It should be remarked that the exponent of x, in the terms 1, 2, -6, 36, and 72, is equal to 0; hence, each of those terms disappears in the following derived polynomial.

2. Let it be required to cause the second term to disappear in the equation

and the operation is reduced to finding the values of the coefficients

Now, it follows from the preceding law, for derived polynomials, that

X' =3

(3)-12. (3)3+17. (3)2-9. (3)1+7, or X-110;

Y' =4. (3)3—36. (3)2+34. (3)1—9, or

Y' = 123;

into another equation, the roots of which shall exceed those of

We have, from the principles established,

X' = 4.(−1)3 – 4. (1) 5. (− 1)2 + 7. (−1)1-9, or X' = −25;

4. What is the transformed equation, if the second term be made to disappear from the equation

Ans. u533u3 — 118u2 -152u - 73 = 0.

5. What is the transformed equation, if the second term be made to disappear from the equation

into another, the roots of which shall be less than the roots of

Properties of Derived Polynomials.

266. We will now develop some of the properties of derived polynomials.

be a given equation, and a, b, c, d, &c., its m roots. We shall then have (Art. 248),

xm + Pxm-1+ Qxm2. . . = (x − a) (x — b) (x -- c) . . . (x − 1).

or omitting the accents, and substituting x + u for x, and we have (x + u)m + P(x + u)m−1 + . . . = (x + u − a) (x + u — b) . . .; or, changing the order of x and u, in the second member, and regarding xa, xb,... each as a single quantity,

(x + u)m+P(x + u)m−1 . . . = (u + x − a) (u+x—b)... (u+x−1). 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, &c., the derived polynomials of this member.

With respect to the second member, it follows from Art. 251:

1st. That the term involving uo, or the last term, is equal to the product ( a) (x − b) . . . (x-1) of the factors of the proposed equation.

2d. The co-efficient of u is of these m factors, taken m

...

equal to the sum of the products 1 and m - 1.

equal to the sum of the products

3d. The co-efficient of u2 is of these m factors, taken m - 2 and m

Moreover, since the two members of the last equation are identical, the co-efficients of the same powers of u in the two members are equal. Hence,

X = (x − a) (x − b) (x − c) . . .

which was already shown.

(x-1),

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 algebraic 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,

Ꮓ

Also, that is, the second derived polynomial, divided by 2,

2

is equal to the sum of the products of the m factors of the first member of the proposed equation, taken m2 and m2; or equal to the sum of the quotients obtained by dividing X by each of the different factors of the second degree; that is,

267. An equation is said to contain equal roots, when its first member contains equal factors of the first degree with respect to the unknown quantity. When this is the case, the derived polynomial, which is the sum of the products of the m factors taken m - 1 and m 1, contains a factor in its different parts, which is two or more times a factor of the first member of the proposed equation (Art. 266): hence,

There must be a common divisor between the first member of the proposed equation, and its first derived polynomial.

It remains to ascertain the relation between this common divisor and the equal factors.

268. Having given an equation, it is required to discover whether it has equal roots, and to determine these roots if possible. Let us make

and

b, n' factors equal to x

C...

suppose that the second member contains n factors equal to x ·a, n' factors equal to x and also, the simple factors x then have,

P, x

q, x-r

.; we shall

X = (x − a)” (x — b)n (x — c)n"... (x − p) (x − q) (x − r) (1).

We have seen that Y, or the derived polynomial of X, is the sum of the quotients obtained by dividing X by each of the in factors of the first degree in the proposed equation (Art. 266).

Now, since X contains n factors equal to x -

a, we shall

and the same reason

x

ing applies to each of the repeated factors, ab, x-c..... Moreover, we can form but one quotient for each simple factor, which is of the form,

therefore, the first derived polynomial is of the form,

By examining the form of the value of X in equation (1), it is plain that

(x − a)n-1,

(x — b)n'−1,

(x — c)n''-1...

are factors common to all the terms of the polynomial Y; hence the product,

(x − a)n−1 × (x — b)n/−1 × (x — c)n''-i.........

is a divisor of Y. Moreover, it is evident that it will alsc divide X: it is therefore a common divisor of X and Y; and it is their greatest common divisor.

For, the prime factors of X, are x −α, x —

now, .;

- b, x — c..., and

x

x-p, x q, x-r
x-P, X
divide Y, since some one of them will be wanting

- r, cannot

in some of other parts.

the parts of Y, while it will be a factor of all the Hence, the greatest common divisor of X and Y, is (x — a)n−1 (x — b)n'—1 (x — c)n''—1

D =

.;

that is,

The greatest common divisor is composed of the product of those factors which enter two or more times in the given equation, each raised to a power less by 1 than in the primitive equation.

269. From the above, we deduce the following method for finding the equal roots.

To discover whether an equation,

X = 0,

contains any equal roots: