3. Given x4-4 x3 — 19 x2 + 106 x—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.) x+y+8y+24y2+ 32y+ 16 4x3-4 y3-24 y2 - 48 y— 32 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=0, 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? and -3. Ans. 1,−2, 7. What are the roots of x+2x2-19x-20=o? Ans.—1, -4, and +5. 8. Required the roots of 2-14x2+31x+126=o? Ans. -2,+7, and +9. 9. What are the roots of r1- 15 x2+10x +24=0? 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 x. 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 for r, the equation amounts to -24, which is the result in this case; this I put in the second column, and its divisors 1, 2, 3, 4, 6, &c. in the third. Secondly, I substitute 1 for x, and the whole equation amounts to -25, which is the second result, and its divisors are 1, 5, and 25. Thirdly, by substituting 0 for x, the result is —24, and its divisors 1, 2, 3, 4, 6, &c. as in the first case. Fourthly, by substituting -I for x, the result is -21, and its divisors are 1, 3, 7, and 21. Fifthly, by substituting -2, the result is -16, the divisors of which are 1, 2, 4, 8, and 16. Sixthly, I try to obtain a progression, by taking one number out of each line of the divisors: and first I try for an increasing one; the only one that can be found is 4, 5, 6, 7, and 8, viz. 4 out of the first line, 5 out of the second, 6 out of the third, 7 out of the fourth, and out of the fifth; these numbers constitute the fourth column. Seventhly, I try for a decreasing progression, and (proceeding as before) find that 6, 5, 4, 3, and 2, which constitute the fifth column, is the only one that can be obtained. Eighthly, the number 6 and 4, standing opposite the 0 in the assumed progres sion, divided by the common difference 1, gives 6 and 4 for the roots of the equation. The former being a term of the increasing progression, must have + prefixed to it; the latter being a term of the decreasing progression, must have prefixed; wherefore the roots are + 6 and −4. 2. Given x3-6 x2 -7x+60=0, to find the roots. Proceeding as before, I obtain three progressions, two increasing, and one decreasing, and the numbers 4, 5, and 3, standing opposite the 0, being divided by 1 the common difference, the quotients are the roots, viz. + 4 and +5 in the increasing progressions, and -3 in the decreasing one. 3. Given x3- x2 — 10x+6=o, to find the roots. Substitutions. Results. Divisors. Progressions. Here we can derive only one progression, and that a decreasing one; wherefore the only root discovered by this method is -3: but by means of this root the given equation may be depressed to a quadratic, (Art. 33.) and the two remaining roots found by the known rule for quadratics; thus, since x+3=0, dividing the proposed equation by this, we obtain x+3 2=0, the two roots of which are (2± √2.=) 3.4142135624 and .5857864376. 4. Required the roots of 6 xa—20 x3 — 12 x3 — 11 x—20=o ? Substit. Results. Divisors. 2 -154 |1, 2, 7, 11, 14, 22, 77, 154 Prog. 2 +210 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, &c. 6 2 Here we obtain only one progression, consequently +4 is the only root found. 5. Given x+x3 — 29 x2 —9x+180=0, to find the roots. Here are four progressions, two increasing and two decreasing, and the roots are 3, 4,-3, and — 5. 6. Required the roots of x2-x-12=o? Ans.+4 and −3. 7. Required the roots of x3+2x2-23x-60=o? Ans. +5, -4, and -3. 8. What are the roots of 2 x3-5 x2+4x-10=o? swer, one root +24. An 9. Required the roots of x3-3x2-46x-72-o? Ans. +9, -2, and -4. 10. To find the roots of r3—6 x2+10x-8=o? RECURRING EQUATIONS. 49. A recurring equation is one having the sign and coefficient of any term, reckoning from the beginning of the equation, the same with those of the term equally distant from the end; and its roots are of the form a, or the recipro cals of one another. - 1 b, 50. If the recurring equation be of an odd number of dimensions, +1 or is a root; and the equation may be depressed to one of an even number of dimensions. (Art. 33.) Thus, let x3-2x2+1=0; + 1 is evidently one root; This equation x-x-1=0, being resolved by the rule for quadratics, its roots will be found to be 1±√5 2 COR. Hence, a cubic equation of the form x+px2+px+1 may always be reduced to a quadratic, and its roots found. 51. If the given equation be of even dimensions above a quadratic, its roots may be found by means of an equation of half the number of dimensions. Thus, by supposing the equation to be the product of the fac 1 1 tors x-a.x— — x-b.x ———, &c. by actual multiplication, and putting m=a+· a' b =0+, α &c. we obtain x2 -mx+1, x2 —nx +1, &c. wherefore by multiplying these quadratic factors together, and equating the coefficients of each term of the product, with that of the corresponding term of the given equation, the values of m and n will be readily found: and since for every single value of m there will be two values of x, it follows that the equation for finding m will be of but half the number of dimensions necessary for finding the value of x by other methods. EXAMPLES.-1. Let x-3x 3+2x2−3x+1=o be the proposed equation. Assume the product (x2—mx+1.x2—nx+1=) x1 —m+n.x3 + mn+2.x2 — m+n.x+1=the proposed equation: then making the coefficients of like powers of x in this product and the given equation equal, we shall have m+n=3, and mn+2=2, or mn=o; wherefore, if n=o, then m=3, and the two equations x3-mx +1=0, and x2-nx+1=0, become respectively x2-3x+1 3+ √5 =0, and x3+1=0; from the former of these x= 2 =) |