Elements of Algebra

584 ADDITIONS. CHAP. IX. $(x + \alpha y) \times (x + \beta y)$, which must be a real product; being equal to $x^2 + (\alpha + \beta)xy + \alpha\beta y^2$. Now, we have $\alpha + \beta = a$, and $\alpha\beta = b$, from the nature of the equation, $s^2 - as + b = 0$; therefore we shall have this formula of the second degree, $x^2 + axy + by^2$, which is composed of the two factors, $x + \alpha y$, and $x + \beta y$. It is evident, that if we have a similar formula, $x^{\prime 2} + ax'y' + by^{\prime 2}$, and wish to multiply them, the one by the other, we have only to multiply together the two factors $x + \alpha y$, $x' + \alpha y'$, and also the other two factors $x + \beta y$, $x' + \beta y'$, and then the two products, the one by the other. Now, the product of $x + \alpha y$ by $x' + \alpha y'$ is, $xx' + \alpha(xy' + yx') + \alpha^2 yy'$; but since $\alpha$ is one of the roots of the equation, $s^2 - as + b = 0$, we shall have $\alpha^2 - a\alpha + b = 0$; whence, $\alpha^2 = a\alpha - b$; and, substituting this value of $\alpha^2$, in the preceding formula, it will become, $xx' - byy' + \alpha(xy' + yx' + ayy')$; so that, in order to simplify, making $X = xx' - byy'$ $Y = xy' + yx' + ayy'$, the product of the two factors $x + \alpha y$, $x' + \alpha y'$, will be $X + \alpha Y$; and, consequently, of the same form as each of them. In the same manner, we shall find, that the product of the two other factors, $x + \beta y$, and $x' + \beta y'$, will be $X + \beta Y$; so that the whole product will be $(X + \alpha Y) \times (X + \beta Y)$; that is, $X^2 + aXY + bY^2$, which is the product of the two similar formulæ, $x^2 + axy + by^2$, and $x^{\prime 2} + ax'y' + by^{\prime 2}$. If we wished to have the product of these three similar formulæ, $x^2 + axy + by^2$, $x^{\prime 2} + ax'y' + by^{\prime 2}$, $x^{\prime\prime 2} + ax''y'' + by^{\prime\prime 2}$, we should only have to find that of the formula, $X^2 + aXY + bY^2$, by the last, $x^{\prime\prime 2} + ax''y'' + by^{\prime\prime 2}$; and it is evident, from the foregoing formulæ, that, by making $x' = Xx'' - bYy''$, $y' = Xy'' + Yx'' + aYy''$,