3+5+7|2

6 +22

11 + 29

6

17

Then, since 29 corresponds to T1 in (E), and 17 to S", and that 3 is the coefficient of y2, we have 3y+ 17y+ 29 for the transformation, as required.

Ex. 2. To transform -3x+2x-11, when we put x = y + 2.

Here n = 3, m = 2, A = 1, B = − 3, C = 2, D = — 11, and we have, by proceeding as in the scheme, (E), the following work.

1-32-11|2

Hence, since - 11 corresponds to T, and 2 to S", 3 to RI in the scheme, and that the coefficient of a3 in the given expression is 1, we shall have y3 + 3y2 + 2y - 11 for the sought transformation.

Ex. 3. To transform - 10+6x+1, by putting x=y-3.

-

Here we have n = 5, m = — 3, A = 1, B = 0, C= — 10, D= 0, E6, F = 1; hence, proceeding as in (E), we get

Since 10, 141, 180, 80, 15, severally correspond to T', S", R", Q, Py in the scheme, and that A = 1, we shall have y — 15y + 80y3 — 180y3 + 141y + 10 for the required transformation.

Ex. 4. To transform 2, when xy+m.

In this case we have n = 2, A = 1, B = 0, C=0; consequently we get

1+0+0m

m + m2

m

m

2m

hence y2+2my + m2 is the required transformation; and we have (y + m)2 = y2 + 2my + m2, a well-known x2 = result.

Ex. 5. To transform a3 when xy- m.

In this instance n = 3, and for m we must use m; also A = 1, B = 0, C = 0, D= 0; consequently, we shall get

Hence, since - m3, 3m3, - 3m severally correspond to T, S", R in the scheme, and that A = 1, we get a3 = (y · — m)3 = y3 — 3my2 + 3m3y — m3, a well-known result.

Ex. 6. Transform a 36, when xy + 6.

Here we have n = 2, m = 6, A = 1, B = 0, C=-36; and by (E) we get

1+0-366
6 + 36

6+0

6

12

Hence, since 0, 12 correspond to T, S" in (E), we get

y+12y for the sought transformation; and we shall have y2+12y= x2 - 36.

=

If a 360, or a 36, then we must have y2+12y =(y+12)y=0, which is satisfied either by putting y = 0, or y+120. When y = 0, we get x = y +6=0+6= 6, and since 236, it follows that = 6 is the square root of 36. Also, y + 120 gives y=12, and since y +6, we must have 12+6= 6; consequently, Hence, we see that

x =

6 must also be a square root of 36. the square root of 36 may be either + 6

or 6.

Ex. 7.-Transform - 27 when xy + 3.

x =

Here m = 3, n = 3, A = 1, B = 0, C= 0, D= 27, and by (E) we get

1+0+ 0 2713

3+ 9 +27

3+9+0

3 + 18

6 +27

=

3

9

Hence, since 0, 27, 9 correspond to T, S", R", in (E), we get y+9y+27y for the sought transformation. Hence we shall have y3 + 9y2+ 27y=23-27, and if a3 — 27=0, or a 27, we must have y+9y2 + 27y = (y2 + 9y+27) y= 0, which is satisfied by putting y = 0, or y2+9y+27 =0; the expression y = 0 gives xy + 3 = 0 + 3 = 3. Consequently, since a3 27, we must have a = 3 = the third or cube root of 27.

Remark. We see that our rule may be

of numbers.

from this and the preceding example applied to the extraction of the roots

Ex. 8.-Transform - 7x + 12, when x = y + 4.

Here n = 2, m = 4, A = 1, B = 7, C= 12, and by (E) we shall get

1-712/4
4-12

- 3+ 0

4

1

Hence we have y2+y for the sought transformation; and of course we must have y + y = x2 · 7x+12. Consequently, if a 7x+12=0, we must have y2+y=(y+1) y = 0, which is satisfied either by y = 0, or y + 1 = 0. The value y = 0 gives x = y + 4 = 4, and the value y + 10 gives y = − 1, and x = y + 4 = 3, and we have

4 = 0, or x 30. Hence x 4 and x 3 must be the factors of a 7+12; indeed, we have (x-4) (∞ — 3) = (x x2-7x+12.

=

Ex. 9.-Transform - 3x2+2x-6, when xy + 3. Here we have n = 3, m3, A = 1, B3, C=2, D= 6; hence, by (E), we get

Consequently, since 0, 11, 6, correspond to T, S", R", and that the coefficient of a3 is 1, we have y3 + 6y2+11y for the required transformation.

If a3-3x2 + 2x-6= 0, then y3+6y2 + 11y = (y2 + 6y +11)y 0; which is satisfied by putting y = 0, or y2+ by +11: = 0; the value y = 0 reduces x = y + 3 to x = 3 or -30; and consequently 2-3 is a factor of a3-3x2 + 2x-6; indeed we have a3-3x2 + 2x − 6 = (x − 3) (x2+2).

Remark. It is clear from this, and the preceding examples, that our rule can be applied to the resolution of equations; or, which comes to the same thing, to the resolution of quantities of the form (A) into their factors. And the principle is, so to assume m in x = y +m, that the term which does not involve y in the transformed equation shall equal 0, or differ from 0 by a quantity that may be rejected as of no consequence in the required degree of exactness of x.

EXAMPLES FOR EXERCISE.

1. Transform 5x-7x-50, when xy + 4.

Ans. 5y+33y+ 2 is the transformation.

2. Transform 23

9, when xy + 2.

Ans. y3+by+ 12y-1.

Ans. y3 + 9y3 + 20y — 1.

3. Transform a 7-7, when xy + 3.

4. Transform 2+11x-102x + 181, when xy- 17. Ans. y3 — 40y2 + 391y + 181.

5. Transform a + 2x3-7x2-8x+12, when xy + 1.

7. Transform x2+2ax - b, when xy-a.

(11.)

APPENDIX TO MULTIPLICATION AND DIVISION.

We have, in Multiplication and Division, supposed the multiplier and divisor (each) in every operation to retain the same form. But it is easy to see that we may, if we please, continually change the form of the multiplier or divisor (at will) in any case or during the same operation; provided always, the value of the multiplier or divisor is not changed.

We will now give some examples for the purpose of showing that it will often be useful to change the forms of the multiplier or divisor (continually), in order to transform expressions that may occur in calculation.

EXAMPLES.

....

Ex. 1.-Let m stand for any positive integer, and suppose it is required to find the sum of the series (m − 1) + (m − 2) +(m-3) + (m 4) + . . . . + 1, (1), which is called a decreasing arithmetical progression; 1, the difference between any two successive terms, being called the ratio of decrease; also (m-1) and 1 are called its first and last terms, or the extremes of the series, and since (m 1) contains as many units as there are terms in the series, (m-1) is called the number of terms in the progression.

To find the sum of the series (1), we shall multiply each