+do, which, properly contracted, becomes y2+b—· 3 a2 4 3. Given x2-4x+8=0, to exterminate the second term. 4 Thus, == 2 =-2; then let y+2=x, and proceed as before. 4. Given x2+10x-100=0, to destroy the second term. 10 Thus, -- +5; let y−5=x, and proceed. 5. To exterminate the second term from x3-3x2+4x-5=0, Thus, -3 3 -=—1, let y+1=x, and proceed. 6. Let the second term be taken away from the equation x++24x3-12x2+4x-30=0". 7. To take away the second term from the equation x350 x2+40 x3-30 x2 +20 x—10—o. 38. To multiply the roots of an equation by any given quantity, that is, to transform it into another, the roots of which will be any proposed multiple of those of the given equation. RULE 1. Take some new letter as before, and divide it by the given multiplier. II. Substitute the quotient and its powers, for the unknown quantity and its powers, in the given equation, and an equation equation, but these transformations being less useful and more difficult than the above, we have in the text omitted the rules: in general, to take away the second term requires the solution of a simple equation; to take away the third term, a quadratic; the fourth term, a cubic; and the nth term requires the solution of an equation of n-1 dimensions. See the note below. 8 This contraction consists in the reducing of the fractional coefficients of the same powers of y to a common denominator, and then adding or subtracting, according to the signs; putting the coefficients of the same power of y under the vinculum, &c. &c. b In like manner, to take away the third term from the equation x3-ax3 +bx-c=o, we assume y+e=x, where e must be taken such that (supposing n=the index of the highest power of x) n. e2 —n-1. n-1 ae+b=o. In which case a quadratic is to be solved; and in general, to take out the mth term, by this method, an equation of m-1 dimensions must be solved, as was observed in a preceding note. See Wood's Algebra, p. 141. will thence arise, whose roots are the proposed multiple of those of the given equation. RULE I. Assume some new letter as before, and place the given quantity under it, for a denominator. II. Substitute this fraction and its powers, for the unknown quantity and its powers respectively, in the given equation, and a new equation will arise, having its roots respectively equal to the given equation multiplied by the given quantity'. EXAMPLES.-1. To transform the equation x2+5x-2=0, into another, the roots of which are 10 times as great as those of the given equation. y2 y 2 Whence x2+5x −2=; + --2=0, that is, y2+50y-200 100 2 =0, the equation required. 2. Let the roots of 3 x3-12x2+15x-21=o, be multiplied by 3. i This rule requires neither proof nor explanation; it is sometimes useful for freeing an equation from fractions and radical quantities. * Hence it appears, that to multiply the roots of an equation by any quantity, we have only to multiply its terms respectively by those of a geometrical progression, the first term of which is 1, and the ratio the multiplying quan IV. When the last term admits of a great number of divisors, it will be convenient to transform the given equation into another, (Art. 35, 36.) the last term of which will have fewer divisors. EXAMPLES.-1. Let x3-2x2-5x+6=0, be given, to find its integral roots by this method. First, the divisors of the last term 6, are +1,−1, +2,−2, +3,−3,+6, and −6; now +1 being substituted for x in the given equation, it becomes +1-2-5+6=0; wherefore +1 is a root. Next, let-1 be substituted, and the equation becomes —1—2 +5+6=8; wherefore -1 is not a root. Thirdly, let +2 be substituted, and the equation becomes 8-8-10+6=-4; wherefore +2 is not a root. Fourthly, let-2 be substituted, and the equation becomes −8−8+10+6=0; wherefore -2 is a root. Fifthly, let +3 be substituted, and the equation will then beeome+27—18—15+6=0; wherefore +3 is likewise a root. Thus, the three roots of the given equation are +1,−2, and +3; and it is plain there can be no more than three roots, since the equation arises no higher than the third degree; consequently there is no necessity to try the remaining divisors. 2. Given ** — 6 x2. 16x+21=0, to find the roots. The divisors of the last term 21, are +1,−1, +3,−3, +7, −7, +21, and −21; these being successively substituted for x, we shall have -7 +21 -21 +2401- ·294+112+21=2240 +194481-2646—336+21=191520 +194481-2646+336+21=192192| Wherefore +1 and +3 are the only roots which can be found by this method; the two remaining roots are therefore impossible, -being-2+√3. 3. Given - 4 x3 — 19 x2 + 106x-120=o, to find the roots. Since the last term 120 has a great number of divisors, it will be proper to transform the equation into another, whose absolute term will have fewer divisors; in order to which, let x=y+2, then (Art. 36.) Here the last term vanishing, the number assumed, viz. +2, is one of the roots of the original equation, (Art. 33. note,) and the transformed equation being divisible by y, will thereby be reduced one dimension lower: thus, y3 +4y2-19y+14=0; the divisors of the last term 14, are +1,-1,+2,−2, +7,−7, +14,-14; each of these being substituted for y in the last equation, +1, +2, and -7 are found to succeed, they are therefore the roots of the transformed equation y3 +4 y2—19 y +14=0; wherefore, since x=y+ 2, three of the roots of the original equation will be (1+2=) 3, (2+2=) 4, and (−7+2=)—5, which with the number 2 assumed above, give +2, +3,+4, and -5, for the four roots required. 4. Given x3-3 ax2-4 a2x + 12 a3=o, to find the roots. The numeral divisors of the last term are +1,−1,+2,−2, +3,-3, +4,-4,+6,−6,+12, and -12; and of these,+2,-2 and-3 are found to succeed; wherefore the roots are +2 a,−2 a, and -3 a. 5. Required the roots of x2+x-12=o? Ans. 3, and —4. 6. What are the roots of x3+4x2+x-6=o? Ans. 1,-2, and -3. 7. What are the roots of r3 + 2x2-19x-20=o? Ans. -1, -4, and +5. 8. Required the roots of -14 x2+31x+126=o? -2,+7, and +9. Ans. 9. What are the roots of x^-15 x2+10x +24=o? Ans. -1, +2,+3, and 4. 10. Required the roots of x+4x2-7x-10=o? 48. SIR ISAAC NEWTON'S METHOD OF DISCOVERING THE ROOTS OE EQUATIONS BY MEANS OF DIVISORS. RULE I. For the unknown quantity in the given equation, substitute three or more terms of the arithmetical progression 2, 1, 0,−1,—2, &c. and let these terms be placed in a column one under the other. II. Substitute each number in this column successively for the unknown quantity in the proposed equation; collect all the terms of the equation arising from each substitution into one sum, and let this sum stand opposite the number substituted from whence it arises: these sums will form a second column. III. Find all the divisors of the sums, and place them in lines opposite their respective sums: these will form a third column. IV. From among the divisors collect one or more arithmetical progressions, the terms of which differ either by unity, or by some divisor of the coefficient of the highest power of the unknown quantity, observing to take one term only (of each progression) out of each line of the divisors: each of these progressions will form an additional column. V. Divide that term of the progression thus found, (or of each progression, if there be more than one,) which stands against O in the assumed progression, by the common difference of the terms of the former; and if the progression be increasing, prefix the sign + to the quotient; but if it be decreasing, prefix the sign - this quotient will be a root of the equation. Hence there will be as many roots found by this method, as there are progressions obtained from the divisors. EXAMPLES.-1. Given x2-2x-24=0, to find the values of r. The left hand column is the assumed progression, the terms of which are substituted successively for x in the given equation: first, by substituting 2 |