Elements of Algebra

CHAP. VIII. OF ALGEBRA. 367 $1 - x$, that the quotient will be $1 - x^2$, that this quotient is composed of the factors $(1 + x) \times (1 - x)$; and, lastly, that our formula, $1-x-x^2+x^3=(1-x)\times(1+x)\times(1-x)=(1-x)^2\times(1+x)$. Now, as it must be a $\square$ [square], and as a $\square$, when divisible by a $\square$, gives a $\square$ for the quotient*, we must also have $1+x = \square$; and, conversely, if $1 + x$ be a $\square$, it is certain that $(1-x)^2 \times (1+x)$ will be a square; we have therefore only to make $1+x=n^2$, and we immediately obtain $x=n^2-1$. If this circumstance had escaped us, it would have been difficult even to have determined only five or six values of $x$ by the preceding methods. 125. Hence we conclude, that it is proper to resolve every formula proposed into factors, when it can be done; and we have already shewn how this is to be done, by making the given formula equal to $0$, and then seeking the root of this equation; for each root, as $x=f$, will give a factor $f-x$; and this inquiry is so much the easier, as here we seek only rational roots, which are always divisors of the known term, or the term which does not contain $x$. 126. This circumstance takes place also in our general formula, $a + bx + cx^2 + dx^3$, when the first two terms disappear, and it is consequently the quantity $cx^2 + dx^3$ that must be a square; for it is evident, in this case, that by dividing by the square $x^2$, we must also have $c + dx$ a square; and we have therefore only to make $c + dx = n^2$, in order to have $x = \frac{n^2-c}{d}$, a value which contains an infinite number of answers, and even all the possible answers. 127. In the application of the first of the two preceding methods, if we do not choose to determine the letter $p$, for the sake of removing the second term, we shall arrive at another irrational formula, which it will be required to make rational. For example, let $f^2 + bx + cx^2 + dx^3$ be the formula proposed, and let its root $= f + px$. Here we shall have $f^2 + bx + cx^2 + dx^3 = f^2 + 2fpx + p^2x^2$, from which the first terms vanish; dividing, therefore by $x$, we obtain * The mathematical student, who may wish to acquire an extensive knowledge of the many curious properties of numbers, is referred, once for all, to the second edition of Legendre's celebrated Essai sur la Theorie des Nombres; or to Mr. Barlow's Elementary Investigation of the same subject.