Elements of Algebra

CHAP. XIII. OF ALGEBRA. 273 754. But whenever the equation contains all the terms, it may be considered as the product of four factors. In fact, if we multiply these four factors together, $(x-p) \times (x-q) \times (x-r) \times (x-s)$, we get the product $x^4 - (p+q+r+s)x^3 + (pq+pr+ps+qr+qs+rs)x^2 - (pqr+pqs+prs+qrs)x + pqrs$; and this quantity cannot be equal to $0$, except when one of these four factors is $=0$. Now, that may happen in four ways; 1. when $x=p$; 2. when $x=q$; 3. when $x=r$; and 4. when $x=s$. Consequently, these are the four roots of the equation. 755. If we consider this formula with attention, we observe, in the second term, the sum of the four roots multiplied by $-x^3$; in the third term, the sum of all the possible products of two roots, multiplied by $x^2$; in the fourth term, the sum of the products of the roots combined three by three, multiplied by $-x$; lastly, in the fifth term, the product of all the four roots multiplied together. 756. As the last term contains the product of all the roots, it is evident that such an equation of the fourth degree can have no rational root, which is not a divisor of the last term. This principle, therefore, furnishes an easy method of determining all the rational roots, when there are any; since we have only to substitute successively for $x$ all the divisors of the last term, till we find one which satisfies the terms of the equation: and having found such a root (for example, $x=p$), we have only to divide the equation by $x-p$, after having brought all the terms to one side, and then suppose the quotient $=0$. We thus obtain an equation of the third degree, which may be resolved by the rules already given. 757. Now, for this purpose, it is absolutely necessary that all the terms should consist of integers, and that the first should have only unity for the coefficient; whenever, therefore, any terms contain fractions, we must begin by destroying those fractions; and this may always be done by substituting, instead of $x$, the quantity $y$, divided by a number which contains all the denominators of those fractions. For example, if we have the equation $x^4 - \frac{1}{2}x^3 + \frac{1}{3}x^2 - \frac{3}{4}x + \frac{1}{18} = 0$, as we find here fractions which have for denominators 2, 3, and multiples of these numbers, let us suppose $x = \frac{y}{6}$, and we shall then have $\frac{y^4}{6^4} - \frac{1}{2}\frac{y^3}{6^3} + \frac{1}{3}\frac{y^2}{6^2} - \frac{3}{4}\frac{y}{6} + \frac{1}{18} = 0$, T