coefficient of the term to the power in question, and multiplying the exponents of the letters of the term by the exponent of the power in question. Thus, (a3)2 = a3 × a3· 3+3 a3 = a3 × 2 (am)n = am × am xam...... ton factors = am+m+m+... ton terms = amn = 33. Development of the third, fourth, and fifth powers of a + b. We know (Art, 26, I.) that (a + b)2 = a2 + 2 ab + b2. •'. (a + b)3 = (a2 + 2 ab + b2) (a + b) = a3 + 3 a2b + 3 ab2 + b3; also, (a + b)* (a3 + 3 ab2 + ab2 + b3) (a + b) 4a3b + 6a2b2 + 4 ab3 + b1; = = a* + and (a + b)5 = (a1 + 4 a3b + 6 a2b2 + 4 ab3 + b3) (a + b) = a* + 5 a1b + 10 a3b2 + 10 a2b3 + 5 ab1 + b3. The following law of the formation of the terms is evident: Law of Formation of Terms. 1. The first term contains a raised to the given power, and the power of a decreases by unity in each successive term, while the power of b (which first appears in the second term) increases by unity in each successive term, till it reaches the power of the given quantity. 2. The first coefficient is unity, and the coefficient of any term is found by multiplying the previous coefficient by the exponent of a in the previous term, and dividing the product by the number of terms hitherto developed. Ex. 1. (2x + 3y)3 = (2x)3 + 3 (2x)2 (3 y) + 3 (2 x) (3 y)2 + (3 y)3 = 8 x2 + 36 x y + 54 xy + 27 y3. = = (a + b + c)3 (a + b)3 + + 3 ab2 + b3) Ex. 2. (a + b + c)3 3 (a + b)2 c + 3 (a + b) c2 + c23 + 3 (a2 + 2 ab + b2) c + 3 (a + b) c2 + c3 + 3a2b+ 3 a2c + 3 ab3 + 3b2c + 3 ac2 + 3 bc2 + 6 abc. = a3 + b3 + c3 (In the following examples the above law may be assumed as generally true.) Ex. X. 1. Find the values of (a2b3)2, ( — 3 ab3)3, (2·a2bc)1, ( − x3yz2)3, Expand 2. (a + 3b), (2 a + b), (a - b), (3 a – 4b)3. 3. (2 m + 1), (5 x + 4. (x2 + x + 1)2, (3 a (3 a + 3b+3c)3. 5. (1 + x 2)3, (3 a − 4 c), (ab)3. − b + 4 c - d)2, (a + 2 b - c)2 x2)3, (ax + by + cz)3, { (a + b) x − (c + d) y } : 6. (1 + x)7, (1 + x + x2), (a + bx + cx2)3. 7. (ax – by)3, (3 x + y)3 (3x − y)3, (x2 + xy + y2)3 (x − y)3. 8. (x2 + x2y2 + y1)2 (x + y)2 (x {(a − Simplify y)2, (a - b)2 + 4 34. Evolution is the operation by which we obtain the roots of quantities. Since the square or second power of a3 is (a3)2 or a3, we call (Art. 17) a3 the square root or the second root of ao. And so, since the cube or third power of a2 is (a2)3 or ao, we call a2 the cube root or the third root of ao. So, generally, since the nth power of am call am the nth root of amn. = (am)n = Thus, we have a = a3, JaR = a2, an = am. amn, we Hence, in the case of quantities consisting of a single letter with a given exponent, when the given exponent contains as a factor the number indicating the root, we must divide the given exponent by this number, the quotient being the exponent of the root. (We shall see farther on that this rule holds when the given exponent is not so divisible, the root in this case being called a surd). 35. Since the product of an even number of negative factors must give a positive result, and the product of an odd number of negative factors a negative result, it follows that— I. When the root is indicated by an even number— 1. The root of a positive quantity may be written either with a + or sign. Thus, 9 a =3a, 166 = ±2b. 2. The root of a negative quantity is impossible. Thus, a, ab1, &c., are impossible quantities. II. When the root is indicated by an odd number, the root has always the sign of the given quantity. (It may be remarked that the theory of impossible quantities forms an important branch of Algebra, which the student cannot yet enter upon. According to that theory, all quantities have as many roots as the number indicating the root.) Square Root. 36. We shall now develop the method of finding the square root of a given quantity. Ex. 1. Find the square root of a2 + 2 ab + b2. = (a + b)2. We know that a2 + 2 