-- quantities, a, b, c, etc.; A, for the sum of the products of every two of the same quantities; A, for the sum of the products of every three of the same quantities; A for the sum of the products of every four of the same quantities, and so on; and if An is put for the absolute term, or for the product of all the quantities, a, - b, C, d, -e, etc., then we shall have (x− a) (x —b) (x c) (x d) × etc., = x2 + Â ̧æ ̃−1 + Aqxn−2 + A3x2A1xn−1 Agen -2 - 3 + A4x2 - 4 +, etc.... +An for the product, as required. - -2 × = = -2 Hence, if we put xa, we have x- - a naught, and the product ( − a) (x — b) (x — c) x etc. naught; consequently, we shall have "+ A1ær −1 + Av2 - 2 + Azur -3, etc.,... + An = 0; which is called an equation of the n degree. ― It is easy to see that the product (x − a) (x —b) (x — c) (x − d) ×, etc., becomes naught by putting equal to any one of the quantities a, b, c, d, etc.; so that there are n different values of x that will reduce the product (x — a) (x − b) (x — c) (x-d) ×, etc., to naught, or which will satisfy the aforesaid equation, which values are called the oots of the equation; so that when a, b, c, d, etc., are known, we see how to find the roots of the equation, or the values of a that satisfy it. Reciprocally, having the equation + A-1 + Agen - 2 + Aga-3 + A4 +, etc.,.... + An = 0; then to find a, b, c, d, etc., is to find the roots of the equations, or the values of a that will reduce a + A1an-1 + Agn−2+ A sin−3 +, etc.,.... + An, to naught, or make it equal 0; which is called resolving or solving the equation. -1 It is evident, since a, b, c, d, etc., are the values of a that reduce A1x2-1 + A2 + A3+, etc., . . . 1 + An, to zero, or the roots of the equation + A-1 + Agan −2 +, etc., + An = 0, that, having found the roots of the equation, we, of have the quantities ∞ d, etc., whose product (a) (x —b) (x — c) (x — d) ×, etc., = x2 + A1x2-1 + An-2+, etc., ... + An; so that to resolve any quantity into factors, we put it equal to naught, and thence find the particular values of the unknown quantity which reduce the quantity to naught, and thence find the factors, as has been said. -2 If the roots of the equation are all real, positive quanti ties, it is evident that a, b, c, d, etc., will each be a real positive quantity, and of course a, b, c, d, etc., will each be a real negative quantity; consequently the equation + A12-1 + A2-2 + Azn -3 +, etc., ... + An0 will have n + 1 terms; since A1, which is put for (a + b + c + d +, etc.), can not become equal to naught, but must be a negative quantity; also A, which is the sum of the products of every two of the quantities — a, – b, c, d, etc., must be a positive quantity greater than naught; and A,, which is the sum of the products of every three of the quantities a, b, c, d, must be a negative quantity; and so on, for all the coefficients. Hence, regarding x as being a positive quantity, ∞ will be a positive quantity, and putting down the signs of the first, second, etc., terms of our equation, we have + − + + − + +, and so on, until we have n + 1 signs; but the number of changes (or variations) of signs, when we count the signs (by twos) from the first which is to the last will equal n, which is the same as the number of real, positive roots in the equation; since the roots are by supposition all real, positive quantities. If n is an even number, then n + 1 is an odd number, and an inspection of the series of signs (above) shows that the sign of the absolute term An of our equation is positive; but if n is an odd number, then the sign of the absolute term is When n is an even number, if we put a negative quantity for a, it is evident that the signs of all the terms of the equation will be +; since 2-1 is negative, and A, negative, and of course their product A-1 is positive; also A, being positive, and a positive, their product is, of course, positive; but A, being negative, and -3 being also negative, their product must be positive, and so on; so that the sign of each term of the equation is +. In like manner, if n is an odd number, and x a negative quantity, it may be shown that the sign of each term is -. -2 Hence if a negative quantity is put for a, the equation does not admit of any changes of signs; or, as it is generally said, the signs are permanences. If the roots of the equation are equal to each other, then, since An is the product of n equal roots, if we extract the n' root of An regarded as positive, we shall have the numerical value of each of the equal roots expressed by VAn. If the signs of the terms of the equation are alternately + and -, we must for a put An in the equation, and it will be reduced to naught; but if all the terms of the