to real terms, so that i can occur only in connection with odd powers of b; thus

x3=(a+bi)3 = a3 +3a2 · bi + 3 a (bi)2 + (bi)s

=(a3-3 ab2) + (3 b a2 — b3) i.

By hypothesis the coefficients of f(x) are real and hence i can not occur except with some odd power of b. The result of substituting abi for x in fc) is therefore obtained by changing the sign of b in the expression found on substituting a + bi for x in f(x); the result is therefore P-Qbi. Since we had

P+ Qbi = 0,

and, since P and Q are real, therefore, P=0, Q=0. [3394, COR.] PQ bi0;

Hence,

therefore, abi is also a root of f(x)=0.

800. According to the preceding article, if a + ib is a root of f(x)=0, a bi is also a root and f(x) is divisible by x −(a + ib), also by x(abi), that is by

[x-(a+bi)] [x-(a−bi)]=[(x—a)+bi] [(x—a)—bi]

= (x − a)2 + b2,

a quadratic factor which is always positive for real values of x. Let abi, cdi, efi, be the imaginary roots of f(x) = 0, and f(x) be the product of the quadratic factors which correspond to the pairs of imaginary roots; then

f(x) = [(x−a)2 +b2] [(x−c)2 + d2] [(x − e)2+ƒ3].

Since each of these factors is positive for all real values of x, therefore f(x) is always positive for real values of x.

801. A proof exactly similar to that given in 2799 shows that some roots of the form ab enter in pairs into equations whose coefficients are rational.

EXAMPLE.-Solve the equation x+2x3-5x+6x+2=0, which has the root 2+1/3.

Since 213 is a root, then

therefore the equation is divisible by

2-13 is also a root; and

[x—(−2+1 ́3)] [x—(−2−√3)]=[(x+2)−√3] [(x+2)+√3] =(x+2)-3x+4x+1.

Hence the remaining roots of the equation are obtained by solving the equation

802. Fractional Roots.-A rational fraction can not be a root of an equation whose coefficients are integers and the coefficient of the first term is unity.

For, let the rational fraction, which is in its lowest term, be a root of the equation

+Pn-1
an

bn−2 a+ Pn bn-1.

b

This equation requires that a fraction, which is in its lowest term, is equal to an integer, which is impossible. Therefore a rational fraction can not be a root of the given equation.

TRANSFORMATION OF EQUATIONS

803. A given equation may be transformed into another whose roots are related in some definite manner to those of the given equation, and, as will be seen as our discussion is developed, we need not know the roots of the given equation. Later illustrations will be given, showing that such transformation may be used in the solutions of equations.

804. First Transformation.-To transform a given equation into another whose roots are equal to those of the given equation with opposite signs. Let f(x) = 0, 761, be the given equation and put x = y, so that for every value of x, y has an equal value with contrary signs; hence, substituting xy in f(x) = 0, we have

f(x) —ƒ(—y)=(—y)"+p,(—y)"−1+P2(—y)"−2+ ...

Pn-1Y+Pn=0,

Pn-1 Y = Pn = 0.

that is Thus the transformed equation is obtained from the given equation. by changing the sign of every other term, beginning with the second.

805. The preceding rule is applicable to a complete equation. If the equation is not complete, for example suppose that some of the coefficients are 0, as in the equation

then to transform this equation into another whose roots have signs contrary to those of the given equation, we may write it thus

x+4x50x1-5.x3±0.x2-3.x + 50.

Therefore, according to the rule, 804, we have

or

x® 4.x3±0.x + 5.x3±0.x2+3x+5=0,

х 4x+5x3 + 3x + 5 = 0.

806. Second Transformation. To transform an equation into another whose roots are equal to those of the given equation multiplied by a given number.

