3. Find the greatest common divisor of the two polynomials 5a4b2+2a3b3 + ca2 -3a2b4bca, 262. The object of a transformation, is to change an equation from a given form to another, from which we can more readily determine the value of the unknown quantity. First. To change a given equation involving fractional co-efficients to another of the same general form, but having the co-efficients of all its terms entire. If we have an equation of the form in which y is a new unknown quantity, and k entirely arbitrary; we shall have, after substituting this value for x, and multiplying every term by km, ym + Pkym-1+ Qk2ym-2 + Rk3ym¬3 + ... + Tkm-1y + Ukm = 0, an equation in which the co-efficients of the different powers of y are equal to those of the same powers of x in the given equation, multiplied respectively by 4o, k1, k2, k3, ka, &c. It is now required to assign such a value to k as will make the co-efficients of the different powers of y entire. To illustrate, let us take, as a general example, the equation 1st. Where the denominators b, d, f, h, are prime with respect to each other. In this case, as k is altogether arbitrary, take kbdfh, the product of the denominators, the equation will then become, y1 + adfh. y3 + cb2dƒ2h2. y2 + eb3d3ƒ2h3 .y + gbadaf4h3 =0, in which the co-efficients of y are entire, and that of the first term is 1. 2d. When the denominators contain common factors, we shall evidently render the co-efficients entire, by making k equal to the least common multiple of all the denominators. But we can simplify still more, by giving to k such a value that k1, k2, k3, . . . shall contain the prime factors which compose b, d, f, h, raised to powers at least equal to those which are found in the denominators. First, if we make k and reducing the terms. 9000, which is a multiple of all the other denominators, it is clear that the co-efficients become entire numbers. But if we decompose 6, 12, 150, and 9000, into their prime factors, we find, 6=2x3, 12=22 x 3, 150=2 × 3 × 52, 900023 x 32 x 53; and by making k = 2 × 3 × 5, the product of the different prime factors, we obtain k2 = 22 × 32 × 52, whence we see that the values of k, k2, k3, k4, contain the prime factors of 2, 3, 5, raised to powers at least equal to those which enter into 6, 12, 150, and 9000. k = 2 × 3 × 5, Hence, making is sufficient to make the denominators disappear. Substituting this value, the equation becomes or, y5.5y3+5.3.52y2-7.22.32.5y-13.2.32.5= 0; yo 25y3+ 375y2—1260y1170 = 0. Hence, we perceive the necessity of taking k as small a number as possible: otherwise, we should obtain a transformed équation, having its co-efficients very great, as may be seen by reducing the transformed equation resulting from the supposi tion k9000. Having solved the transformed equation, and found the values of y, the corresponding values of a may be found from the equation, y x= k' by substituting for y and k their proper values. Making =, and we have, y3 - 14y2+11y — 75 = 0. y-65y+1890y3-30720y2 — 928800y-9720000. Second. To make the second or any other term disappear from an equation. 263. The difficulty of solving an equation generally diminishes with the number of terms involving the unknown quantity. Thus the equation x2=q, gives immediately, x=±. while the complete equation x2 + 2px + q = 0, requires preparation before it can be solved. Now, any given equation can always be transformed into an incomplete equation, in which the second term shall be wanting. For, let there be the general equation, u being a new unknown quantity, and x' entirely arbitrary. By substituting u+x for x, we obtain (u + x)+P(u + x)m−1 + Q (u + x)m-2... +T (u+x') + U = 0. Developing by the binomial formula, and arranging with refer Since is entirely arbitrary, we may dispose of it in such way that we shall have P mx + P=0; whence, x= m Substituting this value of in the last equation, we shall obtain an incomplete equation of the form, um + Q'um-2 + R'um−3 + in which the second term is wanting. T'u + U' = 0, If this equation were solved, we could obtain any value of a corresponding to that of u, from the equation We have, then, in order to make the second term of an equation disappear, the following RULE. Substitute for the unknown quantity a new unknown quantity minus the co-efficient of the second term divided by the exponent which expresses the degree of the equation. Let us apply this rule to the equation, and by performing the indicated operations and transposing, we find u2 = p2 + q. 263*. Instead of making the second term disappear, it may be required to find an equation which shall be deprived of its third, fourth, or any other term. This is done, by making the co-efficient of u, corresponding to that term, equal to 0. For example, to make the third term disappear, we make, in the transformed equation, (Art. 263), from which we obtain two values for x', which substituted in the transformed equation, reduce it to the form, |