Elements of Algebra

280 ELEMENTS SECT. IV. 769. Let us illustrate this rule by an example, and suppose that the equation $x^4 - 10x^3 + 35x^2 - 50x + 24 = 0$ was given. If we compare it with our general formula (at the end of Art. 767.), we have $a = -10$, $b = 35$, $c = -50$, $d = 24$; and, consequently, the equation which must give the value of $p$ is $8p^3 - 140p^2 + 808p - 1540 = 0$, or $2p^3 - 35p^2 + 202p - 385 = 0$. The divisors of the last term are 1, 5, 7, 11, \ldots; the first of which does not answer; but making $p = 5$, we get $250 - 875 + 1010 - 385 = 0$, so that $p = 5$; and if we farther suppose $p = 7$, we get $686 - 1715 + 1414 - 385 = 0$, a proof that $p = 7$ is the second root. It remains now to find the third root; let us therefore divide the equation by 2, in order to have $p^3 - \frac{35}{2} p^2 + 101p - \frac{385}{2} = 0$, and let us consider that the coefficient of the second term, or $\frac{35}{2}$, being the sum of all the three roots, and the first two making together 12, the third must necessarily be $\frac{11}{2}$. We consequently know the three roots required. But it may be observed that one would have been sufficient; because each gives the same four roots for our equation of the fourth degree. 770. To prove this, let $p = 5$; we shall then have, by the formula, $\sqrt{(\frac{1}{4}a^2 + 2p - b)}$, $q = \sqrt{(25 + 10 - 35)} = 0$, and $r = \frac{-50+50}{0} = \frac{0}{0}$. Now, nothing being determined by this, let us take the third equation, $r^2 = p^2 - d = 25 - 24 = 1$, so that $r = 1$; our two equations of the second degree will then be: 1. $x^2 = 5x - 4$, 2. $x^2 = 5x - 6$. The first gives the two roots $x = \frac{5}{2} \pm \sqrt{\frac{9}{4}}$, or $x = \frac{5 \pm 3}{2}$, that is to say, $x = 4$, and $x = 1$. The second equation gives $x = \frac{5}{2} \pm \sqrt{\frac{1}{4}} = \frac{5 \pm 1}{2}$, that is to say, $x = 3$, and $x = 2$. But suppose now $p = 7$, we shall have