Let the given equation be f(x) = 0 and the given multiplier be k; hence, if we put y kx, then for any value of x the corresponding value of y is k times as large.

k

Solving y=kx for x, we get a = and substituting this value for e in f(r) = 0, the transformed equation is

or on multiplying by " we have

y"+p,ky"¬1+p.k2y"−2+ . · · · +Pn-¡k”¬1y+P„k”=0.

....

Hence, the transformed equation is derived from the given equation by multiplying the coefficient of the second term by k, that of the third term by k2, and so on.

807. The Transformation of an Equation with Fractional Coefficients into another whose Coefficients are Integers is one of the most valuable applications of the preceding principle.

EXAMPLE.-Transform the equation

k

in (1), then, according to the rule in 2806, the trans

Put x=
formed equation is

Put in (2) k

=

=2·3=6, .. k38 27 and equation (2) becomes

5

=

2

y3 — 3 (2 · 3) y2 + ¦ ¦ (22 · 3°) y — † (2 · 3)3— 0,

3

2

4

9

y9y+45 y 48 0.

The roots of equation (3) are thus six times as large as those of equation (1).

808. Third Transformation.-To diminish or increase the roots of an equation by a given number.

Let the given equation be f(r)=0; then, to transform this equation into another whose roots are diminished by a given number, k, put y=x-k. Therefore, for any value of x, y is less than x by k. From y=x-k we have xy+k, and the transformed equation is f(y + k) = 0.

(1)

Similarly, in case the roots are to be increased by k, put y whence xy-k and the transformed equation is

=

= x + k,

809. When the degree of the equation f(x) : 0 is greater than 3, the calculation indicated in the preceding article becomes laborious and a more simple mode of effecting the transformation is desirable.

n-2

(1) Let f(x) = x2 + Р1 x2¬1 + P2x2¬2 +

+ Pn-1 x + Pn = 0,

and suppose that the transformed polynomial in y is, by 2808, (1), (2) f(y + k) =y" + Q1y" ~ 1 + Į2 Y" -2 + ... + In-1 Y +Yni since y=x-k, expression (2) is equivalent to

(3) (x−k)"+q, (x−k) "~1 + q2 ( x −k)"−2+ . . . +9π-1 (x−k) +In which must be identical with the given polynomial in (1). Hence if the given polynomial is divided by x k the remainder is qn, and the quotient

(x — k)n−1 + q ̧ (x − k)n−2 +

and if this polynomial is divided by x and the quotient

Thus, it follows that by the repetition of these arithmetical operations, the successive remainders are

In In-1 In-2

Hence, the transformed equation may be determined by the rule:

Divide f(x) by xk according as the roots of the given equation are to be diminished or increased by k, and the remainder is the last term of the transformed equation. Divide the quotient just found by

X

k, and the remainder is the coefficient of the term next the last of the required equation; and so on.

The numerical calculation involved in the application of this rule is much abbreviated by synthetic division.

810. Synthetic Division.-Divide x2-5x+6x3 + 3x2 −8x − 3 by x-3x+4x+5, using detached coefficients.

1+0-3+4+5) 1+0-5+0+6+3-8-3(1+0—2—4

NOTE.-The signs of the partial dividends are changed and then added.

SYNTHETIC DIVISION

Using the last example, the work is arranged as follows:

11+0 50+ 6 + 3 8 3

-

+06+ 8 + 10

NOTE. The coefficients of all the powers of the unknown number are written down and the signs of the terms of the divisor excepting the first are changed. EXAMPLE. Find the quotient of x+4x3- x2 + 11 by x — 3.

Hence the quotient is x4+3x3 + 13x2 + 38x + 114 and the remainder 353.

EXPLANATION.-The coefficients of the dividend are written in the first row. To the left is written the second term of the divisor with its sign changed. The first term of the divisor is not written. The first term, 1, of the third horizontal row is the quotient of 1, the first term in the dividend, by 1, the first term of the divisor. The first term, 3, of the second row is the product of the divisor 3 by 1, the first term of the third row; it is added to the term above it for the second term in the third row and so on.