Elements of Algebra

CHAP. XI. OF ALGEBRA. 257 roots must be divisors of the last term 56, they must be included in the numbers $\pm 1, 2, 4, 7, 8, 14, 28, 56$. Let us, therefore, make $x=2$, and we shall have $8+4-68+56=0$; whence we conclude that $x=2$ is a positive root, and that therefore $x-2$ is a divisor of the equation; by means of which we easily find the two other roots: for, actually dividing by $x-2$, we have $x-2) x^3 + x^2 - 34x + 56 (x^2 + 3x - 28$ $x^3 - 2x^2$ —————— $3x^2 - 34x$ $3x^2 - 6x$ —————— $-28x + 56$ $-28x + 56$ —————— $0$. And making the quotient $x^2+3x-28=0$, we find the two other roots; which will be $x = -\frac{3}{2} \pm \sqrt{(\frac{9}{4}+28)} = -\frac{3}{2} \pm \frac{11}{2}$; that is, $x=4$; or $x=-7$; and taking into account the root found before, namely, $x=2$, we clearly perceive that the equation has two positive, and one negative root. We shall give some examples to render this still more evident. 727. Question 1. There are two numbers, whose difference is 12, and whose product multiplied by their sum makes 14560. What are those numbers? Let $x$ be the less of the two numbers, then the greater will be $x+12$, and their product will be $x^2+12x$, which multiplied by the sum $2x+12$, gives $2x^3+36x^2+144x=14560$; and dividing by 2, we have $x^3+18x^2+72x=7280$. Now, the last term 7280 is too great for us to make trial of all its divisors; but as it is divisible by 8, we shall make $x=2y$, because the new equation, $8y^3+72y^2+144y=7280$, after the substitution, being divided by 8, will become $y^3+9y^2+18y=910$; to solve which, we need only try the divisors 1, 2, 5, 7, 10, 13, &c. of the number 910: where it is evident, that the three first, 1, 2, 5, are too small; beginning therefore with supposing $y=7$, we immediately find that number to be one of the roots; for the substitution gives $343+441+126=910$. It follows, therefore, that $x=14$; and the two other roots will be found by dividing $y^3+9y^2+18y-910$ by $y-7$, thus: s