and by performing the multiplication here indicated, we have, when x +α ̧α2 =0 n=2 By continuing the multiplication to the last, the equation will be found whose roots are those proposed; and from what has been done we learn that (1) The coefficient of the second term in the resulting polynomial will be the sum of all the roots with their signs changed. (2) The coefficient of the third term will be the sum of the products of every two roots with their signs changed. (3) The coefficient of the fourth term will be the sum of the products of every three roots with their signs changed. (4) The coefficient of the fifth term will be the sum of the products of every four roots with their signs changed, and so on; the last or absolute term being the product of all the roots with their signs changed. Cor. 1. If the coefficient of the second term in any equation be 0, that is, if the second term be absent, the sum of the positive roots is equal to the sum of the negative roots. Cor. 2. If the signs of the terms of the equation be all positive, the roots will be all negative, and if the signs be alternately positive and negative, the roots will be all positive. Cor. 3. Every root of an equation is a divisor of the last or absolute term. Cor. 4. No equation, whose coefficients are all integers, and that of the highest power of the unknown quantity unity, can have a fractional root. This will be obvious by transposing the absolute terin in any equation, and substituting for the unknown quantity a fraction in its lowest terms, which will give a fraction in its lowest terms equal to an integer, showing that such equation cannot have a fractional root. Cor. 5. In any equation whose roots are all real, and the last, or absolute term very small when compared with the coefficients of the other terms, then will the roots of such an equation be also very small. EXAMPLES. (1.) Form the equation whose roots are 2, 3, 5, and -6. Here we have simply to perform the multiplication indicated in the equation, (x-2) (x-3)(x-5)(x+6)=0; f(x) must be divisible by x-a2, for a-a, is not divisible by x-a,; hence, if f(x) represent the quotient of f (x) divided by x-a,, we have In like manner, if a,, α, α5, nomial ƒ(x) is divisible by x—a ̧, x—ɑμ9 therefore, assume the form (x—a1) (x—a2) (x—α3) and, consequently, there are as many roots as factors, that is, as units in the highest power of X, the unknown quantity; for the last equation will be verified by any one of the n conditions, x=α, x=ɑ2, x=α3, x=α49 .... x=α: and since the equation contains n factors, there are n roots. Cor. When one root of an equation is known, the depressed equation containing the remaining roots is readily found by synthetic division; and if two or more roots are known, the equation containing the remaining roots is found by two or more corresponding divisions. EXAMPLES. (1.) One root of the equation a1—25x2+60x—36=0 is 3; find the equation containing the remaining roots. is the equation containing the remaining roots. (2.) Two roots of the equation x1—12x3+48x2—68x+15=0, are 3 and 5; find the quadratic containing the remaining roots. 1-12 +48-68+15 (3 is the equation containing the two remaining roots. (3.) One root of the cubic equation æ3—6x2+11x—6=0 is 1; find the quadratic containing the other roots. Ans. x2-5x+6=0. (4.) Two roots of the biquadratic equation 4x1—14x3—5x2+31x+6=0 are 2 and 3; find the reduced equation. Ans. 4x2-6x+1=0. (5.) One root of the cubic equation x3+3x2—16x+12=0 is 1; find the reinaining roots. Ans. 2 and—6. (6.) Two roots of the biquadratic equation xa—6x2+24x—16=0, are 2 and -2; find the other two roots. Ans. 3+5. PROPOSITION IV. an To form the equation whose roots are a1, a2, ɑз, ɑ4; The polynomial, f(x), which constitutes the first member of the equation required, being equal to the continued product of x—ɑ1, x—ɑ2, x—A3, a-a, by the last proposition, we have (x—a,) (x—a2) (X—αs) (x—a1)=0; ... and by performing the multiplication here indicated, we have, when By continuing the multiplication to the last, the equation will be found whose roots are those proposed; and from what has been done we learn that (1) The coefficient of the second term in the resulting polynomial will be the sum of all the roots with their signs changed. (2) The coefficient of the third term will be the sum of the products of every two roots with their signs changed. (3) The coefficient of the fourth term will be the sum of the products of every three roots with their signs changed. (4) The coefficient of the fifth term will be the sum of the products of every four roots with their signs changed, and so on; the last or absolute term being the product of all the roots with their signs changed. Cor. 1. If the coefficient of the second term in any equation be 0, that is, if the second term be absent, the sum of the positive roots is equal to the sum of the negative roots. Cor. 2. If the signs of the terms of the equation