Elements of Algebra

286 ELEMENTS SECT. IV. Let there be proposed the general equation $y^4 + ay^3 + by^2 + cy + d = 0$. If we add to $y$ the fourth part of the coefficient of the second term, or $\frac{1}{4}a$, and write, instead of the sum, a new letter $x$, so that $y + \frac{1}{4}a = x$, and consequently $y = x - \frac{1}{4}a$: we shall have $y^2 = x^2 - \frac{1}{2}ax + \frac{1}{16}a^2$, $y^3 = x^3 - \frac{3}{4}ax^2 + \frac{3}{16}a^2x - \frac{1}{64}a^3$, and, lastly, as follows: $y^4 = x^4 - ax^3 + \frac{3}{8}a^2x^2 - \frac{1}{16}a^3x + \frac{1}{256}a^4$ $ay^3 = ax^3 - \frac{3}{4}a^2x^2 + \frac{3}{16}a^3x - \frac{1}{64}a^4$ $by^2 = bx^2 - \frac{1}{2}abx + \frac{1}{16}a^2b$ $cy = cx - \frac{1}{4}ac$ $d = d$ Or, $\left. \begin{aligned} & x^4 + 0 - \frac{3}{8}a^2x^2 + \frac{1}{8}a^3x - \frac{5}{256}a^4 \\ & \qquad \quad bx^2 - \frac{1}{2}abx + \frac{1}{16}a^2b \\ & \qquad \qquad \quad cx - \frac{1}{4}ac \\ & \qquad \qquad \qquad \quad d \end{aligned} \right\} = 0$. We have now an equation from which the second term is taken away, and to which nothing prevents us from applying the rule before given for determining its four roots. After the values of $x$ are found, those of $y$ will easily be determined, since $y = x - \frac{1}{4}a$. 780. This is the greatest length to which we have yet arrived in the resolution of algebraic equations. All the pains that have been taken in order to resolve equations of the fifth degree, and those of higher dimensions, in the same manner, or, at least, to reduce them to inferior degrees, have been unsuccessful: so that we cannot give any general rules for finding the roots of equations, which exceed the fourth degree. The only success that has attended these attempts has been the resolution of some particular cases; the chief of which is that, in which a rational root takes place; for this is easily found by the method of divisors, because we know that such a root must be always a factor of the last term. The operation, in other respects, is the same as that we have explained for equations of the third and fourth degree. 781. It will be necessary, however, to apply the rule of Bombelli to an equation which has no rational roots. Let there be given the equation $y^4 - 8y^3 + 14y^2 + 4y - 8 = 0$. Here we must begin with destroying the second term, by adding the fourth of its coefficient to $y$, supposing $y - 2 = x$, and substituting in the equation, instead of $y$, its new value $x+2$, instead of $y^2$, its value $x^2 + 4x + 4$; and doing the same with regard to $y^3$ and $y^4$, we shall have,