1st. Form Y, or the derived polynomial of X; then seek for the greatest common divisor between X and Y.

2d. If one cannot be obtained, the equation has no equal roots, or equal factors.

If we find a common divisor D, and it is of the first degree, or of the form x h, make x h 0, whence xh.

We then conclude, that the equation has two roots equal to h, and has but one species of equal roots, from which it may be freed by dividing X by (xh)2.

If D is of the second degree with reference to x, solve the equation D 0. There may be two cases; the two roots will be equal, or they will be unequal.

1st. When we find D = (xh)2, the equation has three roots equal to h, and has but one species of equal roots, from which it can be freed by dividing X by (xh)3.

2d. When D is of the form (x − h) (x — h'), the proposed equation has two roots equal to h, and two equal to h', from which it may be freed by dividing X by (x − h)2 (x — h')2, or by D2.

Suppose now that D is of any degree whatever; it is necessary, in order to know the species of equal roots, and the number of roots of each species, to solve completely the equation,

D= 0.

Then, every simple root of the equation D0 will be twice a root of the given equation; every double root of the equation D = 0 will be three times a root of the given equation and so on.

As to the simple roots of

X = 0,

we begin by freeing this equation of the equal factors contained in it, and the resulting equation, X′ = 0, will make known the simple roots.

EXAMPLES.

1. Determine whether the equation,

2x4 12x3+ 19x2 6.x + 9 = 0,

-

contains equal roots.

We have for the first derived polynomial,

8x3 36x2 + 38x

6.

Now, seeking for the greatest common divisor of these poly nomials, we find

= 3:

D=x-3= 0, whence x=

hence, the given equation has two roots equal to 3. Dividing its first member by (x-3)2, we obtain

The equation, therefore, is completely solved, and its roots are

2. For a second example, take

2,5 2x + 3x3. 7x2+8x3 = 0.

The first derived polynomial is

5x48x3+9x2 - 14x+8;

hence, the proposed equation has three roots equal to 1.

Dividing its first member by

3. For a third example, take the equation

x+5x+6x5 - 6x4 - 15x3 — 3x2 + 8x + 4 = 0.

The first derived polynomial is

7x6 + 30x5 + 30x4

24x345x2

6x+8;

and the common divisor is

x4 + 3x3 + x2. 3x-2.

The equation,

x2 + 3x3 + x2 - 3x-2=0,

cannot be solved directly, but by applying the method of equal roots to it, that is, by seeking for a common divisor between its first member and its derived polynomial,

we find a common divisor, x + 1; which proves that the square of x1 is a factor of

and the cube of x + 1, a factor of the first member of the given equation.

Dividing

x + 3x3 + x2 -3x-2 by (x + 1)2 = x2 + 2x + 1,

we have 2-2, which being placed equal to zero, gives the two roots x = 1, x =

x+2. Hence, we have

2, or the two factors, x-1 and

x + 3x3 + x2 - 3x − 2 = (x + 1)2 (x − 1) (x + 2).

Therefore, the first member of the proposed equation is equal to

(x + 1)3 (x − 1)2 (x + 2)2 ;

that is, the proposed equation has three roots equal to -1, two equal to 1, and two equal to -2.

4. What is the product of the equal factors of the equation x7x610x5 + 22x4

43x335x2 + 48x + 36 = 0 ?

Ans. (x2)2(x − 3)2 (x + 1)3.

5. What is the product of the equal factors in the equation,

x7. 3x69x5-19x4 + 27x3- 33x2 + 27x 9=0?

Ans. (x1)3 (x2 + 3)2.

Elimination.

270. We have already explained the methods of eliminating Ole unknown quantity from two equations, when these equations are of the first degree with respect to the unknown quantities.

When the equations are of a higher degree than the first, the methods explained are not in general applicable. In this case, the method of the greatest common divisor is considered the best, and it is this method that we now propose to investigate.

One quantity is said to be a function of another when it depends upon that other for its value; that is, when the quantities are so connected, that the value of the latter cannot be changed without producing a corresponding change in the former.

271. If two equations, containing two unknown quantities, be combined, so as to produce a single equation containing but one unknown quantity, the resulting equation is called a final equation; and the roots of this equation are called compatible values of the unknown quantity which enters it.

Let us assume the equations,

in which P and Q are functions of x and y of any degree whatever; it is required to combine these equations in such a manner as to eliminate one of the unknown quantities.

If we suppose the final equation involving y to be found, and that ya is a root of this equation, it is plain that this value of y, in connection with some value of x, will satisfy both equations.

If then, we substitute this value of y in both equations, there will result two equations containing only x, and these equations will have at least one root in common, and consequently, their first members will have a common divisor involving x (Art. 246).

This common divisor will be of the first, or of a higher degree with respect to r, according as the particular value of y = a cor responds to one or more values of x.

Conversely, every value of y which, being substituted in the two equations, gives a common divisor involving x, is necessarily a compatible value, for it then satisfies the two equations at the same time with the value or values of x found from this common divisor when put equal to 0.

272. We will remark, that, before the substitution, the first members of the equations cannot, in general, have a common divisor which is a function of one or both of the unknown quantities. For, let us suppose, for a moment, that the equations P0 and 2 = 0,

are of the form

PXR0 and QX R= 0,

R being a function of both ≈ and y.

Placing R = 0, we obtain a single equation involving two unknown quantities, which can be satisfied with an infinite number of systems of values. Moreover, every system which renders R equal to 0, would at the same time cause P'. R and Q'. R to become 0, and consequently, would satisfy the equations

P=0 and Q=0.

Thus, the hypothesis of a common divisor of the two polynomials P and Q, containing x and y, brings with it, as a consequence, that the proposed equations are indeterminate. Therefore, if there exists a common divisor, involving a and y, of the two polynomials P and Q, the proposed equations will be indeterminate, that is, they may be satisfied by an infinite number of systems of values of x and y. Then there is no data to determine a final equation in y, since the number of values of y is infinite.

Again, let us suppose that R is a function of a only. Placing R = 0, we shall, if the equation be solved with reference to x, obtain one or more values for this unknown quantity.

Each of these values, substituted in the equations

P'. R 0 and Q'. R = 0,