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This theorem will now be given, and the student can verify its results by actual multiplication in the mean time, as the general proof would not be understood at this stage of the pupil's advancement.

BINOMIAL THEOREM.

57. Let it be required to raise (a+x) to the nth power, where n may be any number, and a and x any quantities, either simple or compound. The first term will be a", and the second will be obtained from it, by multiplying by

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a

and will therefore be na-x. The third will be ob

tained from the second, by multiplying by "-12

2

x and in the same manner, the 4th, 5th, 6th, 7th, will be obtained by multiplying the 3d, 4th, 5th, and 6th successively by n-2 x R-3 x n-4 x

3

n-5 x

and

and so on, so that

5 a

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the resulting series will be (a+x)"=a”+na"¬1x+

u(n-1) (n−2) (n−3) ɑn→2*+

1.2.3.4

n(n−1) (n−2) (n−3) (n−4)-5+&c., where, by substituting

1 • 2

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4

instead of n, 2, 3, 4, 5, successively it becomes (a+x)2=a2+2x+x2.

(a+x)3=a3+3a2x+3αx2+x3.

22

(a+x)1a1+4a3x+6a2x2+4ɑx3 +x1, (a+x)=a5+5a1x+10a3x2+10a2x3 +5ax1 +x3. From the above, it is evident, (1st), That the power of a in the first term is the same as the power to which the binomial is raised. (2d), That the powers of a decrease by unity in each successive term, whereas those of x increase by unity, till the last term, where it is equal to the power to which the binomial is raised. (3d), That the number of terms in the expansion is always one more than the power of the binomial. (4th), That the co-efficient of the second term is always equal to the power to which the binomial is raised, and that the successive co-efficients can be obtained by multiplying the co-efficient of the previous term into the power of a in that term, and dividing by the number of terms from the beginning of the expansion; thus 10, the co-efficient of the third term in the expansion of (a+x) can be obtained by multiplying 5, the co-efficient of the second term, into 4, the power of a in that term, and dividing by 2, the number of the term from the begin

ning of the expansion. In the same manner, may all the other co-efficients be obtained.

If the sign of the second term of the binomial were minus, since the odd powers of a minus quantity are minus, and the even powers are plus, the terms which contain an odd power of the second term will be minus, (Art. 55), and those which contain even powers of that term will be plus. Thus,

(α—x)¤—गбa3x+15aax2—20a3x3 +15a2xa—6ax3+xo. If it were required to expand (a+b-c)", it might be effected by first considering (b-c) as one quantity, and then raising it to the power denoted by its index in each term, and separating into single terms. Thus,

(a+b—c)3=a3 +3a2(b—c)+3a(b—c)2+(b—c)3

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3a2 (b-c)

3a(b-c)2=

(b—c)3 =

+3a2b-3a2c

+3ab2-6abc+3ac
b3-3b2c+3bc2-c3

••{a+(b—c)}3—a3+b3—c3+3a2b—3a2c+3ab2+3ac2 —

3b2c+3bc2-6abc.

1. What is the 4th power of (a-b)?

Ans. a44a3b+6a2b2-4ab3+64.

2. What is the 3d power of (4a-2x)?

Ans. 64a5-96a2x+48ax2-8x3.

3. What is the 9th power of /x+y?

Ans. x3+3x2y+3xy2+y3.

4. What is the 5th power of (2a—x)?

Ans. 32a5-80a4x+80a3x2-40а2x3 +10αx1—x5. 5. What is the 3d power of {a—(x+y)}?

Ans. a3-x3-y3—3a2x-3a2y+3ax2+3ay2-3x2y -3xy2+6axy.

EVOLUTION.

58. CASE I. When the given quantity is simple. RULE. Extract the given root of the numerical coefficient for the coefficient of the root, then divide the exponents of each of the literal factors by the name of the root, and the several results connected by the sign of multiplication will be the root sought.

EXAMPLE. Extract the 4th root of 81a4x6.

The 4th root of 81 is 3, the 4th root of a1 is a1-a, and the 4th root of x is xa=x3. ..3×a×x3-3ax is the

x6

root sought. In the same manner may the roots of the following quantities be found.

1. What is the 2d or square root of 16a2b1c ?

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59. CASE II. When the given quantity is compound, and it is required to extract its square root.

RULE. Arrange the terms according to the powers of the same letter, so that the highest power may be first, and the next highest in the second term, and so on. Extract the square root of the first term by Case I., and place its root for the first term of the root. Subtract its square from the first term, and there will be no remainder. Bring down the next two terms for a dividend, and for a divisor take twice the part of the root already found. Divide the first term of the dividend by the divisor, and place the quotient both in the root and in the divisor. Multiply the divisor thus completed by the term last placed in the root, and subtract the product from the dividend; to the remainder, if any, bring down the next two terms, and proceed thus till the root is found.

