с The value of in the last step is also readily ascertained in 2 all equations of the form x2+(2b+1)x2+(b2+b+c)x+(b+1)c=0...... (B), which is what the above general form becomes when a=b+1, for in this case is found by dividing the final term by the co 2 efficient of x2 increased by 1. In the class of equations comprehended in the form (A) the coefficient of x2, when a is an integer, is always even: in the class (B) this coefficient, when b is an integer, is always odd. Several other forms besides those here deduced may be obtained from the more general one discussed in the foregoing example. We have selected those here noticed for the purpose of showing that, as c is often discoverable by inspection, a variety of distinct classes of equations may be easily resolved into rational factors by the present method: this, however, as already observed, is not the chief advantage of it; it is a process necessary to facilitate the solution of numerical equations in general, as we shall in some degree show in the next chapter. EXAMPLES FOR EXERCISE. Resolve the following equations into rational factors. (1) x3-x2+2=0. (9) x3-6x2+10x-3=0. (2) x3—1x2+1=0. (4) x3-3x2+4x-12=0. (6) x3-3x2+5x-3=0. (8) x+2x2+3x+2=0. (10) 2x+8x2+9x+9=0. (11) 4x3+16x+23x+20=0. (12) x3-px2+qx-pq=0. 88. We shall now give an illustration or two of the application of the foregoing process to the solution of irrational equations. (1) Given x2-2x-11±6√√(x2-2x+5)=0 to find a. x2—2x—11—6.2±6√ (11+36 +5)=0; that is, x2-2x-29±6.5=0..x2—2x=59, or x2—2x——1. From the first of these, x=1±√60=1±2√/15; From the second, x=1 and x=1. In this example the proposed equation is in reality a combination of two equations, its twofold character being indicated by the double sign before the radical; and, in consequence, each of the four resulting values of x is admissible. If the square root were restricted by previous stipulation to one of these signs in examples of this kind, only one, or at most two of the resulting values will frequently satisfy the equation, and sometimes no value whatever will satisfy it. The resulting values are always such as to satisfy the rational equation to which we are led upon clearing the original of radicals; that is, upon multiplying our two component factors together; and whatever reduces either factor to zero will, of course reduce the product of both to zero. Whenever, therefore, the sign of the original radical is restricted, we must, in general, actually substitute the resulting values, one after another, in the proposed equation, before we can determine which are admissible, under the stipulated restriction, and which not. (See the "Treatise on Algebra," p. 130.) The learner will not fail to observe that the successive pairs of factors, derived in the way here taught, are all different from one another, while the product of each pair is invariable: and therefore that a value of x, which renders one factor zero only when the radical is plus, may render a factor in another pair zero, only when the radical is minus: so that, although the sign of the radical may be fixed at the outset, it will not do to exclude the other sign from the several steps of the work: all that can be affirmed, without actual substitution, is, that a value which renders one or other of the factors of a pair zero, will always render one or other of the factors of every pair zero. The following is an example of this: (2) Given 2x-5±√(x2—7)=0, to find x. .. 3x-8=0, or x-4=0 .. x=§, or x=4. The first of these values of x arises from taking the sign of the radical plus, but it satisfies the proposed equation only when the sign of the radical is minus; the second value also satisfies the proposed only when the sign of the radical is minus; consequently, no value whatever will satisfy it if the radical be exclusively plus. The two values referred to are both of them roots of the quadratic equation 3x2-20x+32=0, obtained by multiplying the original pair of equations, or either of the derived pairs together, or by clearing the radical from either in the ordinary way. (3) 2x2+3x+3±5√(2x2+3x+9)=0 -5? — ±5√ (−3+25 +9)= 2x2+3x+3—5?-±5√ -3. 4 that is, 2x2+3x+8=0, or 2x2+3x-27=0; two equations, which the learner may easily solve. CHAPTER IX. ON THE GENERAL SOLUTION OF NUMERICAL EQUATIONS OF THE THIRD AND FOURTH DEGREES. 