Elements of Algebra

248 ELEMENTS SECT. IV. the second degree, $x^2 \pm ax \pm b$, must always be resolvible into two factors, such as $(x \pm p) \times (x \pm q)$. For, if we took three factors, such as these, we should come to a quantity of the third degree; and taking only one such factor, we should not exceed the first degree. It is therefore certain, that every equation of the second degree necessarily contains two values of $x$, and that it can neither have more nor less. 704. We have already seen, that when the two factors are found, the two values of $x$ are also known, since each factor gives one of those values, by making it equal to $0$. The converse also is true, *viz.* that when we have found one value of $x$, we know also one of the factors of the equation; for if $x=p$ represents one of the values of $x$, in any equation of the second degree, $x-p$ is one of the factors of that equation; that is to say, all the terms having been brought to one side, the equation is divisible by $x-p$; and farther, the quotient expresses the other factor. 705. In order to illustrate what we have now said, let there be given the equation $x^2 + 4x - 21 = 0$, in which we know that $x=3$ is one of the values of $x$, because $(3 \times 3) + (4 \times 3) - 21 = 0$; this shews, that $x-3$ is one of the factors of the equation, or that $x^2+4x-21$ is divisible by $x-3$, which the actual division proves. Thus, $x-3) x^2+4x-21 (x+7$ $x^2-3x$ --- $7x-21$ $7x-21$ --- $0$. So that the other factor is $x+7$, and our equation is represented by the product $(x-3) \times (x+7) = 0$; whence the two values of $x$ immediately follow, the first factor giving $x=3$, and the other $x=-7$. --- CHAP. X. Of Pure Equations of the Third Degree. 706. An equation of the third degree is said to be *pure*, when the cube of the unknown quantity is equal to a known