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4. What is the equation whose roots are each of those of +18x2+99x+81-0 taken negatively?

Ans. y3—6y2+11y—3=0.

5. Transform the equation x3-2x2+3x-10=0, into one having integral coefficients. Ans. y3-by2+3y-270=0.

ART. 406. PROP. II.- To transform an equation into one whose roots are greater or less by any given quantity than the corresponding roots of the proposed equation.

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The relation between x and y will be expressed by the equation y=x+r. As the principle is the same, whether x is increased or diminished, we will consider the case where y=x-r. This gives x=y+r, and by substituting this for x in the proposed equation, we have

(y+r)"+A(y+r)2+B(y+r)2.... +T(y+r)+V=0. Developing the different powers of y+r by the Binomial theerem, and arranging the terms according to the powers of y, we have

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ART. 407. Cor.- By means of the preceding transformation we may remove any intermediate term of an equation. Thus, to transform an equation into one which shall want the second term, r must be assumed, so that nr+A=0, or r=—- To take

A

n

away the third term, {n(n−1)r2+(n−1)Ar+B must be put =0, and the value of r derived from the solution of this equation.

The pupil may solve the following examples by the preceding principles:

1. Transform the equation x3-7x+7=0 into another whose roots shall be less by 1 than the corresponding roots of this equation. Ans. y3+3y2-4y+1=0.

2. Find the equation whose roots are less by 3 than those of the equation

-3x3-15x2+49x-12=0.

Ans. y1+9y3+12y2—14y=0.

3. Transform the equation 3—6x2+8x—2—0 into another whose second term shall be absent.

Here A=-6, n=3,

r=2; hence, x=y+2.

Ans. y3-4y-2=0.

4. Transform the equation x2+2px-q=0 into another wanting the second term. Ans. y2-p2-q=0.

ART. 408. There is a more easy and elegant method of performing the operation of transformation, so as to increase or diminish the roots of an equation, than by direct substitution, which we will now proceed to explain.

Let the proposed equation be

Ax1+Bx3+Сx2+Dx+E=0,

(1)

and let it be required to transform it into another, whose roots shall be less by r; then y=x-r and x=y+r.

By substituting y+r, instead of x, we have

A(y+r)*+By+r)3+C(y+r)2+D(y+r)+E=0.

By developing the powers of y+r, and arranging the terms according to the powers of y, as in Art. 406, the transformed equation will take the form

Ay+By+C1y2+D1y+E,=0.

1

(2)

where the coefficient A must evidently be the same as in equation (1), while the coëfficients B1, C1, D1, and E1, are unknown quantities to be determined from the calculation. For y substitute its value x-r, and equation (2) becomes

A(x—r)1+B ̧ (x—r)3+C1(x—r)2+D,(x—r)+E ̧=0 (3) Now since the values of x are the same in equations (1) and (3), it is evident these equations are identical. Hence, whatever operation is performed on equation (1), the result will be the same as if this process had been applied to equation (3). Now as the object is to obtain the values of the coëfficients B1, C1, &c., let equation (3) or (1) be divided by x-r, and it is evident that the quotient will be

A(x)+B, (—r)2+C, (x—r)+D1,

and the remainder will be the last coëfficient E,; hence, E, is determined.

be

Again, divide this quotient by x-r, and the next quotient will

A(x)+B,(x-r)+C, with a remainder D, ; hence, D, is determined. Dividing again by x-r we get the remainder C,; and lastly, by another division, we obtain the remainder B,; and thus we find all the coefficients of equation (2).

To illustrate this method we will now proceed to solve Ex. 1, Art. 407; that is, to find the equation whose roots are less by 1, than those of the equation x3-7x+7=0.

Here y=x-1, and we proceed to divide the proposed equation and the successive quotients, by x-1. The successive remainders will be the coefficients of y in the transformed equation, except that of the highest power, which will have the same coëfficient as the highest power of x in the proposed equation.

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Here the last quotient is 1, and the successive remainders are +3,-4, and +1. Comparing these with the general equation, we have A=1, B1=+3, C‚——4, and D1=+1. Placing these as coëfficients to the respective powers of y, the transformed equation is y+3y2-4y+1=0.

This method of transforming an equation, would not, in general, be shorter than direct substitution, were it not that the successive divisions may be greatly shortened by a process, called from its discoverer, HORNER'S SYNTHETIC METHOD OF DIVISION, which we shall now proceed to explain.

ART. 409. SYNTHETIC DIVISION.-This may be considered as an abridgment of the method of division by Detached Coëfficients (Art. 77). To explain the process, we shall first divide 5x-12x3 +3x2+4x-5 by x-2, by detached coëfficients.

Divisor 1-2)5-12+3+4-5(5-2-1+2 Quotient.
or 5x3-2x2x+2.

5-10

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By changing the signs of the terms of the divisor, except that of the first term, which must not be changed, as by means of that we determine the signs of the respective terms of the quotient, and adding each partial product instead of subtracting it, except the first term, which being always the same as the first term of each dividend, may be omitted, the operation may be represented thus :

1+2)5-12+3+4-5(5-2-1+2

*+10
-2+3
*4

-1+4
*2
+25

Let it be observed that the figures over the stars are the coëffi cients of the several terms of the quotient. It will also be seen that it is unnecessary to bring down the several terms of the dividend. Hence, the last operation may be represented as follows:

+2)5-12+3+4-5
+10-4-2+4
-2-1+2-1

In this operation 5 is the first term of the quotient, +10 is the product of +2, the divisor, by 5; the sum of +10 and -12 gives 2, the second term of the quotient, -4 is the product of +2, the divisor, by -2, the second term of the quotient, and the sum of 4 and +3 gives-1, the third term of the quotient, and so on. The last term, -1, is the remainder.

Supplying the powers of x, the quotient is 5x3-2x2—x+2, with a remainder -1.

A similar method may be used when the divisor contains three terms; and if the coefficient of the first term of the divisor is not unity, it may be made unity by dividing both dividend and divisor by the coefficient of the first term of the divisor. The method, however, is rarely used except when the divisor is a binomial, the coefficient of whose first term is 1.

In the application of Synthetic division, when any term of an equation is absent, its place must be supplied with a zero.

ART. 410. We shall now illustrate the use of Synthetic division in the transformation of equations, by the method described in Art. 408.

1. Let it be required to find the equation whose roots are less by 1 than those of the equation x3-7x+7.

To effect this transformation, it is required to find the successive remainders which arise from dividing x3-7x+7, and the successive quotients by x-1.

Since the second term is wanting, its place must be supplied with 0. Also, in arranging the operation, it is customary to place the second term of the changed divisor on the right, as in division.

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Hence, the required coëfficients are 1, +3,-4, and +1.

.. y3+3y2—4y+1=0 is the transformed equation required. 2. Transform the equation 5x1+28x3+51x2+32x-1=0, into another having its roots greater by 2 than those of the given equation.

Here, y=x+2; hence, to find the coefficients of the transformed equation, we must find the successive remainders arising from dividing the proposed equation and the successive quotients by x+2. Changing the sign of +2, the operation is as follows:

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