be all positive, the roots will be all negative, and if the signs be alternately positive and negative, the roots will be all positive. Cor. 3. Every root of an equation is a divisor of the last or absolute term. Cor. 4. No equation, whose coefficients are all integers, and that of the highest power of the unknown quantity unity, can have a fractional root. This will be obvious by transposing the absolute terin in any equation, and substituting for the unknown quantity a fraction in its lowest terms, which will give a fraction in its lowest terms equal to an integer, showing that such equation cannot have a fractional root. Cor. 5. In any equation whose roots are all real, and the last, or absolute term very small when compared with the coefficients of the other terms, then will the roots of such an equation be also very small. EXAMPLES. (1.) Form the equation whose roots are 2, 3, 5, and —6. Here we have simply to perform the multiplication indicated in the equation, (x-2) (x-3)(x-5)(x+6)=0; and this is best done by detached coefficients in the following manner: .. x1—4x3-29x2+156x-180=0 is the equation sought. (2.) Form the equation whose roots are 1, 2, and -3. (3.) Form the equation whose roots are 3, −4, 2+√/3, and 2—√3. (2.) x3—7x+6=0. ANSWERS. (3.) x1—3x3—15x2+49x—12=0. (4.) x3-32x+24=0. (5.) x®+3x5—41x1—87x3+400x2+444x—720=0. (7.) 8x1—54x3+101x2-54x+8=0. PROPOSITION V. If the signs of the alternate terms in an equation be changed, the signs of all the roots will be changed. be an equation; then changing the signs of the alternate terms, we have クロー2 n-2 .... +AA=0 (2). (3). or—x”+ Â ̧ã”—'—A11212+ But equations (2) and (3) are identical, for the sum of the positive terms in each is equal to the sum of the negative terms, and therefore they are identical. Now if a be a root of eq (1), and if a and -a be substituted for x in equation (1) and (2) respectively, the results will be the very same; and since the former is verified by such substitution, a being a root, the latter is also verified, and therefore —a is a root of the identical equations (2) and (3). Cor. If the signs of all the terms are changed, the signs of the roots remain unchanged. EXAMPLES. (1.) The roots of the equation x3—6x2+11x—6=0 are 1, 2, 3. What are the roots of the equation 3+6x2+11x+6=0? Ans. 1, -2, −3. (2.) The roots of the equation x2-6x2+24x-16=0 are 2, -2, 3±√√5. Express the equation whose roots are 2, −2,-3+ √5, and −3 −√5. Ans. x1-6x2—24x-16=0. PROPOSITION VI. Surds and impossible roots enter equations by pairs. .... A1+A=0, be an equation, having a root of the form a+b√−1; then will a—b√√—1 be also a root of the equation. For, let a+b√—1 be substituted for x in the equation, and we have (a+b√−1)"+A,(a+b√—I)'~'+ . . . . A...„-1(a+b√√−1)+A„=0. Now, by expanding the several terms of this equation, we shall have a series of monomials, all of which will be real, except the odd powers of b√−1, which will be imaginary. Let P represent the real, and Q√1 the imaginary terms of the expanded equation; then P+Q√−1=0, an equation which can exist only when P=0, and Q=0. Again, let a—b√−1 be substituted for a in the proposed equation; then the only difference in the expanded result will be in the signs of the odd powers of b√1, and the collected monomials, by the previous notation, will assume the form P-Q=1; but we have seen that P=0, and Q=0; ... P_Q√=I=0, and hence a-b√-1 also verifies the equation, and is therefore a root. In a similar manner, it is proved that if a +√b be one root of an equation, a-b will also be a root of that equation. Cor. 1. An equation which has impossible roots is divisible by {x−(a+b√=1)} {x—(a−b√−1)} or x2-2ax+a2+b2, and therefore every equation may be resolved into rational, simple, or quadratic factors. Cor. 2. All the roots of an equation of an even degree may be impossible, but if they are not all impossible, the equation must have at least two real roots. Cor. 3. The product of every pair of impossible roots being of the form a2+b2, is positive; and, therefore, the absolute term of an equation whose roots are all impossible must be positive. Cor. 4. Every equation of an odd degree has at least one real root, and that root must, necessarily, have a contrary sign to that of the last term. Cor. 5. Every equation of an even degree, whose last term is negative, has at least two real roots; the one positive, and the other negative. PROPOSITION VII. An equation cannot have a greater number of positive roots than there are variations of signs in the successive terms from + to —, or from to +, nor can it have a greater number of negative roots than there are permanencies, or successive repetition of the same sign in the successive terms. Let an equation have the following signs in the successive terms, viz.:+ − + + +· Now, if we introduce another positive root, we must multiply the equation by 1-a, and the signs in the partial and final products will be + + +++ −, or + |