Taking this last equation, and substituting for its first member the second, in the next preceding equation, and thus continuing the process of substitution until the first equation of the series is arrived at, the result will be the following identical equation: (x − a) (x − b) (x — c) . . . . } = { ( x − a) (x (x − p) (x − q). The second member of this equation vanishes for any one the m values, .... x = p, x = 9, x = α, x = b, x = c,.. and consequently these values are severally roots of the equation, x3 + Axm¬1 + Bxm-2 . . . . Tx + U = 0. Moreover, no value of x that differs from some one of these values, can satisfy the equation; for no such value will cause any one of the factors in the second member of the identical equation to be zero, a condition requisite to make the product zero. The equation therefore has m roots and no more. 428. From the foregoing principles we conclude, 1. That in an equation in which the second term does not appear, that is, the term containing the next to the highest power of the unknown quantity,—the algebraic sum of the roots is 0. 2. If an equation has no absolute term, at least one of its roots is 0. 3. The absolute term being the continued product of all the roots of an equation, it must be exactly divisible by each of them. 4. An equation may be constructed, which shall have any assumed roots. 5. The degree of an equation may be reduced by 1 for each f its known roots. EXAMPLES. 1. What is the equation having +2, 3 for its roots? 2. What is the equation having the roots + 1, -2, -4? Ans. xs+5x2 + 2x 8=0. 3. What is the equation having for its roots 3, -2, -1, 5x3-7x2+29x+30= 0. +5? Ans. x4 4. What is the equation of which the roots are 1 +√5, 1−√−5, +√5, −√/5? Ans. x-2x+x2+10x-30 = 0. 5. What is the equation of which the roots are -1, -2, +3, 2 + √−3, 2−√—3? Ans. 25-4x+22x2-25x-42=0. 6. One root of the equation, 0, Ans. x2-2x+7= €2 X3 5x2 + 13x 21= is +3; what is the reduced equation? 7. One root of the equation, x+2x3-34x2 + 12x + 35=0, is; what is the depressed equation? Ans. x35x2+x+5=0. 8. Two of the roots of the equation, 24. 3x3 4x2 + 30x36 = 0, are + 2, -3; what is the depressed equation, and what are its roots? Ans. The depressed equation is x2-4x+6=0; and its roots are 2+√ −2, 2−√√ —2. 429. Any equation having fractional coefficients can be transformed into another in which the coefficients are entire, that of the first term being unity. If the coefficient of the first term of the given equation is not unity, make it so by dividing through by this coefficient. Then the equation will be of the form x+Axm-1+ Bxm-2+.... Tx + U = 0, in which it is supposed that some or all the coefficients, A, B, etc., are fractional. value of x in the equation; it then becomes a being entirely arbitrary, and substitute this am ym Whence, by multiplying through by am, ym + Aaym1+ Ba2ym-2 + .... Now since a is arbitrary, its value may be so selected that it and α + T +U=0. ... Tam-ly + Vam = 0. its powers will contain the denominators of the fractional coefficients of the original equations. We present the following examples for illustration. into another which shall have no fractional coefficients, and which shall have unity for its first coefficient. Make x = У mnp ; substituting this value of x, the equation becomes ay? + by с = 0. + Multiplying every term of this by m3n3p3, y3 + anpy2 + bm2np2y + cm3n3p2 = 0. When the denominators of the coefficients have common factors, we may make x equal to y divided by the least common multiple of the denominators. ax2 bx 2. Transform the equation, 23 + + + = pm m p 0, into the first term be unity. another which shall have no fractional coefficients, and that of into another having no fractional coefficients. Ans. ya + 20y3 + 18. 24y2 + 7 (24)2 y + 2 (24)3 = 0. In transforming an equation having fractional, into another with entire coefficients, in terms of another unknown quantity, it is important to have the transformed equation in the lowest possible terms. The least common multiple of the denominators will not necessarily be the least value of a that will give the required equation. If, in each case, the denominators be resolved into their prime factors, it will be easy to decide upon the powers of these factors to be taken as the factors of a. The following illustration will render further explanation unnecessary. into another of the same form with the smallest possible entire coefficients. Writing y for x and multiplying the second, third and fourth terms, by a, a3, a3, respectively, we have The denominators, resolved into their prime factors, are 7.5, 72.52.2, 73.52.23; and assuming a 752, the equation may be written y3 — 6y2 + 26у — 85 = 0. In this example, the least common multiple of the denominators is 73 · 52 · 23; and had this value been taken for a, instead of 752, the coefficients of the transformed equation would have been much larger than they are, as found above. When a root of the transformed equation is known, the corresponding root of the original equation will be given by the relation COMMENSURABLE ROOTS. 430. A number is commensurable with unity when it can be expressed by an exact number of units or parts of a unit; a number which cannot be so expressed is incommensurable with unity. 431. Every equation having unity for the coefficient of the first term, and for all the other coefficients, whole numbers, can have only whole numbers for its commensurable roots. This being one of the most important principles in the theory of equations, its enunciation should be clearly understood. Such equations may have other roots than whole numbers; but its roots cannot be among the definite and irreducible fractions, such as, 3, 4, etc. Its other roots must be among the incommensurable quantities, such as √√2, (3), etc.; ¿. e., surds, indeterminate decimals, or imaginary quantities. a To prove the proposition, let us suppose ō› a commensurable but irreducible fraction, to be a root of the equation, xm + Аxm¬1 + Bxm−2.... Tx + U = 0. A, B, etc., being whole numbers. Substituting this supposed value of x, we have Transpose all the terms but the first, and multiply by bm-1, and we have am b = (Aam-1 + Bam-2b.... Tabm-2 + Ubm-1). Now, as a and b are prime to each other, b cannot divide a, and it cannot possibly divide any power of a; because being a a irreducible, xa is also irreducible, as the multiplier a will not b be measured by the divisor b; therefore a2 b cannot be expressed am in whole numbers. Continuing the same mode of reasoning, |