If the square of a+b+c had been developed, the operation would have been essentially the same, as will be seen by inspecting the operation. a2 +2ab+b2 + 2ac + 2bc + c2 || a+b+c 2a+b2ab+b2 2ab+b2 2a+2b+c+2ac+2bc + c2 +2ac+2bc+c2 If the root have four terms, or any number of terms whatever, the same reasoning will hold true; for we always have (41), (a+b+c+d+e.... +i+k)2=(a+b+c+d+e+ . . . . i)2 +2(a+b+c+d+e+ . . . . i)k+k2; and hence it will appear, that whatever be the number of terms in the root, we shall always find the next term by doubling the root already found for a divisor. EXAMPLES.. 1. Extract the square root of ao — 4a3x+8a3x2 + 4a2x2. a-4a3x+8a3x3 +4a2x2 || a3 — 2a3x-2ax2 2. Extract the square root of 4a® +4a3x+α2x2—‚1ax3 4a+4a5x+15a2x2—‚1ax3+ ̧2+x® ||2a3+a2x—‡ax2+}x2 64 - a2x2 + 15 % a2x2 4. Extract the square root of 4x1—16x3 +24x2—16x+4. Ans. 2x2-4x+2. 5. Extract the square root of 16x*+24x3+89x2+60x+100. Ans. 4x2+3x+10. 6. Extract the square root of 4x-16x+8x2+16x+4. 7. Extract the square root of 4a2b2-2ab2c+4b2c2+4a2bc -abc2 + a2c2. Ans. 2ab-bc+ac. 8. Extract the square root of 16xa—16x2+28x2-12x+9. Ans. 4x-2x+3. 9. Extract the square root of 16x2" 162"-1 + 4x2--2 +24x-12x+9. Ans. 4x-2x-1+3. EXTRACTION OF THE CUBE ROOT OF POLYNOMIALS. RULE. (47.) 1. Arrange the polynomial with reference to the powers of some letter. Then find the cube root of the first term, which must always admit of an exact root, and place the root, thus found, in the quotient, and subtract its cube root from the first term. 2. Bring down the next three terms for a dividend, and take three times the square of the root already found for a divisor. Divide the first term of the dividend by this divisor; the quotient will be another term of the root. 3. Complete the divisor by adding to it three times the product of this last term of the root by those which precede it, and the square of this last term. Multiply the divisor, thus increased, by the last term of the root, and subtract the product, and to the remainder bring down the next three terms, and proceed in the same manner, till all the terms are brought down, and the whole root obtained. The reason for this rule will be seen by a reference to the third power of a+b, which is a3 +3a2b+3ab? +b3. a3 +3a2b+3ab2 + b2 || a + b The first term of the root is, evidently, the root of the first term of the power; and since the second term of the power is 3ab, it is evident that we shall find the second term of the root by dividing this term by 3a2, which is three times the square of the root already found. The two last terms of the power 3ab2+b3, or (3ab + b2)b, being three times the product of the two terms of the root, and the square of the last multiplied by this last term, it will be seen why this product, and the square of the last, should be added to complete the divisor. The same reasoning will apply when the root is a trinomial; since a +b can first be obtained, and then c obtained from a + b, in the same manner as b was obtained from a. EXAMPLE 1. Root. —6a3x + 15a1x2 — 20a3x3 + 15a3x* — 6ax3 +x3 ]] a3 ‚—2ax +2a 3a* - 12a3x + 15a2x2. 3a4x2 6ax3+x* | 3a2x2 12a3x+15a2x2 — 6ax 12a3x3 +15a2x1 — 6ax 3(a1)2 = 3a1, first partial divisor. 3a1×-2ax=-6a3x, triple product of the root by (2ax)2 = 4a2x2, square of — 2ax. 3a1 — 6a3x + 4a2x2, complete divisor. 2ax. 3(a2-2ax)2 = 3a1-12a3x+12a2x2, 2d partial divisor. 3(a2 — 2ax) x2 = 3a2x2 — 6ax3, triple product of the root by x2. (x2)2= x1, square of the last term. 3a1- 12a3x+15a2x2-бax+x', complete divisor. |