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To CUBE ANY MULTINOMIAL.
176. Consider a trinomial : (x+y+z) 8 = [x + (y+z)]
= 28+3.co(y+z) + 3x(y+z)2 + (y+z) 8
=x8+48 + 8 + 3x'(y+z) + 3y(+)+3z2(x+y)+ 6xyz. It can be readily shown that the law of formation holds for any multinomial, hence we obtain the following rule:
RULE. To cube a multinomial : take the cube of each term, three times the square of each term by every other term, and six times the product of every three different terms. The signs are determined by the law of signs for multiplication.
EXAMPLES XXII. d.
Write the cube of each of the following expressions : 1. 1+x+22 2. 1+x-22.
3. 1-2x+3x2. 4. a+bx+x2. 5. 2a+bx— cx2. 6. 3x + 2x2 — 23.
2c2 7. î+ g -28. 8. 1+3+x2+28. 9. 2-3x+x2+2x3.
177. THE root of any proposed expression is that quantity which being multiplied by itself the requisite number of times produces the given expression. [Art. 15.]
The operation of finding the root is called Evolution: it is the reverse of Involution.
178. By the Rule of Signs we see that
(1) any even root of a positive quantity may be either positive or negative;
(2) no negative quantity can have an even root;
(3) every odd root of a quantity has the same sign as the quantity itself.
NOTE. It is especially worthy of notice that every positive quantity has two square roots equal in magnitude, but opposite in sign. Example.
(4) V(81x12)=3.03. 179. Hence we obtain a general rule for extracting any proposed root of a simple expression:
RULE. (1) Find the root of the coefficient by Arithmetic, and prefix the proper sign found by Art. 35.
(2) Divide the exponent of every factor of the expression by the index of the proposed root. Examples. (1) ( - 64x6)= – 4.x2. (2) (16a8)=2a.
5c2• It will be seen that in the last case we operate separately upon the numerator and the denominator.
EXAMPLES XXIII. a. Write down the square root of each of the following expressions : 1. 4a264. 2. 9.267 3. 25x4y6. 4. 16a46oc. 5. 81a88, 6. 100.08. 7. 220616c4. 8. a8b2c12.
81.210,18 Write down the cube root of each of the following expressions : 17. 270°63c".
- 8a1239, 19. 64xby3z12 20. - 343a12718
64763 Write down the value of each of the following expressions : 25. Var12). 26. V(x14,21). 27. (32w5y10). 28. \(729a186). 29. \(256a82/4). 30. V-2-10y15). 128 10 /a30,50
To FIND THE SQUARE ROOT OF A COMPOUND EXPRESSION.
180. Since the square of a +b is a2 + 2ab+ , we have to discover a process by which a and b, the terms of the root, can be found when a2 + 2ab + b2 is given.
The first term, a, is the square root of aạ.
Arrange the terms according to powers of one letter a. The first term is a’, and its square root is a. Set this down as the first term of the required root. Subtract az from the given expression and the remainder is 2ab +62 or (2a+b) x b.
Thus, b, the second term of the root, will be the quotient when the remainder is divided by 2a+b.
This divisor consists of two terms:
a2 + 2ab + b2 (a +6
Example 1. Find the square root of 9x2 – 42xy +49y.
9x2 – 42xy +49y2 (3x – 7y
- 42xy + 49y?
Explanation. The square root of 9x2 is 3x, and this is the first term of the root.
By doubling this we obtain 6x, which is the first term of the divisor. Divide - 42xy, the first term of the remainder, by 6x and we get - Ty, the new term in the root, which has to be annexed both to the root and divisor. Next multiply the complete divisor by – 7y and subtract the result from the first remainder. There is now no remainder and the root has been found.
The process can be extended so as to find the square root of any multinomial. The first two terms of the root will be obtained as before. When we have brought down the second remainder, the first part of the new divisor is obtained by doubling the terms of the root already found. We then divide the first term of the remainder by the first term of the new divisor, and set down the result as the next term in the root and in the divisor. We next multiply the complete divisor by the last term of the root and subtract the product from the last remainder. If there is now no remainder the root has been found; if there is a remainder we continue the process.
Example 2. Find the square root of
25xoa– 12xa3 + 16x4 + 4a* – 24x'a. Rearrange in descending powers of æ.
16.c* – 24x'a + 5x+ao – 12xa3 + 4a* (4x2 – 3xa + 2a2
- 24x’a + 9.c?a?
16x?a? – 12xa + 4a4 16x?a? – 12.xa3+424
Explanation. When we have obtained two terms in the root, 4x2 – 3xa, we have a remainder
16x?a? – 12xa3+4a4. Double the terms of the root already found and place the result, 8x2 – 6.xa, as the first part of the divisor. Divide 16xưa”, the first term of the remainder, by 8xạ, the first term of the divisor; we get + 2a” which we annex both to the root and divisor. Now multiply the complete divisor by 2a2 and subtract. There is no remainder and the root is found.
EXAMPLES XXIII. b.
Find the square root of each of the following expressions : 1. 2? + 4.xy+4y?
2. 9a2 + 12ab +462. 3. 22 – 10xy+25y2.
4. 4.°-12xy +9y. 5. 81.22 +182y+ya.
6. 25x2 – 30xy +9y2. 7. 24-2.cya+y
8. 1-20+ al. 9. a4-203+3a” – 2a +1. 10. 4x4 – 12x3 +29.x2 – 30x+25. 11. 9.44 – 12.x3 – 2x2 + 4x +1. 12. x4 – 4.x3 + 6x2 - 4x +1. 13. 494 +4a3 – 7a? – 4a +4. 14. 1 - 10x+27x2 – 10x3 + x4, 15. 4.x2 +9y2 + 2522 +12.cy - 30yz-20.wz. 16. 16.06 +1627 – 428 — 4x? + x10. 17. GO – 22x4+34x3 + 121x2 – 374x + 289. 18. 23.x4 – 30ax2 +49a-x2 – 24a’x+16a4. 19. 4.24+4x=yja – 12x2z2+y4 – 6y2z2+924. 20. Cab?c – 4abc+a-b2 +4a2c2 +962c2 – 12abc-. 21. - 662c2 +9c4 +64 – 12c?a+404+ 4a2b2. 22. 4.14+9y4+ 13x-ya-6xy) - 4.0-%y. 23. 67x2+49+9x4 – 70x - 30x3. 24. 1 - 4x+1022 – 20.x3 + 25x4 – 24.x+ 16x6. 25. Cacxó +462x4 + a2x10 +962 – 12bcx? — 4abx7.
181. When the expression whose root is required contains fractional terms, we may proceed as before, the fractional part of the work being performed by the rules explained in Chap. xv.
182. There is one important point to be observed when an expression contains powers of a certain letter and also powers of its reciprocal. Thus in the expression
5 2x + +4 +23+-+7x2 +. the order of descending powers is
5 1 23 +7.x2+2x+4+
and the numerical quantity 4 stands between x and het
The reason for this arrangement will appear in Chap. XXXI.