5. Then 6 is the coefficient to the third term, the index of a decreasing 1, and the index of x increasing 1, the third term will be 6a2x2. 6. The terms thus found are a4+4a3x+6a2x2. 7. Multiply 6, the coefficient of the third term, by 2, the index of a in this term, and divide the product 12 by 3, the number of terms already found, the quotient 4 will be the coefficient of the fourth term; therefore, the fourth term will be 4ax3. The index of a having decreased 1, and the index of x having increased 1. 8. The terms already found are a1+4a3x+6a2x2+4ax3. 9. In 4ar3, the index of a is 1, (every letter in its primitive state having 1 for an index,) therefore, 1x4-4, for a product, this product being divided by 4, the number of terms already found, gives 1 for the coefficient to the fifth term: the a in this term will vanish, and the index of x will increase as above; therefore, the fifth term will be a4. 10. Hence, the fourth power of a+x is a1+4a3x+6a2x2+ 4ax3+x4. NOTE. If the second quantity in the root, or quantity to be raised, be the terms will be + and alternately; thus, if it be required to raise a-x, to the fourth power, it will be +a-4a3x+6a2x2—4ax3+x1. 2. Required the fourth power of 3—y.* EXAMPLES FOR PRACTICE. 1. Required the 3rd power of x+y. 2. Required the 5th power of a+x. 3. Required the 4th power of 5+x. 4. Required the 4th power of x+2. * The 3 in this example is placed in a similar situation to that of a in the first example, and is operated with in every respect the same. EVOLUTION. EVOLUTION, or the extraction of roots, is the reverse of Involution, and is commonly divided into three cases, as follows: CASE I. To find the roots of simple quantities. RULE.-Extract the required root of the coefficient for the numeral part, and divide the index of the latter by 2 for the square root, by 3 for the cube, by 4 for the fourth root, &c. This root being annexed to the numeral part, will be the root required. The square root, the fourth root, or any other even root of an affirmative quantity, may be either + or —.(a) Any odd root of a quantity will have the same sign as the quantity itself. EXAMPLES. 1. Required the square root of 16x2.. Here, 16-4, and xxx; . 4x is the root required. 2. Required the cube root of 27x3. 3 3 Here 27=3, and 3⁄4æ3 = x23= x2= 3. Required the fourth root of 16x3. 4 4 8 Here 16-2, and x = x+x2 4. Required the cube root of 8x1. 3 3 3— Here /8=2, and xx or x.x or xx. 4 3 ... 2x or (a) Here 16x2=+4x or -4x. For by note fourth, Case I. in Multiplication, -4x-4x=16x2. EXAMPLES FOR PRACTICE. 1. Required the square root of 36x*. 2. Required the square root of 5x2y2. 3. Required the cube root of 27x6. 4. Required the square root of 4.x2 5. Required the fifth root-32x10y. Ans. -2x2y. CASE II. To find the square or cube root of a compound quantity. RULE. Arrange the quantities according to their dimensions, as in Case III. in Division, and proceed as in Arithmetic. EXAMPLES. 1. Required the square root of x1-4x3a+6x2a2— 4xa3+a1. x-4x3a+6x2a2-4xa3+a1(x2-2xa+a2 2x2-2xa) -4x3a+6x2a2 -4x3a+4x2a2 2a2-4xa+a2) 2x2a2-4xa3+a1 Any even root of a negative quantity is impossible; for +xx+x will not make -x2; nor will-xx-x make-x2. But the square, or any root of a quantity, may be nominally extracted, by placing over it its proper sign; the square root of -3x2, is—3x2; the cube root of -3x2 is -3x2, and so on. 3 2. Required the cube root of a3+3a2x+3ax2+x3. BY HILL'S RULE. The cube of your first period take, &c. a3+3a2x+3ax2+x3(a+x EXAMPLES FOR PRACTICE. 1. Required the square root of x2-2xy+y2. Ans. x-y. 2. Required the square root of 9+6x+x2. Ans. 3+x. 3. Required the cube root of 27+27x+9x2+x2. Ans. 3+x. 4. Required the cube root of x+3x3y3+3x3y3+y. CASE III. To find the roots of powers in general. I Ans. x+y RULE.-1. Arrange the terms as directed in the third case of division, then find the required root of the first term, and place it in the quotient. 2. Raise this root to the power of the first term, and subtract it from the first term. 3. Bring down the second term for a dividend. 4. Divide this dividend, or second term, by twice the root last found, for the square root; by three times the square of the root last found, for the cube root; by four times the cube of the root last found, for the fourth root, &c. and the quotient thus found will be the second term in the quotient or root. 5. Involve the whole quotient or root thus found to the power to be extracted, then subtract this power from the given quantity of which the root is to be extracted. 6. Divide the greatest power of the remainder by the divisor already found, and proceed as above until the whole of the terms have been brought into operation. EXAMPLES. 1. Required the square root of x2+4x3y+6x2y2+ 4xy3+ys. x2+4x3y+6x2y2+4xy3+y1(x2+2xy+y2 x2+4x3y+6x2y2+4xy3+y1=(x2+2xy+y2)2 2. Required the cube root of x-6x+15x1-20x3 +15x2-6x+1. x-6x+15x1-20x+15x2-6x+1(x2-2x+1 xo—6x3+15x1—20x3+ 15x2-6x+1=(x2—2x+1)3 |