Elements of Algebra

CHAP. XI. OF ALGEBRA. 255 for $x=2$ being one of the roots, $x-2$ is a factor of the equation, and we have only to seek the other factor by means of division as follows: $x-2) x^3 - x - 6 (x^2 + 2x + 3$ $\underline{x^3 - 2x^2}$ $2x^2 - x - 6$ $\underline{2x^2 - 4x}$ $3x - 6$ $\underline{3x - 6}$ $0$. Since, therefore, the formula is represented by the product $(x-2) \times (x^2 + 2x + 3)$, it will become $=0$, not only when $x-2=0$, but also when $x^2+2x+3=0$: and, this last factor gives $x^2+2x=-3$; consequently, $x = -1 \pm \sqrt{-2}$; and these are the other two roots of our equation, which are evidently impossible, or imaginary. 723. The method which we have explained, is applicable only when the first term $x^3$ is multiplied by 1, and the other terms of the equation have integer coefficients; therefore, when this is not the case, we must begin by a preparation, which consists in transforming the equation into another form having the condition required; after which, we make the trial that has been already mentioned. Let there be given, for example, the equation $x^3 - 3x^2 + \frac{11}{4}x - \frac{3}{4} = 0$. As this contains fourth parts, let us make $x = \frac{y}{2}$, which will give $\frac{y^3}{8} - \frac{3y^2}{4} + \frac{11y}{8} - \frac{3}{4} = 0$, and, multiplying by 8, we shall obtain the equation $y^3 - 6y^2 + 11y - 6 = 0$, the roots of which are, as we have already seen, $y=1, y=2, y=3$; whence it follows, that in the given equation, we have $x = \frac{1}{2}, x=1, x=\frac{3}{2}$. 724. Let there be an equation, where the coefficient of the first term is a whole number but not 1, and whose last term is 1; for example, $6x^3 - 11x^2 + 6x - 1 = 0$.