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

Ν

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.

277. Having given an equation, we can always transform it into another of which the roots will be equal to a given multiple or sub. multiple of those of the proposed equation.

Take the equation

2+Рx-1+Qx-3+... Tx+U=0,

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

yTM+PKy"-1+QK3yTM-2+RK3yTM¬3+... +TK-y+UK"=0.

an equation of which the co-efficients are equal to those of the pro. posed equation multiplied respectively by Ko, K1, K2, K3, K*, &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+cb3dƒ3h2. y2+eb3d3ƒah3 •y+gb*d*f*h3=0, an equation the co-efficients of which are entire, and that of its first term unity.

We have besides, the equation x=

of a corresponding to those of y.

y

"bdfh'

to determine the values

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 ir

such a manner that K', 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 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

k=2X3X5, l3=23X33X5*, k=24X34X5*,

whence we see that the values of k, k2, k3, k1, 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+5.3.5y-7.22.39.5y-13.2.3.5=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

the transformed equation resulting from the supposition k=9000 in the preceding equation.

y3— 65y^+1890y3-30720y2 — 928800y+972000=0.

278. The preceding transformations are those most frequently used; there are others very useful, of which we shall speak as they present themselves; they are too simple to be treated of separately.

In general, the problem of the transformation of equations should be considered as an application of the problem of elimination between two equations of any degree whatever, involving two unknown quantities. In fact, an equation being given, suppose that we wish to transform it into another, of which the roots have, with those of the proposed equation, a determined relation.

Denote the proposed equation by F(x)=0, (enunciated function of r equal to zero), and the algebraic expression of the relation which should exist between x and the new unknown quantity y, by F' (x, y)=0; the question is reduced to finding, by means of these two equations, a new equation involving y, which will be the required equation. When the unknown quantity x is only of the first degree in F'(x, y)=0, the transformed equation is easily obtained, but if it is raised to the second, third . . . power, we must have recourse to the methods of elimination.

...

Elimination.

279. To eliminate between two equations of any degree whatever, involving two unknown quantities, is to obtain, by a series of operations, performed on these equations, a single equation which contains but one of the unknown quantities, and which gives all the values of this unknown quantity that will, taken in connection with the corresponding values of the other unknown quantity, satisfy at the same time both the given equations.

This new equation, which is a function of one of the unknown quantities, is called the final equation, and the values of the unknown quantity found from this equation, are called compatible values.

Of all the known methods of elimination, the method of the common divisor, is, in general, the most expeditious; it is the method which we are going to develop.

Let F(x, y)=0 and F'(x, y)=0 be any two equations whatever, or, more simply,

A=0, B=0.

Suppose the final equation involving y obtained, and let us try to discover some property of the roots of this equation, which may serve to determine it.

Let y=a be one of the compatible values of y; it is clear, that since this value satisfies the two equations, at the same time as a certain value of x, it is such, that by substituting it in both of the equations, which will then contain only x, the equations will admit of at least one common value of x; and to this common value there will necessarily be a corresponding common divisor involving x. Art. 262. This common divisor will be of the first, or a higher degree with respect to x, according as the particular value of y=a corresponds to one or more values of x.

Reciprocally, every value of y, which, substituted in the two equations, gives a common divisor involving x, is necessarily a compatible value, because it then evidently satisfies the two equations at the