second term of the root and subtract the product from the first remainder. 3rd. Again find how often the trial divisor is contained in the first term of the remainder, this will give the third term of the root. Then form a complete divisor as before, by adding together three times the square of the first and second terms, plus three times the product of the first and second terms by the third, plus the square of the third. Multiply these by the third term of the root and subtract the product from the last remainder. 4th. Continue this process till all the terms of the root are found. NOTE. The remainder in each case, is all the terms left after each subtraction. 1. Find the cube root of x-6x3+12x1+3a2x1—8x3-12a2x3 +12a2x2+3a2x2-6x+a. x6—6x3+12x1+3a2x1—8x3-12a2x3±12a2x2+3a2x2—6aax+ao 3x4—12x3+12x2+3a2x2—6a2x+a1)+3a2x1—12a2x2+12a2x2 To bring the work within the page, the last remainder and subtrahend are each written in two lines. +3a4x2-6a1x+a®. +3a2x2-12a2x2+12a2x2 +3a2x2-6x+x+ao. We first extract the cube root of x6, which gives x2 for the first term of the required root. Then 3 times the square of this, 3(x2)2=3x4, constitutes the trial divisor for finding the remaining terms. To find the second term of the root we divide 3x4 into -6x, the first term of the remainder, which gives —2x, the second term of the root. We then form the complete divisor by adding together 3(x2)2+3(x-2x)+(-2x)2=3x2-6x3+4x2. Multiplying this by the second term, -2x, and subtracting the product from the first remainder, the first term of the second remainder is +3a2x4, which, divided by the trial divisor, gives +a2, for the third term of the root. We next find the complete divisor by adding together 3(x2-2x)2+3(x2 —2x)a2 + (a2)2=3x1 —12x3 +12x2+3a2x2-6a2x+a. Multiplying this by a2 and subtracting, there is no remainder; hence the root obtained is exact. Find the cube root 2. Of a3+24a2b+192ab2+512b3. 3. Of 8a3-84a2x+294ax2-343x3. 4. Of a6-6a5+15a4-20a3+15a2—6a+1. Ans. a+8b. Ans. 2a-7x. Ans. a2—2a+1. 5. Of x-9x+39x 99x3-156x2-144x+64. Ans. x2-3x+4. 6. Of (a+1)6nx3-бca2(a+1)11x2+12c2a2(a+1)21x—8c3a3r Ans. (a+1)2-2ca". 7. Find the first three terms of the cube root of 1-x. XC 3 9 IV. EXTRACTION OF THE FOURTH ROOT, ROOT, NTH ROOT, & C. -&c. SIXTH ART. 192. The fourth root of a number is one of four equal factors, into which the number may be resolved; and in general, the nth root of a number is one of the n equal factors into which the number may be resolved. When the degree of the root to be extracted is a multiple of two or more numbers, as 4, 6, &c., the root can be obtained by extracting the roots of more simple degrees. To explain this we remark, that (a3)1=a3Xa3Xa3×a3=a3+3+3+3=—a3×4=α12 and in general Hence, the nth power of the mth power of a number, is equal to the mnth power of the number. Reciprocally, the mnth root of a number, is equal to the nth root of the mth root of that number; that is raising both members to the nth power, we have "Ja=a'n; and by raising both members of the last equation to the ma power, a=a'mn; extracting the mnth root of both members, It may be proved similarly, that am/Ja. From this it follows that a=√√√ā; and ƒã=√√√ãos in like manner Wa√√√√√a, and so on. 13 9. Find the 4th root of a++-4a3bx+6a2b2x2+4ab3x3+b^xa. Ans. a+bx. x4, 4x2 x2 4y2+6. x2 X4 11. Find the 4th root of x-4x+10x2-16x2+19-16+10 12. -6 Find the 6th root of a+ -20. Ans. a - a ART. 193. It has been shown already (Arts. 182, 183,) that the square root of a monomial, or a polynomial, may be preceded either by the sign+, or; we shall now explain the law in regard to the roots generally. If we take the successive powers of +a, we have +a, +a2, +2, the successive powers of —a, are +a2, —a3, +a1, From this we see that every even power is positive, and that an odd power has the same sign as the root. In general, let n be