56 X3 seventh, X a266=28a26. 6 28 X 2 eighth, Xabi=8ab7. 7 8 X 1 minth, Х 88= 68. 8 And thus we have, (a+b)8=a+sa+b+28a%b2+56a563+70 a484+5605+28a266 +8ab7+68. 288. From this example and the foregoing Table the whole number of terms will evidently be one more than the index of the given power; after having calculated therefore as many terms as there are units in the index, of the given power, we may immediately proceed to the last term. And in like manner it may be observed, that when the number of terms in the resulting quantity is even, the coefficients of the two middle terms is the same ; and that in all cases the coefficients increase as far as the middle term, and then decrease precisely in the same manner until we come to the last term. By attending to this law of the coefficients, it will be necessary to calculate them only as far as the middle term, and then set down the rest in an inverted order. Thus in the above example, the middle term is 70a4b", and we have, The first four coefficients, 1, 8, 28, 56. The last four 56, 28, 8, 1. 289. But we are not yet arrived at the most general form in which this Rule may be exhibited. Suppose it was required to raise the binomial a+b to any power denoted by the num Proceeding with n as we have done with the several indices in the preceding examples; it appears that, The first term would be an. nan-16. ber (n). . n(n-1)01-26. 2 n The third, . The fourth, 2X3 The fifth, (n-1)x(–3)^(n-3) – . 2 X3 X4 -565, The sixth, 2 X3 X4 X5 The last, b. n n(n-1)×(n—2) an–2697 n(n-1)05-22 Or, (a+b)n=atnam–16+(n-1) an=24+ 2 2.3.4 an-474 &c. +6n. By the same process, (a−b)=an. (a−b)=an_nan-1+ n(n-1) X(1-2) an_363 + 2 2.3 n(n-1)×(1-2)(n-3) lan_464—&c.; the signs of the terms 2.3.3. being alternately + and -1; and the sign of the last term is + or -1, according as n is even or odd; we have the last term in the former case, ton, and in the latter-b". This general and compendious method of raising a binomial quantity to any given power, is called from the name of its celebrated inventor, Sir Isaac Newton's " Binomial Theorem." The demonstration of this Theorem, with its application to the finding the powers and roots of compound quantities, forms the subject of another Chapter. Its present use will appear from the following Example. Ex. 2. Required the fifth power of x + saj?. Substituting these quantities for a, b, n, in the foregoing general formula, it appears, that =x. (**) term, 2nd, (nan-16) . is 5 X (22)4 X 3yo . =15x®y. 4 n(n-1) 3d, an2 5X5X(x)ex(3y?)a=90x®yt . in(n-1)*(n -2) 4 3 4th, -an--383 is 5 X-X-X 2.3 (34*): =270x*y®. -i)(n—2)(n-3) 4 3 2 5th, -an-464 is 5 X 2.3.4 X-X-X 2012 X(3ya) =405x+y®. Last, is (3y2) =243y So that x2+372)5=° +15x®y2 +S0x®y4 +270x°y® +105x*y* 10 +243y 290. By means of this Theorem, we are enabled to raise a trinomial, or quadrinomial quantity to any power, without the process of actual multiplication. Ex. 3. Required the square of a+b+c? Here, including a +b in a parenthesis (a+b), and considering it as one quantity, we should have (a+b+c)=f(a+b) ?2; and comparing them with the general formula ; The first } (an) (61) we have, (an)=(a+b)a=a+2ab +62 (nan-ib)=2(a+b) Xc=2ac+2bc (bn)=( Hence, (a+b+c)-(a+b)+2(a+b)xc+c=a?+2ab+ba + 2act-2bc+ca. Ex. 4. Required the seventh power of a--b. Ans. 72928 +2916xby _4860x*ya + 4320x?y9+2160x®y4 +576 xy*+-64y. Ex. 6. Required the square of x+y+3z. Ans. 22 +-2xy+ya+6x2+6yart-9z. Ex. 7. Required the fifth power of 1+2x. Ans. 1 + 10x +-40x2 +80x3 +80x4 +32x". Ex. 8. Required the cube of x-2xy+ya. Ans. 6–6xy+1 5x4y—20x?y?+15xy* -6xy+ye. § II. EVOLUTION OF ALGEBRAIC QUANTITIES. 