in the exponent which denotes its degree, and no more. Since every equation is known to have at least one root (Art. 245), if we denote that root by a, the first member will be divisible by xa, and we shall have the equation, we obtain a new equation, which has at least one root. Denote this root by b, and we have (Art. 246), Substituting the second member, for its value, in equation (1), we have, æm + Pæm−1 + . . . = (x − a) (x − b) (xm−2 + P2 xm-3 + ...) · · (2). : — · Reasoning upon the polynomial, 2-1 n-3 xTM - Pxm-1·+ ... = (x -- a) (x -- b) (x — c) (xm−3+P'''x→1) - - - (3). · By continuing this operation, we see that for each binomial factor of the first degree with reference to x, that we separate, the degree of the polynomial factor is reduced by 1; therefore, after m 2 binomial factors have been separated, the polynomial factor will become of the second degree with reference to x, which can be decomposed into two factors of the first degree (Art. 115), of the form k, x 7. Now, supposing the m2 factors of the first degree to have already been indicated, we shall have the identical equation, ¿em + Påm−1 + . . = (x − a) (x — b) (x − c) . . (x − k) (x − 1) = 0 ; from which we see, that the first member of the proposed equation may be decomposed into m binomial factors of the first degree. As there is a rooi corresponding to each binomial factor of the first degree (Art. 246), it follows that the m binomial factors of the first degree, x b, x -c give the m roots, a, b, c..., of the proposed equation. But the equation can have no other roots than a, b, c For, if it had a root a', different from a, b, c. 7, it would .... which is impossible; therefore, Every equation of the mth degree has m roots, and can have по more. 249. In equations which arise from the multiplication of equal factors, such as (x − a)1 (x — b)3 (x − c)2 (x — d) = 0, the number of roots is apparently less than the number of units in the exponent which denotes the degree of the equation. But this is not really so; for the above equation actually has ten roots, four of which are equal to a, three to b, two to c, and one to d. It is evident that no quantity a', different from a, b, c, d, can verify the equation; for, if it had a root a', the first mem ber would be divisible by xa', which is impossible. Consequence of the Second Property. 250. It has been shown that the first member of every equation of the mth degree, has m binomial divisors of the first degree, of the form If we multiply these divisors together, two and two, three and three, &c., we shall obtain as many divisors of the second, third, &c. degree, with reference to x, as we can form different combinations of m quantities, taken two and two, three and three, Now, the number of these combinations is expressed by &c. 251. If we resume the identical equation of Art. 248, ... xm+Pxm-1+Qxm-2, +U=(x-a) (x —b) (x — c)... (x − 1)... and suppose the multiplications indicated in the second member to be performed, we shall have, from the law demonstrated in article 135, the following relations: The double sign has been placed before the product of a, b, c, &c. in the last equation, since the product ах bx C. x - 1, will be plus when the degree of the equation is even, and minus when it is odd. By considering these relations, we derive the following conclusions with reference to the values of the co-efficients: 1st. The co-efficient of the second term, with its sign changed, is equal to the algebraic sum of the roots of the equation. 2d. The co-efficient of the third term is equal to the sum of the different products of the roots, taken two in a set. 3d. The co-efficient of the fourth term, with its sign changed, is equal to the sum of the different products of the roots, taken three in a set, and so on. 4th. The absolute term, with its sign changed when the equation is of an odd degree, is equal to the continued product of all the roots of the equation. Consequences. 1. If one of the roots of an equation is 0, there will be no absolute term; and conversely, if there is no absolute term, one of the roots must be 0. 2. If the co-efficient of the second term is 0, the numerical sum of the positive roots is equal to that of the negative roots. 3. Every root will exactly divide the absolute term. It will be observed that the properties of equations of the second degree, already demonstrated, conform in all respects to the principles demonstrated in this article. EXAMPLES OF THE COMPOSITION OF EQUATIONS. 1. Find the equation whose roots are 2, 3, 5, and - 6. We have, from the principles already established, the equation (x − 2) (x − 3) (x − 5) (x + 6) = 0; whence, by the application of the preceding principles, we obtain the equation, 24 4x3 29x2+156x - 180 = 0. |