(6.) 2a-x.. Ans. 16a4-32a3x+24a2x2-8ax3 +xa.

44. By examining the successive developments of a binomial a+b; viz.:

2d power, a2+2ab+b2

3d power, a +3a2b+3ab2+b3

4th power, a1+4a3b+6a2b2+4ab3+b1

5th power, a +5a1b+10a3b2 +10a2b5+5ab1+b5

it will be observed that the powers of a decrease, and those of b increase, by a unit in each successive term, and that the highest power of each is the same as that to which the binomial is raised. Also, if the coefficient of any term be multiplied by the index of a in that term, and the product divided by the number of terms reckoned from the commencement (that is, if it be the third term, divide by 3,-if it be the seventh, by 7, &c. &c.), the result is the coefficient of the next term. If both terms of the binomial be positive, all the terms of the expansion will be positive; but if one term of the binomial be negative, such as b, the terms of the expansion which contain the odd powers of b will be negative.

Thus to raise a + b to the sixth power, the powers of a and b in the successive terms would be ao, a5b, a1b2, a3b3, a2b1, ab3, and bo. The coefficient of the first term a is 1, and by multiplying the index 6 by 1, and dividing by the number of terms 1, we get 6 for the coefficient of the second term; or as the multiplier and divisor are each unity, it is plain that the coefficient of the second term will always be the same as the index of a in the first. By multiplying 6 the coefficient of the second term by 5, the index of a in that term, and dividing the product by 2, the number of terms, the result is 15, which is the next coefficient. Again, multiplying 15 the coefficient of the third term by 4, the index of a in that term, and dividing the product by 3, the number of terms, the result is 20, which is the coefficient of the fourth term. Proceeding in the same manner, we would find that

(a + b) = a + 6a5b + 15a4b2 + 20a3b3 +

15a2b1 + 6ab5 + bo,

the signs being all positive. Or in general terms,

This is the celebrated binomial theorem; and by means of it, a binomial may be raised to any power, without the trouble of the continued multiplication. As this theorem can be shown to hold for all powers, positive or negative, whole or fractional, it is capable of very extensive application.

EXERCISES.

Expand the following by means of the Binomial Theorem. (1.) (a + b)®.

Ans. a6+6a5b+15a1b2 +20μ3b3 + 15a2b1+6ab5+bo.

45. Evolution is the extraction of any root of a given quantity.

To extract any root of a simple quantity, divide the index of the quantity by the index of the root required.

NOTE 1. Any odd root of a quantity will have the same sign as the quantity itself; but any even root of a positive quantity may be either positive or negative, and no even root of a negative quantity can be extracted.

NOTE 2. Any root of a product may be found by taking that root of each factor; and any root of a fraction by taking that root of both numerator and denominator.

EXERCISES.

Extract the square root of each of the following quantities:

Extract the cube root of each of the following quantities:

46. To extract the square root of a compound quantity. (1.) Arrange the terms according to the powers of some one

of the letters, commencing either with the highest or lowest power. (2.) Find the root of the first term, for the first term of the required root; and subtract its square from the given quantity. (3.) For a new divisor, take the double of the part of the root found, and divide the first term of the remainder by it, setting down the quotient for the next term of the required root, and annex this term also to the divisor, then proceed as in common Division; and repeat the same process as often as may be necessary.

EXERCISES.

Extract the square root of each of the following quantities:

(10.) 9+24x-68x2 — 112x3 +196x1.

Ans. a+b.

47. To extract the cube root of a compound quantity. (1.) Arrange the terms as directed above, and take the cube root of the first term for the first term of the answer. (2.) Subtract the cube of this term from the given quantity, and divide the first term of the remainder by three times the square of the part of the root found; setting down the quotient for the next term of the answer. (3.) Subtract the cube of the binomial thus found from the given quantity; and if there be a remainder, divide its first term by the same divisor as before, for the third term of the answer. (4.) Subtract the cube of the trinomial thus found from the given quantity, and if there still be a remainder, proceed in exactly the same manner as often as may be necessary.

EXERCISES.

Find the cube root of each of the following quantities:

(13.) ao+6a5+9a1—4a3—9a2+6a—1. a2+2a-1.

(14.) 8x6—36ax3+30a2x2+45a3x3-30a4x2-36a5x—8a".

2x2-3ax-2a2.

47

CHAPTER VI.

SURDS OR RADICALS.

48. Any root of a quantity which cannot be exactly determined is called a Surd or Radical, and sometimes an Irrational Quantity." Thus 2 and 3/4 are surds.

Since the extraction of any root of a product may be performed by extracting that root of each of the factors, many surds may be much simplified by extracting the root of such of the factors as admit of it. Thus Jab=a2 √b; √108= √36x3=√36/3=6/3.

49. Any quantity may be reduced to the form of a surd by raising it to any power, and then indicating the corresponding root. Thus, a√a2 = 3/a3, &c. &c. 2√5=√4 √5=√20. 25/3 = 5/8 5/3 = 3/24.

EXERCISES.

Reduce each of the following quantities to the form of the square root, cube root, and nth root: