Elements of Algebra

CHAP. XI. OF ALGEBRA. 261 $= 40x^2 + 8240$. Now, with this sum they gain $x$ per cent; so that the whole gain is $\frac{40x^3}{100} + \frac{8240x}{100} = \frac{2}{5}x^3 + \frac{824}{10}x = \frac{2}{5}x^3 + \frac{412}{5}x$. From which sum each receives $10x$, and consequently they all together receive $10x^2$, leaving a remainder of 224; the profit must therefore have been $10x^2 + 224$, and we have the equation $\frac{2x^3}{5} + \frac{412x}{5} = 10x^2 + 224$. Multiplying by $5$ and dividing by $2$, we have $x^3 + 206x = 25x^2 + 560$, or $x^3 - 25x^2 + 206x - 560 = 0$: the first however, will be more convenient for trial. Here the divisors of the last term are $1, 2, 4, 5, 7, 8, 10, 14, 16, \ldots$ and they must be taken positively; because in the second form of the equation the signs vary three times, which shews that all the three roots are positive. Here, if we first try $x = 1$, and $x = 2$, it is evident that the first side will become less than the second. We shall therefore make trial of other divisors. When $x = 4$, we have $64 + 824 = 400 + 560$, which does not satisfy the terms of the equation. If $x = 5$, we have $125 + 1030 = 625 + 560$, which likewise does not succeed. But if $x = 7$, we have $343 + 1442 = 1225 + 560$, which answers to the equation; so that $x = 7$ is a root of it. Let us now seek for the other two, by dividing the second form of our equation by $x - 7$. $x - 7) x^3 - 25x^2 + 206x - 560 (x^2 - 18x + 80$ $\underline{x^3 - 7x^2}$ $-18x^2 + 206x$ $\underline{-18x^2 + 126x}$ $80x - 560$ $\underline{80x - 560}$ $0$. Now, making this quotient equal to nothing, we have $x^2 - 18x + 80 = 0$, or $x^2 - 18x = -80$; which gives $x = 9 \pm 1$, so that the two other roots are $x = 8$; or $x = 10$. This question therefore admits of three answers. According to the first, the number of merchants is $7$; according to