Now it follows from the preceding law, that = (3)-12. (3)3+17. (3)-9. (3)1+7, or 4. (3)3-36. (3)+34. (3)1-9, or 6.(3)2-36. (3)'+17, or 2 V' 2.3 Therefore the transformed equation becomes u4-37u2-123u-110=0. Again, transform the equation 4x3-5x+7x-9=0 into another, the roots of which exceed the roots of the proposed equation by unity Take u=x+1; there will result x=-1+u, which gives the transformed equation Z X' 4.(-1)3- 5.(-1)+7.(-1)-9, or = Therefore the transformed equation becomes 4u3-17u2+29u-25=0. The following examples may serve the student for exercises: Make the second term vanish from the following equations. 1st. x5-10x+7x3+4x-9=0. Ans. u5-33u3-118u2-152u-73-0. Transform the equation 3x-13x2+7x2-8x-9=0 into another, the roots of which shall be less than the roots of the proposed by We shall frequently have occasion for the law of formation of derived polynomials. 298. These polynomials have the following remarkable properties. Let X or a+Pam-1+Qam-2... =0, be the proposed equation, and a, b, c, l, its m roots, we shall then have (Art. 291), x+Px-1+ =(x-a) (x-b) (x−c) . . . (x-1). Substituting +u (or to avoid the accents), x+u in the place of x; it becomes, -1 (x+u)+P(x+u)m-1+ =(x+u-a) (x+u—b) . . . ; or changing the order of the terms in the second member, and regarding x-a, x-b, . . . each as a single quantity, (x+u)TM+P(x+u)m-1... =(u+x—a) (u+x—b) ... (u+x—b). Now, by performing the operations indicated in the two members, we shall, by the preceding Article, obtain for the first member, X being the first member of the proposed equation, and Y, Z... the derived polynomials of this member. With respect to the second member, it follows from Art. 294, 1st. That the part involving uo, or the last term, is equal to the product (x-a) (x—b)... (x−7) of the factors of the proposed equation; 2d. The co-efficient of u is equal to the sum of the products of these m factors taken m. .1 and m-1. 3d. The co-efficient of u2 is equal to the sum of the products of these m factors taken m-2 and m-2; and so on. Moreover, the two members of the last equation are identical; therefore, the co-efficients of the same powers are equal. Hence X=(x-a)(x-b) (x−c). . . (x−1), which was already known. Hence also, Y, or the first derived polynomial, is equal to the sum of the products of the m factors of the first degree in the proposed equation, taken m-1 and m-1; or equal to the sum of all the quotients that can be obtained by dividing X by each of the m factors of the first degree in the proposed equation; that is, Ꮓ 2 or the second derived polynomial, divided by 2, is equal to the sum of the products of the m factors of the proposed equation taken m-2 and m—2, or equal to the sum of the quotients that can be obtained by dividing X by each of the factors of the second degree; that is, To make the denominators disappear from an equation. 299. Having given an equation, we can always transform it into another of which the roots will be equal to a given multiple or suh. multiple of those of the proposed equation. Take the equation and denote by y the unknown quantity of a new equation, of which the roots are K times greater than those of the proposed equation. y If we take y=Kx, there will result x=- ; whence, substituting K and multiplying every term by K”, we have ym+PKym-1+QK3ym▬2+RK3ym−3+...+TKm-1y+UKm=0. an equation of which the co-efficients are equal to those of the proposed equation multiplied respectively by Ko, K1, K2, K3, K1, &c. This transformation is principally used to make the denominators disappear from an equation, when the co-efficient of the first term is unity. To fix the ideas, take the equation of the 4th degree 1st. Where the denominators b, d, f, h, are prime with each other; in this hypothesis, as K is altogether arbitrary, take K=bdfh, the product of the denominators, the equation will then become y+adfh.y3+cb3dƒah2. y2+eb3d3ƒah3.y+gb1d1ƒah3=0, an equation the co-efficients of which are entire, and that of its first term unity. We have besides, the equation a= of a corresponding to those of y. y to determine the values 'bdfh' 2d. When the denominators contain common factors, we shall evidently render the co-efficients entire by taking for K the smallest multiple of all the denominators. But we can simplify this still more, by observing, that it is reduced to determining K is such a manner that K1, K2, K3... shall contain the prime fac tors which compose b, d, f, h, raised to powers at least equal to those which are found in the denominators. First make k=9000, which is a multiple of all the other denominators, it is clear that the co-efficients become whole numbers. But if we decompose 6, 12, 150 and 9000 into their factors, we find 6=2×3, 12=22×3, 150=2×3×52, 9000=23×32×53; and by simply making k=2×3×5, the product of the different simple factors, we obtain k2=22×32X52, k3=23×33×533, k=21×3×54, whence we see that the values of k, k2, k3, ka, contain the prime factors of 2, 3, 5, raised to powers at least equal to those which enter in 6, 12, 150 and 9000. Hence the hypothesis k=2×3×5 is sufficient to make the denominators disappear. Substituting this value, the equation or y-5.5y+5.3.52y2-7.22.32.5y-13.2.32.5=0; y-25y3+375y-1260y-1170=0. Hence, we perceive the necessity of taking k as small a number as possible: otherwise, we should obtain a transformed equation, having its co-efficients very great, as may be seen by reducing |