any whole number, then every power of an even degree, as 2n, may be considered as the nth power of the square, that is, a2n=(a2)". Hence, every power of an even degree is essentially positive, whether the quantity itself be positive or negative. Thus, (3a)+81a1 ;(±2b2)=64612. Again, as every power of an odd degree (2n+1) is the product of a power of an even degree, 2n, by the first power, it follows, that every power of an uneven degree of a monomial has the same sign as the monomial itself. Thus, (+2a)3+8a3, (—2a)3——8a3. Hence, it is evident, 1st. That every odd root of a monomial must have the same sign as the monomial itself. Thus, +8a2a, 3/8a3=-2a, &-32a10-2a2. 2nd. That an even root of a positive monomial may be either positive or negative. Thus, £/81a4b8=±3ab2, §/64a12=±2a2. 3rd. That every even root of a negative monomial is impossible ; since no quantity raised to a power of an even degree can give a negative result. Thus, -a, -6, -c, are symbols of operations which cannot be performed. They are imaginary expressions like a, b. (Art. 182.) TO EXTRACT THE NTH ROOT OF A MONOMIAL. ART. 194. In raising any monomial to the nth power according to the rule, Art. 172, it is obvious that the process consists in raising the numeral coëfficient to the nth power, and multiplying the exponent of each letter by n, thus, (2a2b4)3=2a2b4×2a2b4 X2a2b4=23a2×3b1×3=8a6b12. Hence, conversely, to find the nth root of a monomial, Extract the nth root of the coëfficient, and divide the exponent of each letter by n. REMARK. In the following examples, the pupil is expected to find the root of the numeral coëfficient by inspection, as we have given no rules for extracting the 5th, 7th, &c., roots of numbers. Indeed, in the present state of science such rules are useless, for when the operations are required they are readily performed by means of Logarithms. 1. Find the 5th root of -32a5x1o. 7. Find the mth root of abmcmq. Ans. -2ax2. Ans. 3bc3. Ans. 2xy2. Ans. 3ab2. Ans. -2xz2. Ans. +2bz3. Ans. Ja.bc. Ans. a4bc2. NOTE. These quantities are generally called surds by English writers; while the French more properly term them radicals, from the Latin word radix, a root, because they express the roots of quantities. The Germans also distinguish them by a synonymous term, wurzel grössen, (root quantities). ART. 195. A rational quantity is one either not affected by the radical sign, or of which the root indicated can be exactly ascertained; thus, 2, a, √4, and 2/8 are rational quantities. A radical quantity is one of which the root indicated cannot be exactly expressed in numbers; thus, 5 is a radical; its value is 2.23606797 nearly. Radicals are frequently called irrational quantities, or surds. ART. 196. From Art. 193 it is evident that when a monomial is a perfect power of the nth degree, its numeral coëfficient is a perfect power of that degree, and the exponent of each letter is divisible by n. Thus 4a2 is a perfect square, and 8a6 is a perfect cube; but 6a3 is not a perfect square, because 6 is not a perfect square, and 3 is not divisible by 2; also, 8a4 is not a perfect cube, for, although 8 is a perfect cube, the exponent 4 is not divisible by 3. In extracting any root, when the exact division of the exponent cannot be performed, it may be indicated by writing the divisor under it in the form of a fraction. Thus, as may be written a, and a may be written a3; and in general the n root of the mth root of any quantity, is expressed either by "/aTM, m or an. Since a is the same as a1 (Art. 19), the square root of a may be expressed thus, at; the cube root thus, as; and the nth root thus, a Hence, the following expressions are to be considered equivalent: From this we see, that the numerator of the fractional exponent denotes the power of the quantity, and the denominator the root of that power to be extracted. |