The above rule will be obvious, by observing that the square of a+x is a +2ax+x2, and that consequently the square root of a2+2ax+x2 will be a+x; but after we have subtracted the square of a, the remainder is 2ax+x2 =(2a+x)x, the first term of which remainder, if divided by 2a, will give the quotient x; and a being added to 2a, and then the sum multiplied by x, will leave no remainder. EXAMPLE. Extract the square root of a2+2ab+b2+ 2ac+2bc+c2.

a2+2ab+b2+2ac+2bc+c2 a+b+c root.

a2

Divisor 2a+b 2ab+b2

2a+2b+c

2ab+b2

2ac+2bc+c2
2ac+2bc+c2

Here, after (a+b) has been obtained in the root, it is evident that the remainder can be written 2(a+b)c+c2= {2(a+b)+cc; where a+b takes the place of a in the remainder, and is of the same form as (2a+x)x.

1. Extract the square root of 9a2+6ab+b2.

Ans. 3a+b.

2. Extract the square root of a2+8ax+16x2.

Ans. a+4x.

3. Extract the square root of a1c2+4a2cx+4x2.

Ans. a2c+2x.

4. Extract the square root of 4a2+12ab+962+16ac+ 24bc+16c2. Ans. 2a+3b+4c. 5. Extract the square root of 6+4x+2x+9x2-4x+4. Ans. x+2x2-x+2.

60. CASE III. When the given quantity is compound, and it is required to extract its cube root.

RULE. Arrange the terms as in CASE II. Find the cube root of the first term, which will be the first term of the root. Subtract its cube from the first term, which will leave no remainder. Bring down the next term, and divide it by three times the square of the root already found; the quotient will be the second term of the root. Raise these two terms to the third power, and subtract the result from the given quantity; if there be a remainder, divide its first term by three times the square of the first part of the root as before, and thus proceed till the work is finished.

The third power of (a+x) is a3+3a2x+3ax2+x3 ; hence it is evident, that the cube root of a3+3a2x+3ax2 +x3 is a+x; taking away the cube of a, the first term of the root, the remainder is 3a2x+3ax2+x3; the first term of which divided by 3a2, gives the quotient x.

EXAMPLE. Extract the cube root of ao+6a*x+12a2x2 +8x3.

a6+6a1x+12a2x2+8x3, | a2+2x. Root.

α.

The first term of the remainder is 6a4x, which divided by 3a, which is three times the square of a2, gives 2x for quotient; and a2+2x raised to the third power, gives the quantity whose root was to be extracted, and no remainder; so that a2+2x is the root sought.

1. Extract the cube root of 27a3-27a2x+9ax2-x3.

Ans. 3a-x.

2. Extract the cube root of ao+бa1b2+12a2ba+86o.

Ans. a2+262.

3. Extract the cube root of 39x2y+27xy2-273. Ans. x-3y.

61. Any root whatever may be extracted by the following formula: let n be the name of the root, which will be 2 for the square root, 3 for the cube root, and so on; then having arranged the terms as in the square and cube roots, extract the root of the first term by Case I., and let a represent that root, then the second term divided by nan-1, will give the second term of the root; the first and second terms of the root being raised to the nth power, and subtracted from the given quantity, the remainder, if there be any, will be such, that its first term divided by the same divisor will give the third term. The exercises in the square and cube root may be wrought by this rule.

62. Definitions.

EQUATIONS.

1. An Equation is a proposition which declares the equality of two quantities expressed algebraically.

This is done by writing the two quantities, one before and the other after the sign(): thus 4+x=3x-4 is an equation, which asserts the equality of 4+ and 3x-4.

2. A Simple Equation is one which, being reduced to its simplest form, contains only the first power of the unknown quantity.

3. A Quadratic Equation is one which, being reduced to its simplest form, contains the square of the unknown quantity.

4. When an Equation, after being reduced to its simplest form, contains the third power of the unknown quantity, it is called a Cubic Equation.

5. A Pure Quadratic is one into which only the square of the unknown quantity enters.

6. An Adfected Quadratic is one which contains both the first and second powers of the unknown quantity.

7. The Resolution of Equations is the determining from some given quantities the value or values of those that are unknown.

The resolution of equations is effected by the application of one or more of the following axioms:

AXIOMS.

(1.) If equal quantities be added to equal quantities, the sums will be equal.

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