360. A Binomial with Coefficients. If the terms of the given binomial have coefficients or exponents, we may obtain any power of it by means of the binomial formula, For this purpose, each term must be raised to its proper power denoted by the exponents in the binomial formula. 7. Find the fourth power of 2x+3a. For convenience, let us substitute y for 2x and b for 3a. Then (y+b)*=y*+4y3b+6y2b2+4yb3+64. Restoring the values of y and b, the first term will be (2x)=16x4, the third term will be 6(2x)2 × (3a)2=216x2a2, the second term will be 4(2x)3×3a=96x3a, the fourth term will be the fifth term will be 4(2x) × (3a)3=216xa3, (3a)*=81a*. Therefore (2x+3α)*=16x*+96x3a+216x2a2+216xa3+81a*. It is recommended to write the three factors of each term in a vertical column, and then perform the multiplication as indicated below: Coefficients, Powers of 3a, 1 + 4 + 6 + 4 +1 16x+8x3 + 4x2 + 2x + 1 1+ 3a + 9a2 + 27a3 +81a* (2x+3α)*=16x2+96x3a+216x2a2+216xa3+81aa. 8. Find the fifth power of 2ax-36. (2ax-36)5=32a3x3 —240a*x*b+720a3x3b2 — 1080a2x2b3+810axba—24363. 9. Find the fourth power of 2x+5a2. Ans. 16x+160x3a2+600x2a1 +1000xα +625a3. 10. Find the fourth power of x3+4y2. 11. Find the sixth power of a3+3ab. Ans. a18+18a16b+135a14b2+540a123 +1215a10+1458a8b5+729a*b*. 12. Find the seventh power of 2a-3b. Ans. 128a-1344a b+6048a5b2-15120a13 +22680a3b1-20412a2b5+10206al-21876'. 13. Find the fifth power of 5a2-4x2y. Powers and Roots of Polynomials. 361. If it is required to raise a polynomial to any power, we may, by substituting other letters, reduce it to the form of a binomial. We obtain the power of this binomial by the general formula; then, restoring the original letters, and performing the operations indicated, we obtain the required power of the proposed polynomial. or Ex. 1. Let it be required to raise a+b+c to the third power. If we put b+c=m, we shall have (a+b+c)3=(a+m)3=a3+3a2m+3am2+m3, a3+3a2(b+c)+3a(b+c)2+(b+c)3. Developing the powers of the binomial b+c, and performing the operations indicated, we obtain (a+b+c)3=a3+3a2b+3a2c+3ab2+6abc+3ac2+b3+3b2c +3bc2+c3. Ex. 2. Find the fifth power of x+a+b. Ans. a8—4a7b+10ab2—16a3b3+19a1b1 Ex. 4. Find the fifth power of 1+2x+3x2. Ans. a6+6a5b+6a5c+15a+b2+30a1bc+15a2c2+20a3b3 +60a3b2c+60a3bc2+20a3c3 +15a2b++ 60a2b3c+ 90a2b2c2+60a2bc3 +15a2c++ 6ab5+30ab+c+60ab3c2 +60ab2c3+30abc1+6ac3+b+6b3c+15b*c2+2063c3 +15b2c1+ 6bc3+c®. 362. The binomial theorem will inform us how to extract any root of a polynomial. We know that the mth power of x+a is xm+maxm-1+other terms. The first term of the root is, therefore, the mth root of the first term of the polynomial. Also the second term of the root may be found by dividing the second term of the polynomial by mxm-1; that is, the first term of the root raised to the next inferior power, and multiplied by the exponent of the given power. Hence, for extracting "any root of a polynomial, we have the following RULE. Arrange the terms according to the powers of one of the letters, and take the mth root of the first term for the first term of the required root. Subtract the mth power of this term of the root from the given polynomial, and divide the first term of the remainder by m times the (m-1) power of this root; the quotient will be the second term of the root. Subtract the mth power of the terms already found from the given polynomial, and, using the same divisor, proceed in like manner to find the remaining terms of the root. Ex. 1. Find the fourth root of 32a3) 16a-96a3x+216a2x2-216ax3+81x4. 16a-96a3x+216a2x2-216ax3+81x4 (2a-3x 16a1 - 96а3х 16a-96a3x+216a2x2-216ax3+81x4. Here we take the fourth root of 16a4, which is 2a, for the first term of the required root, subtract its fourth power, and bring down the first term of the remainder, -96a3x. For a divisor, we raise the first term of the root to the third power and multiply it by 4, making 32a3. Dividing, we obtain —3x for the second term of the root. The quantity 2a-3x, being raised to the fourth power, is found to be equal to the proposed polynomial. Ex. 2. Find the fifth root of 80x3+32x5-80x-40x2+10x-1. Ans. 2x-1. Ex. 3. Find the fourth root of 336x+81x8-216x7-56x+16-224x3+64x. Ans. 3x2-2x-2. 363. To extract any Root of a Number.-The preceding method may be applied to the extraction of any root of a number. Let n be the index of the root, n being any whole number. For a reason similar to that given for the square and cube roots, we must first divide the number into periods of n figures each, beginning at the right. The left-hand period may contain less than n figures. Then the first figure of the required root will be the nth root of the greatest nth power contained in the first period on the left. If we subtract the nth power of this root from the given number, and divide the remainder by n times the (n-1)th power of the first figure, regarding its local value, the quotient will be the second figure of the root, or possibly a figure too large. The result may be tested by raising the whole root now found to the nth power; and if there are other figures they may be found in the same manner. In the extraction of the nth root of an integer, if there is still a remainder after we have obtained the units' figure of the root, it indicates that the proposed number has not an exact nth root. We may, if we please, proceed with the approximation to any desired extent by annexing any number of periods of n ciphers each, and continuing the operation. We thus obtain a decimal part to be added to the integral part already found. So, also, if a decimal number has no exact nth root, we may annex ciphers, and proceed with the approximation to any desired extent, dividing the number into periods commencing with the decimal point. Ex. 1. Find the fifth root of 33554432. 335.54432 (32 5.34-405) 925 325- 33554432. Ex. 2. Find the fifth root of 4984209207. Ex. 3. Find the fifth root of 10. Ex. 4. Find the fifth root of ...... Ans. .922 364. When the index of the required root is composed of two factors, we may obtain the root required by the successive extraction of simpler roots, Art. 217. For the mnth root of any number is equal to the mth root of the nth root of that number. Thus we may obtain the fourth root by extracting the square root of the square root. We may obtain the sixth root by extracting the cube root of the square root, or the square root of the cube root. It is, however, best to extract the roots of the lowest degrees first, because the operation is less laborious. We may obtain the eighth root by extracting the square root three times successively. We may obtain the ninth root by extracting the cube root twice successively. Ex. 1. Find the fourth root of 6a2b2+a2-4a3b-4ab3+b*. ̧ Ex. 2. Find the sixth root of 6a5b+15a2b2+a+20a3b3+15a2b*+b+6ab3. Ex. 3. Find the eighth root of 1024x1y+1792xoy2+256x+1120x*y*+1792x*y3+448x3y3 +y3+112x2y®+16xy1. |