ab + b2 Hence, a+b is the square root of a2 + 2 ab + b2 or a2 + (2 a + b) b. Now, it is evident that the first term a of the root is the square root of the first term a of the given quantity; and if this term be subtracted, there remains 2 ab + b2, from which to determine b the second term of the root. Now, b is contained in 2 ab + b2 or (2 a + b)b exactly (2 a + b) times. Hence it follows that the second term of the root is found by dividing the remainder by twice the first term of the root, and, if we wish to arrange our work in a way similar to long division, it is evident that we first take for our divisor 2 a + b, that is, twice the first term of the root added to the second term, which multiplied by b the second term, and subtracted, leaves no remainder. Thus the whole process may be arranged as follows:- Ex. 2. Find the square root of a2 + 2 ab + b2 + 2 ac + 2 bc + c2. Now we know (Art. 26) that— a2 + 2 ab + b2 + 2 ac + 2bc + c2 or (a + b)2 And if we compare the form (a + b)2 + 2 (a + b) c + c2 with the form a2 + 2 ab + b2, it is evident that, having obtained as in the last example the first two terms a + b, we shall by continuing the process obtain the third term. Thus 2a + b a2 + 2 ab + b2 + 2 ac + 2 bc + c2(a + b + c a2 2a + 2b + c 2 ab + b2 2 ab + b2 2 ac + 2 bc + c2 2 ac + 2 bc + c2 We may deduce from the above examples the following general rule : RULE. 1. Arrange the terms of the given quantity according to the ascending or descending powers of some letter, and take the square root of the first term for the first term of the quotient. 2. Subtract the square of the quotient, and bring down the next two terms of the given quantity. 3. Double the quotient, and place the result as a trial divisor; then, dividing the first of the terms brought down by this trial divisor to obtain the second term of the root, add the quotient so obtained to the first term of the root, and also to the trial divisor, to obtain a complete divisor. 4. Multiply the complete divisor by the second term of the root, and subtract the product, as in long division, from the terms brought down. 5. If there be any remainder or more terms to bring down, double the whole quotient for a trial divisor, and divide the remainder by the new trial divisor, to obtain the third term of the root; and so on. Ex. 3. Find the square root of 1 - 4x + 10 x2 - 12x3 + 9x3. The terms are here arranged according to the ascending powers of x. Then proceeding according to rule, we have 1 - 4 x + 10 x2 – 12x3 + 9 x1 (1 − 2 x + 3 x2 1 The student will observe that twice the quotient is most easily obtained by bringing down the previous complete divisor with its last term doubled. Square Root of Numerical Quantities. 37. It is easy to apply the above method to numerical quantities. Since 12 1,102 100, 100210,000, 1,000 1,000,000, &c., it is evident that the square roots of numbers having less than three figures must contain one figure only ; That those having not less than three and less than five must contain two figures and two only; Those having not less than five and less than seven must contain three figures and three only; and so on. Hence it follows that, if a dot be placed over the units' figure, and over every alternate figure to the left, the number of dots will give the number of figures in the square root. Thus, the square roots of the numbers 141376 and 1522756 have three and four figures respectively. In the number 141376 we call 14, 13, 76 respectively the first, second, and third periods. So in the number 1522756, the first, second, third, and fourth periods are respectively 1, 52, 27, 56. It is evident that the number of periods correspond to the number of figures in the square root, and it will be seen that the figures of each period are used in the operation for the corresponding figure of the root. |