b

square of the square root of anything, is, of course, that thing itself; so that the square of a3 is a, and the square of a square is the fourth power: hence the fourth power of a is a2. Again, means the third power of b; so that the fourth power of b* is the twelfth power of b. Now, b is the second power of bt; hence bxbxbxbxbxb must be the twelfth power of b3; in other words, it is the fourth power of b3; so that the fourth power of bis 66, the original quantity. We see, therefore, that as is really the fourth root of a2, and b3 the fourth root of bo. And in a similar way the learner may always satisfy himself that every fractional exponent may be reduced to its lowest terms, without introducing error. But a distinct article will be devoted to the general theory of exponents hereafter.

EXAMPLES FOR PRACTICE.

Express the values of the following quantities:

In some of these examples, the coefficient under the radical, though not admitting of the proposed extraction, may be divided into two factors, one of which does admit of it. Thus, in example (14), though 8 is not a square, yet of its two factors, 4 and 2, the first is a square; so that √8= √4.2=2√2. In like manner, in example (19), 16=/8.2=24/2. The learner is expected to take notice of these peculiarities, and to extricate from the radical sign as much of the proposed expression as he can, leaving under that sign as little as possible in the finished result. We recommend this more for the purpose of encouraging in the learner the habit of looking into the composition of the expressions he is engaged with, than for any other reason. It is customary to call 2/2 a simpler form than 8 and 23/2 a simpler form than 16; but, in reality, such is not the case: the so-called simpler forms are the more inconvenient in practice. It is as easy to extract the square root of 8, and the cube root of 16, approximately, as to extract the square root of 2 and the cube root of 2; and, of course, as easy to take the former roots out of a table as to take out the latter. In the changed forms, in addition to the discovery of the roots, a multiplication by 2 is afterwards necessary

36. II. To extract the square root of a compound quantity:

RULE. Let the proposed polynominal or compound quantity be arranged as if for division, and mark off a place as for a quotient.

Find the square root of the leading term of the polynomial, put it in the quotient's place, and the square of it under the leading term, to which it will, of course, be equal; draw a line, as if for subtraction, and bring down under it the next two terms of the polynomial: these will form a dividend; and a place to the left is now to be marked off for a divisor.

Put twice the root just found in the divisor's place: see how often this is contained in the leading term of the dividend; and connect the quotient, with its proper sign, both to the root already found, and to the quantity in the divisor's place: the divisor will thus be completed.

Multiply this complete divisor by the term last placed in the quotient, subtract the product from the dividend, and to the remainder annex the two next terms of the polynomial, and a second dividend will be obtained. Proceed with this as with the former, marking off a divisor's place; and, as a portion of the complete divisor, putting in that place twice the entire quantity in the quotient, as before: and so on, till all the terms of the polynomial have been brought down.

a2+2ab+b2(a+b

The process here described for finding the square root of a compound quantity, is suggested by the form which the square of a compound quantity takes. The square of the binomial a+b, we know to be a2+2ab+b2; and from this square, we can return to the root thus: The square root of the leading term is a, the first term of the sought 2a+b] 2ab+b2 root: the square of this a being subtracted from the proposed trinomial, leaves 2ab+b2; and we see that we have

a2

2ab+b2

only to divide the leading term of this by 2a, to get the other term b, of the root; and, moreover, that if this b be added to the 2a, the sum 2a+b will be the complete divisor of 2ab+b2, corresponding to the quotient b. The steps, therefore, may be arranged as in the margin, and they are no other than the steps directed in the rule.

Again, the square of the trinomial a+b+c, we know

to be (a+b)+2(a+b)c+c2, since (a+b+c) is the same as [(a+b)+c]; that is the square is,

a2+2ab+b2+2(a+b)c+c2;

and we may return from this square to its root by proceeding exactly as above, namely, by first finding a and then b, as already shown, and afterwards deriving c from the quotient terms a+b, just as b was derived from a; c being found, the second divisor may be completed and the work finished as below, the steps of which are at once seen to be in accordance with, and thus to justify the rule.

a2+2ab+b2+2(a+b)c + c [a+b+c

a2

2a+b]2ab+b2

2ab+b2

2(a+b)+c]2(a+b)c+c2

2(a+b)c+c2

In like manner, by applying the same process to the square of a quadrinomial, (a+b+c+d)2=(a+b+c)2+2(a+b+c)d +d, we should find the rule to be true when the root of the polynomial consists of four terms; and so on, since each square is formed according to the same uniformı law.* Of course, the operation will terminate, as above, only when the proposed polynomial is a complete square: when such is not the case, we shall have a remainder after all the terms have been brought down; and this remainder can never be exhausted, however far the steps be carried, or the approximation to the root extended. (1) 4x-12x+25x-24x+16[2x2-3x+4

Therefore (4x-12x3+25x2—24x+16)=2x2—3x+4.

* The square of a polynomial, of whatever number of terms, may be expressed thus: namely, (a+b+c+d+ . . . . . . . . . . l)2 = a2+ (2a+b)b + {2 (a+b)+c}c + {2(a+b+c)+a} ..+12: a form which suggests the rule in all cases.

d+......

(2) 9x-12x+10x*-28x3+17x2-8x+16[3x3-2x2+x−4

926

6x3-2x-12x+10x1

-12x+4x1

6x34x2+x]6x1-28x+17
6x- 4x3+ x2

6x3-4x2+2x-4] −24x3+16x2-8x+16

-24x16x-8x+16

37. As there is a remainder in this example, we infer that the proposed expression is not a complete square. If, instead of 4, the final term of the polynomial had been 1, there would have been no remainder, and the polynomial would then have been decomposable into two factors, each equal to 2x2—x—1; but, as it is, these factors require correction, which correction, like as in division, the remainder suggests: the corrected factors

are

(2x3-x-1+√-3) (2x2-x-1-√—3);

for by the theorem at page 44, so often referred to, the product of these is

(2 x3—x—1)3—(-3)=(2x2-x-1)+3=4x-4x3-3x2+ 2x+4.

And by thus changing the sign of the remainder, prefixing the radical, and connecting it with the imperfect root, first by one sign and then by the other, we shall always get a pair of unequal factors which will produce the polynomial; and these unequal factors may be considered as the equal but imperfect factors corrected.*

* This decomposition of a polynomial into two factors, which are themselves the sum and difference of the same two quantities, has been recently applied by the author to some improvements in the Analysis of Equations. See "The General Principles of Analysis," Part I. published by S. Maynard, Earl's Court, London; and C. C. Spiller, Holborn Hill. A portion of improvements will be briefly touched upon in a subsequent part of the present work.

What is here called the imperfect root of a polynomial is not a fixed and unchangeable expression like the true root of a polynomial. When a perfect root does not exist, there are innumerable expressions which have claim to be called imperfect roots, each of which, when increased and diminished by the proper correction, furnishes a pair of unequal factors whose product is the proposed polynomial. An example will sufficiently explain our meaning.

In the first mode of working this example, we have proceeded in the usual way to the end; but in the second process, we have departed from the beaten track at the last step, having determined the final term of the quotient, or imperfect root, so as to render the remainder free of x, without regard to the entrance of x2. In the ordinary way of proceeding, every remainder is rendered free of the leading quantity, or power, in the preceding remainder; and this is only in obedience to the prescribed rule: but when a complete root does not exist, we need not restrict ourselves to the rule in this particular: we may render our final remainder free of any power we please; and a departure from the ordinary course will often enable us to decompose an expression into a pair of simpler unequal factors than we could otherwise get. Thus, the two pair of factors of x-4x3+ 7x-4x+1, furnished by the above incomplete roots and remainders, are: