Since the equation can have no more than one positive root, we will only make trial of the remaining negative divisors, thus: Solve the following equations, which have one or more inte gral roots: x+4x3-x2-16x-12=0. x1+4x3-22x2-4x+21=0. 18. x2+2x3 −7x2-8x+12=0. 17. CARDAN'S MEthod for the Solution of CUBIC EQUATIONS. 588. The general type of a cubic equation is x3+Px2+Qx+R=0, but as explained in Art. 556 this equation can be reduced to the simpler form x2+qx+r=0, which we shall take as the standard form of a cubic equation. 589. To solve the equation x3+qx+r=0. Let x=y+z; then x3=y3+z8+3yz(y + z) = y3+z8+3yzx, and the given equation becomes y3+z3+(3yz+q)x+r=0. At present y, z are any two quantities subject to the condition that their sum is equal to one of the roots of the given equation; if we further suppose that they satisfy the equation 3yz+q=0, they are completely determinate. We thus obtain 4 27 We obtain the value of x from the relation x=y+z; thus The above solution is generally known as Cardan's Solution, as it was first published by him in the Ars Magna, in 1545. Cardan obtained the solution from Tartaglia; but the solution of the cubic seems to have been due originally to Scipio Ferreo, about 1505. In this solution we assume x=y+z, and from (2) find 2=- 1, hence to solve a cubic equation of the form x3+qx+r=0 Зу Example. Solve the equation x3-15x=126. Dividing the given equation x3-15x-126-0 by x-6, we obtain the depressed equation x2+6x+21=0, the roots of which are −3+2√--3, and −3-2√−3. Thus the roots of x3-15x=126 are 6, −3+2√3, and −3 -2√-3. 590. The incommensurable roots of an equation cannot be found exactly. If, however, a sufficient number of the initial figures of the root have been found to distinguish it from the other roots we may carry the approximation to the exact value to any required degree of accuracy by a method first published in 1819 by W. G. Horner. HORNER'S METHOD OF APPROXIMATION. 591. Let it be required to solve the equation By Sturm's Theorem there are 3 real roots and one of them lies between 1 and 2; we will find its value to four places of decimals, which will sufficiently illustrate the method. Diminishing the roots of the equation by 1 [Arts. 554, 555], we have Equation (1) has a root between 1 and 2. The roots of equation (2) are each less by 1 than those of equation (1); hence equation (2) has a root between 0 and 1. This root being less than unity the higher powers of y are each less than y. Neglecting them, we obtain an appropriate value of y from -5y+1=0, or y=2. Diminishing the roots of (2), the first transformed equation, by 2, we have Equation (2) has a root between 2 and 3; the roots of equation (3) are less by 2 than those of equation (2); hence equation (3) has a root between 0 and 1. Neglecting the higher powers of z, we have -4.882+0080, or z=·001. Diminishing the roots of (3), the second transformed equa tion, by 001, we have The transformed equation is v3+603v2-4.878797v+003120601. .......... (4). Equation (3) has a root between 001 and 002; the roots of equation (4) are less than those of equation (3) by 001; therefore equation (4) has a root between 0 and 001. Solving -4.878797v+003120601=0, we have v=" =0006. Diminishing the roots of (4), the third transformed equation, by 0006, we find this to be the correct figure for the fourth decimal place; hence the value of x to four places of decimals is 1.2016. Denoting the coefficients of successive transformed equations by (A), (B), (C), etc., the work is more compactly arranged thus: |