89. WE have already noticed (page 185) the distinction between equations with common figures for the coefficients of the terms and those in which the coefficients are general symbols, admitting of any interpretation whatever. The former are called numerical equations, and the solution of them may be cffected by methods sufficiently general to comprehend the entire class without any exception. The solution of equations with literal coefficients depends upon the discovery of an algebraical formula which shall exhibit the values of the unknown quantity in terms of the general coefficients. Beyond literal equations of the second degree no such formula has been obtained adequate to meet all cases and to express the roots, when the literal coefficients are replaced by numbers, in finite terms. The solution of equations with figures for the coefficients depends not upon a formula but upon a purely arithmetical process, by which, as in the process for the square root of a number, the figures of the unknown quantity are determined one after another. The operation, being thus entirely restricted to the figures of arithmetic, could not by any possibility be replaced by an algebraical formula; because, from the very nature of general symbols, a formula of algebra cannot be so narrowed as to include the arithmetical values only of the letters, and to reject the symbolical; and, as all attempts to discover a formula for the general solution of an equation of a high degree have proved unsuccessful, there is strong reason to conclude that, except in the arithmetical sense, the problem is impossible. Although, as just said, the solution of numerical equations in general is the work of pure arithmetic, yet, as in other inquiries, it is algebra that suggests the operations, directs the course they are to take, and prescribes the most efficient way of conducting them. It so happens that this sort of aid, in the matter before us, is chiefly comprised in a single precept or rule, and that in reference only to so simple a thing as the division of a polynomial by a binomial of the form x-a. As we do not propose here to extend our inquiries beyond equations of the fourth degree, we shall confine the investigation of this rule to that limit; but the learner will easily see that its application is general. 90. To divide a polynomial of the fourth degree A,x+А3.æ3 +А ̧x2+А ̧x+N by the binomial x-a; where a, N, A1, A2, &c. are numbers, either positive or negative. Let the quotient and remainder resulting from the division be represented by Q and R respectively: then, since the product of quotient and divisor, with the remainder added on, is always the dividend, we have the identity, A1x2+A ̧3+A, x2+A1x+N=(x—a)Q+R ......(1), which, by-the-bye, immediately leads to the truth established by actual process at pp. 53, 55, namely, that the remainder of the division is always what we should get by putting a for x in the dividend, as we see that the substitution of a for x reduces the above to A1a+Aa3+A,a2+A1a+N=R. As the quotient Q must evidently be an expression of the third degree, we may write Q = A'3 x3+A'2x2 + A',x+N' ......(2) 29 in which N', A'ı, A′2, and A'3, are unknown numbers now to be determined. Multiplying each member of this equation by x-a, and then adding R, we have (x-a) Q+R-A', a1+(A',—a A's) x3 + (A'-a A') x2+(N'-a A',) x-a N'+R. But the first mem(N'—a x—a ber of this is the same as the second member of (1); in other words, the second member of this and the first of (1) are identical, so that the corresponding coefficients in both must be the same in value. Hence the unknown coefficients in (2) become A'1⁄2-a A'3-A3 .: A'1⁄2=a A'3+A3 A'a A'2+A2 3 It follows, therefore, that the first coefficient A', in the quotient (2) is the same as the first coefficient A, in the proposed polynomial (1); that the second is found by multiplying A', by a, and adding the second in (1); and that generally every coefficient in Qis obtained by the same process of multiplying the preceding coefficient in Q by a, and adding to the product the corresponding coefficient in (1), and that this process extended up to N', the last coefficient, or the coefficient of x° in Q, furnishes the remainder R. Such is the mode of proceeding that algebra suggests whenever we have to divide a polynomial in x by x—a, the coefficients of the terms of the polynomial, and the quantity a being numerical. We shall now show the practical convenience of this method of proceeding by an example or two. (1) Required the quotient and remainder arising from the division of 3x-2x3 + 5 x2-7x-19 by x- -2. 3-257-19 6+8+26+38 4+13+19+19 hence the quotient is 3x3+4x2+13x+19, and the remainder 19. (2) Required the quotient and remainder arising from the division of 5x+8x2-6x+12 by x-4. 586+ 12 28+106+436 hence the quotient is 5x2+28x+106, and the remainder 436. (3) Required the quotient and remainder arising from the division of 4x+3x3-2x2-7x+13 by x+3. 4+ 3—2— 7+ 13 -9+25-82+259 hence the quotient is 4x3-9x2+25x-82, and the remainder 259. |