equation have the same sign, we must for a put An, and the equation will be satisfied, as required; as is evident from what has been shown. = Thus, if we have 2 · 2xa + a2 = 0, then n = 2 and An = a2, and va2 = a; and since the signs of the terms of the equation are alternately + and -, we must put x = a, and the equation reduces to a2 - 2a+a2=0, as it ought to do; consequently, we shall have a-a, xa, for factors of x2- 2xa + a2, or, which is the same, we shall have (x — a)2 = x2- 2xa + a2. Again, if we have +2.ca + a2=0, we must for a put the particular value x = Vaa in the equation, and it will be reduced to a2 2a2 + a2 = 2a2 · 2a2 = 0, as it ought to do; consequently, a + a, x+a, are factors of x2 + 2xa + a2, or we have (x + a)2 = x2 + 2xa + a2. - If the roots of the equation are unequal, then the numerical value of An will be a species of mean between the roots of the equation being numerically greater than some of them and less than others. Finally, if n = 1, our equation becomes + A1 = 0, which x is called an equation of the first degree, or a simple equation, since the exponent of x is 1. If n = 2, the equation. becomes + A+ A2 =0, which is called an equation of the second degree, a quadratic equation, or simply a quadratic, since the exponent of the highest power of x is 2. = If n equals 3 (n = 3), the equation becomes a + Âμæ2 + A+ A, 0, which is an equation of the third degree, or a cubic equation, the exponent of the highest power of x being 3. If n = 4, the equation becomes + A+ A2 + Ag? + A1 = 0, which is an equation of the fourth degree, or a biquadratic equation, or simply a biquadratic, since the exponent of the highest power of x is 4. If n = 5, 6, 7, etc., the equation is said to be of the fifth, sixth, seventh, etc., degrees, where it may be noted that an equation of the fifth degree is sometimes called a sur-solid, and an equation of the sixth degree a quadrato-cubic, or square-cubic equation. Ex. 5.-To find the product of a + b and a-b. a + b a b a2 + ab -ab-ba the product, as required. a2 — b3, Reciprocally, we can resolve ab into factors; for by adding ab ab0 to it, we have a2 + ab — ab —b2 — a2 b; but it is easy to see that a is a factor of a + ab, and b a factor of ab+b, so that we have a2+ab = a(a + b) and ab + b2 = b (a + b); consequently, we have a2+ab - abb2 = a (a + b) — b (a + b) = (a − b) (a + b), or we have a2 b2 = (a — b) (a + b) = (a + b) (a − b), so that a2 — b2 has a + b and ab for factors. = Again, if in ab we put b orb for a, it reduces to b2b20; consequently, ab and a+b are factors of a2 — b2, or a2 — b2 = (a + b) (a − b), as evidently follows from what was shown in the preceding remarks. Hence, it follows that the product of the sum and difference of any two numbers or quantities of the same kind is equal to the difference of their squares; and conversely, the difference of the squares of any two numbers or quantities of the same kind is equal to the sum of the numbers or quantities multiplied by their difference. 14, b = Thus, if a = 10, the sum of a and b is a + b = 24, and the difference of a and b is a-b4; then (a + b) (ab) a2 62 changes into 24 x 4 = 142-102 96, and = — b2 × a2b2 = (a + b) (a - b) becomes 142-10224 × 4 = 96; results that agree with what is affirmed above. Ex. 6. To multiply a2 + xy + y2 by x-y, by detaching the coefficients. The coefficients being detached, give 1 + 1 + 1 for the coefficients of the multiplicand, and 1-1 for those of the multiplier. Hence we have to multiply 1+1+1 by 1-1, which gives 1+1+1 = the sum of the products of the coefficients. x Since is the first term of the multiplicand, and a that of the multiplier, it is clear that a, ay, xy, and y3 will correspond to the first, second, third, and fourth coefficients in the sum of the products of the coefficients; consequently, we have 3 + 0. x2y + 0. xy3 — y3 = x23 — y3 = the required product. Hence, since (x2 + xy + y2) (x − y) = 28 y3, it follows that ay has +y+y, and cy for its factors. x2 x Conversely, it is easy to see that y must be a factor of ay; for if we put y for x, x3-y3 becomes y3-y3=0, and consequently, from what has been proved, xy is a factor of y3. = To resolve - y', we add x2y + x2y — xy2 + xy2 = 0 to it, which gives a3 - y3 3 - x2y + x2y — xy2 + xy3 — y3 ; then since is a factor of a-ay, and ay a factor of xyxy, and y a factor of ay-y, we get a-y= x2 (x − y) + xy (x − y) + y2 (x − y), and since a right member of this equation, we get y3) (e — y); and of course 2+ xy + y2 tors of ay, as required. a y is a factor of the y= (x2 + xy + and x y are fac Ex. 7.-To develop (x2 −xy+ y2) (x + y), by detaching the coefficients. Detaching the coefficients, and proceeding as in the last example, we get |