4. Transform the equation

x2+16x3+99x2+228x+144=0,

into another whose roots shall be greater by 3.

Put x=-3+y. Result, y1+4y+9y2—42y=0. '

5. Transform the equation

x8x3x2+82x-60-0,

into another wanting its second term.

Result, y-23y2+22y+60=0.

(Art. 173.) We may transform an equation by division, as well as by substitution, as the following investigation will show. Take the equation

x+Ax+Bx2+Cx+D=0. . . . . .

(1)

If we put x=a+y, in the above equation, it will be transformed (Art. D.) into

in

As xa+y, therefore y=x-a; and put this value of У equation (2), we have

(x-a)+

X"

X"
2

—a)1 + 2.3 (x− +:
2. 3 ( x − a)2 + — — " (x—a)2 + X '(x—a)+X=0. . .(3)

Now it is manifest that equation (3) is identical with equation (1), for we formed equation (2) by transforming equation (1), and from (2) to (3) we only reversed the operation.

Now we can divide equation (3), or in fact equation (1), by (x-ɑ), and it is obvious that the first remainder will be X. Divide the quotient, thus obtained, by the same divisor, (x-a), and the second remainder must be X'.

Divide the second quotient by (x-a), and the third remainder X" 2

must be

The next remainder must be

X!!!
2.3'

&c., &c., according to the

degree of the equation.

Now if we reserve these remainders, it is manifest that they may form the coefficients of the required transformed equation; taking the last remainder for the first coefficient; and so on, in reverse order.

For illustration, let us take the third example of the last article.

For a further illustration of this method, we will again operate

on the first example of the last article.

x-3)x-12x2+17x2-9x+7(x3-9x3-10x-39

Hence y0y-37y2-123y-110=0, must be the transformed equation.

We shall have a 4th remainder, if we operate on an equation of the 4th degree; a 5th remainder with an equation of the 5th degree; and, in general, n number of remainders with an equation of the nth degree.

Y

But to make this method sufficiently practical, the operator must understand

SYNTHETIC DIVISION.

(Art. 174.) Multiplication and division are so intimately blended, that they must be explained in connection. For a particular purpose we wish to introduce a particular practical form of performing certain divisions; and to arrive at this end, we commence with multiplication.

Algebraic quantities, containing regular powers, may be multiplied together by using detached coefficients, and annexing the proper literal powers afterwards.

2. Multiply +xy+y2 by x2-xy+y3.

As the literal quantities are regular, we may take detached coefficients, thus:

Here the second and fourth coefficients are 0; therefore the terms themselves will vanish; and, annexing the powers, we shall have for the full product

4. Multiply —ax3+a2x2—a3x+a by x+a.

1-1+1-1+1

1+1

1−1+1—1+1

+1-1+1-1+1

1+0+0+0+0+1

As all the coefficients are zero except the first and last, therefore the product must be

(Art. 175.) Now if we can multiply by means of detached coefficients, in like cases we can divide by means of them.

Take the last example in multiplication, and reverse it, that is, divide a by x+a.

Here we must suppose all the inferior powers of x3 ́and a3 really exist in the dividend, but disappear in consequence of their coefficients being zero; we therefore write all the coefficients of the regular powers thus:

“