Properties of derived Polynomials.

299. We will now develop some of the remarkable properties

of derived polynomials.

Let

X = xm + Pxm-1 + Qxm-2... Tx + U=0

be a given equation, and a, b, c, d, &c., its m roots. We shall then have (Art. 281),

xm + Pxm-1 + Qxm-2... = (x – a) (x - b) (x - c) . . . (x - 1). Making

x = x' + u, or omitting the accents, and substituting x + u for x, and we have (x + u)m + P(x + u)m-1 + = (x + и - a) (x + u - b)...; or, changing the order of x and u, in the second member, and regarding xа, х

- b,

(x+u)m+P(x + 4)m-1

...

each as a single quantity,

=(u+x-a) (u+x-b)... (и+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 ... the derived polynomials of this member.

With respect to the second member, it follows from Art. 295, 1st. That the part involving uo, or the last term, is equal to the product (x - a) (x - b)... (x - 1) 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, since the two members of the last equation are identical, the co-efficients of the same powers are equal. Hence,

X = (x - a) (x - b) (x - c) ... (x - 7), 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,

Y =

Z

Also,

2'

that is, 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,

300. An equation is said to contain equal roots, when its first member contains equal factors. 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 proposed equation (Art. 299).

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.

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

X = xm + Pxm-1 + Qxm-2 + ... + Tx + U = 0, and suppose that the second member contains n factors equal to

a, n' factors equal to x - b, n" factors equal to x-c..., and also, the simple factors xх - р, х q, x-r...; we shall then have,

X = (x - a) (x - b)/(x - c)n"... (х - р) (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 m factors of the first degree in the proposed equation (Art. 299).

Now, since X contains n factors equal to x a, we shall have

X

a

n partial quotients equal to ; and the same reasoning applies to each of the repeated factors, x-b, x — с.... More over, 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)-1, (x - b)-1, (x - c)n/-1

...

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

(x - a)n-1 x (x - b)n-1 x (x - c)-1...

is a common divisor of Y. Moreover, it is evident that it will also 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 - a, x - b, х- с..., and х - р, х- q, x-r...; now, x - p, x-q, x - r, cannot divide Y, since some one of them will be wanting in each of the parts of Y, while it will be a factor of all the other parts. Hence, the greatest common divisor of X and Y is

D = (x - a)-1 (x - b)n/-1 (x - c)n/-1...; 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 unity than in the primitive equation.

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

To discover whether an equation

X = 0

contains any equal roots, form Y or the derived polynomial of X; then seek for the greatest common divisor between X and Y; 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 x = h. 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, resolve 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 = (x-h), 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 (x - h)3.

2d. When Dis 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 (xh)2 (x - h')2, or by D2. Suppose now that Dis 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 resolve completely the equation

D = 0.

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

As to the simple roots of

we begin by freeing this

X = 0,

equation of the equal factors contained in it, and the resulting equation, X = 0, will make known the simple roots.

We have for the first-derived polynomial (Art. 297),

Now, seeking for the greatest common divisor of these poly

nomials, we find

D = x - 3 = 0, whence x = 3;

hence, the given equation has two roots equal to 3.

Dividing its first member by (x - 3)2, we obtain

2x2 + 1 = 0; whence x = ±

1

2

-2.

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

2. For a second example, take

x5 - 2x4 + 3x3 - 7x2 + 8x - 3 = 0.

The first-derived polynomial is

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

Dividing its first member by

3. For a third example, take the equation,

cannot be resolved 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

4x3 + 9x2 + 2x – 3,

we find a common divisor, x + 1; which proves that the square of x + 1 is a factor of x4 + 3x3 + x2 3x 2, and the cube of x + 1, a factor of the first member of the given equation.