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does not reduce to O by this substitution, we may conclude that there is no common divisor depending upon the principal letter.

Farther, having obtained a remainder of the 2d degree, with reference to a, it is not necessary to continue the operation any farther. For,

Decompose this polynomial into two factors of the 1st degree, which is done by placing it equal to 0, and resolving the resulting equation of the 2d degree.

When each of the values of a thus obtained, substituted in the remainder of the 3d degree, destroys it, it is a proof that the remainder of the 2d degree, simplified, is a common divisor; when only one of the values destroys the remainder of the 3d degree,. the common divisor is the factor of the 1st degree with respect to a, which corresponds to this value.

Finally, when neither of these values destroys the remainder of the 3d degree, we may conclude that there is not a common divisor depending upon the letter a.

It is here supposed that the two factors of the 1st degree with reference to a, are rational, otherwise it would be more simple to perform the division of the remainder of the 3d degree by that of the second, and when this last division cannot be performed exactly, we may be certain that there is no rational common divisor, for if there was one, it could only be of the 1st degree with respect to a, and should be found in the remainder of the 2d degree, which is contrary to the hypothesis.

and

3. Find the greatest common divisor of the two polynomials

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4. Find the greatest common divisor of the polynomials 12x5 + 16x - 15x3 + 14x2 - 15x + 4. 9x3+47x2

and

20x6
15x4

21x + 28.

Ans. 5x2 3x + 4.

5. Find the greatest common divisor of the two polynomials

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Transformation of Equations.

The transformation of an equation consists in changing its form without affecting the equality of its members. The object of a transformation, is to change an equation from a given form, to another form that is more easily resolved.

First Transformation.

To make the Denominators disappear from an Equation.

294. If we have an equation of the form

and make

xm + Pxm-1 + Qxm-2 + ... Tu + U = 0,

x=

y. -; k

we shall have, after substituting this value for x, and multiplying every term by km,

...

+Tkm-ly+Ukm = 0,

ym + Pkym-1 + Qk2ym-2 + Rk3ym-3+ an equation of which the co-efficients are equal to those of the given equation, multiplied respectively by ko, k1, k2, k3, kt, &c.

This transformation is principally used to make the denominators disappear from an equation, when the co-efficient of the first term is unity.

As an example, take the equation of the 4th degree,

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y being a new unknown quantity and k an indeterminate quantity, we have

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Now, there may be two cases

1st. Where the denominators b, d, f, h, are prime with each other. In this hypothesis, ask is altogether arbitrary, take k=bdfh, the product of the denominators, the equation will then become

y2 + adfh. y3 + cb2df2h2 . y2 + eb3d3f2h3. y + gb*d*f*h3 = 0, in which the co-efficients are entire, and that of its first term unity.

We can determine the values of a corresponding to those of y from the equation,

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2d. When the denominators contain common factors, we shall evidently render the co-efficients entire, by making k equal to the smallest multiple of all the denominators. But we can simplify still more, by giving to k such a value that k1, k2, k3, shall contain the prime factors which compose b, d, f, h, raised to pow

...

ers at least equal to those which are found in the denominators.

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after making x =

y k'

and reducing to entire terms. First, if we 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 x 3, 150=2×3×52, 9000 = 23 × 32 × 53; and by simply making

k=2×3× 5,

24 x 34 x 54;

the product of the different simple factors, we obtain k2 = 22 × 32 × 52, k3 = 23 × 33 × 53, k whence we see that the values of k, k2, k3, ht, contain the prime factors of 2, 3, 5, raised to powers at least equal to those which enter into 6, 12, 150, and 9000.

Hence, the hypothesis

k = 2 × 3 × 5,

is sufficient to make the denominators disappear. Substituting this

value, the equation becomes

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5.22.32.52

y3 +

y2

7.23.33.53
2.3.52

y

13.24.34.54

0,

22.3

23.32.53

which reduces to

у - 5.5y3 + 5.3.52y2 - 7.22.32.5у - 13.2.32.50; or у - 25у3 + 375у2 - 1260у - 1170 = 0.

Hence, we perceive the necessity of taking kas 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.

Hence we see, that any equation may be transformed into another equation, of which the roots shall be a multiple or sub-multiple of those of the given equation.

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y5 - 65y + 1890y3 - 30720y2 — 928800y + 972000 = 0.

Second Transformation.

To make the second Term disappear from an Equation. 295. The difficulty of resolving an equation generally diminishes with the number of terms involving the unknown quantity. Thus the equation

x2 = q, gives immediately, x = ± √q

while the complete equation

x2 + 2px+q = 0,

requires preparation before it can be resolved.

Now, any given equation can always be transformed into another equation, in which the second term shall be wanting.

For, let there be the general equation

Suppose

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

x = u + x',

u being unknown, and a' an indeterminate quantity. By substituting u + x' for x, we obtain

(u+x')m+P(u + x')m-1 + Q(u + x')m-2... +T(u+ x) + U = 0.

Developing by the binomial formula, and arranging according to the decreasing powers of u, we have

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Since a' is entirely arbitrary, we may dispose of it in such a

way that we shall have

mx' + P = 0; whence, x =

P

m

Substituting this value of x' in the last equation, we shall obtain

an equation of the form,

um + Q'um-2 + R'um-3 + T'u + U' = 0,

...

in which the second term is wanting.

If this equation were resolved, we could obtain any value of x

corresponding to that of u, from the equation

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Whence, in order to make the second term of an equation disappear,

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Substitute for the unknown quantity a new unknown quantity, united with the co-efficient of the second term, taken with a contrary sign, and divided by the exponent of the degree of the equation. Let us apply the preceding rule to the equation

If we make

we have

x2 + 2px = q.

x = и - р,

(up)2 + 2p (u - p) = q;

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