291. The quantity which has been raised to any power is called the root of that power; thus the mth root of a power, is that quantity which we must continually multiply into itself, till the number of factors be equal to m, m being a positive whole number, in order to produce the power proposed. We may conclude from this definition, and from the Articles in the preceding section, 292. That the mth root of a quantity such as apm, pm being a multiple of p, is obtained by dividing the exponent pin of this quantity, by the index of the required root. Thus the mth root of apm=a 을 =aP; the square root of aø=? =a?, and the cube root of ac-a3=a?. 293. Also that the mth root of a product such as aa mb3n, is equal to the math root of cach of its factors multiplied together. Thus, the mih root of u~my3m is =the mth root of a2m x the 2m mth root of 6311=a Xbm=ab3. 294. And that the mth root of a fraction such as is equal to bm' the mth rooi of the numerator divided by the mth root of its denominator. pm 6 a 295. The square, the fourth root, or any even root of an affirmative quantity may be either + or -1. Thus the square root of a:=a or -u; for tax tarta', and -ax-art a>. In fact, the 2inth root of a2m is equal to ta or -a ; for (+a)am=(+a) Xam=a?m( Art. 280). 296. Any odd root of a quantity, will have the same sign as the quantity itself. Thus the (2m+1)th root of +azm is equal to +u ; for (+a) 2m +1 is equal to + a2mt1 (Art. 281). 297. Evolution, or the rule for extracting the root of an y algebraic Quantity whatever, is divided into the four following Cases. m+1 CASE I. To find any root of a simple algebraic Quantity. RULE. 298. Extract the root of the coefficient for the numeral part, and the root of the quantity subjoined to it for the literal part, by the methods pointed out in the above propositions ; then, these, joined together, will be the root required. Ex. 1. It is required to find the square root of zo. Here, (Art. 295), the square root of x+=+vx'=(Art. 292) +r+=+x Ex. 2. Required the cube root of —27x?q. Here, (Art. 296), the cube root of -27x'= -127x'a®= (Art. $93) –27x/cox/a=3XrXa=-3ax. a x2 Ex. 3. Required the square root of 6c Here, the square root of a?r2=Vax Vr=ar, and the square root of boc=box vco=bc ; ..(Ărt. 294, 995), + is the root required. be Ex. 5. It is required to find the square root of 6ta*r* ? Ans. 8nx, or — Sax”. Ex. 6. It is required to find the cube root of 729 w®ze12. Ans. 9aox. Ex. 7. Required the fourth root of 256a468. Ans. 4ab”, or-4ab2. Ex. 8. Required the fifth root of 32aRx10. Ans. 2ax“. .729a676 3ab Ex. 9. Required the sixth root of 4096x13 Ang. +42 Ans. xya ab 6axa Ex. 10. Required the ninth root of zvoy18 abo Ex. 11. Required the square root of 36aRxc4 4.x'ya Ex. 12. Required the cube root of 6423 272663 4x Ans. 3a21 CASE II. To extract the square root of a compound Quantity. RULE. 299. Observe in what manner the terms of the root may be derived from those of the power ; and arrange the terms accordingly; then set the root of the first term in the quotient ; subtract the square of the root, thus found, from the first term, and bring down the next two terms to the remainder for a dividend. Divide the dividend, thus found, by double that part of the root already determined, and set down the result both in the quotient and divisor. Multiply the divisor, so increased, by the term of the root last placed in the quotient, and subtract the product from the dividend, and to the remainder bring down as many terms as are necessary for a dividend, and continue the operation as before. Ex. 1. Required the square root of a2 +2ab + ba ? a + 2ab +62 aa (att 2a +-6 | 2ab +62 2ab +62 On comparing a tb with aa + 2ab+ba, we observe that the first term of the power (aa) is the square of the first term of the root (a). Put a therefore for the first term of the root, square it, and subtract that square from the first term of the power. Bring down the other two terms 2ab +62, and double the first term (a) of the root; set down 2a, and having divi. ded the first term of the remainder (2ab) by it, we have b, the other term of the root; and since 2ab +69=(22+6) Xb, if to, 2a the term b is added, and this sum multiplied by b, the result is 2ab +62; which being subtracted from the terms brought down, nothing remains. |