Elements of Algebra

258 ELEMENTS SECT. IV. $y-7) y^3 + 9y^2 + 18y - 910 (y^2 + 16y + 130$ $y^3 - 7y^2$ --- $16y^2 + 18y$ $16y^2 - 112y$ --- $130y - 910$ $130y - 910$ --- $0$. Supposing now this quotient $y^2 + 16y + 130 = 0$, we shall have $y^2 + 16y = -130$, and thence $y = -8 \pm \sqrt{-66}$; a proof that the other two roots are impossible. The two numbers sought are therefore 14, and $(14 + 12) = 26$; the product of which, 364, multiplied by their sum, 40, gives 14560. 728. Question 2. To find two numbers whose difference is 18, and such, that their sum multiplied by the difference of their cubes, may produce 275184. Let $x$ be the less of the two numbers, then $x+18$ will be the greater; the cube of the first will be $x^3$, and the cube of the second $x^3 + 54x^2 + 972x + 5832$; the difference of the cubes $54x^2 + 972x + 5832 = 54(x^2 + 18x + 108)$, which multiplied by the sum $2x + 18$, or $2(x+9)$, gives the product $108(x^3 + 27x^2 + 270x + 972) = 275184$. And, dividing by 108, we have $x^3 + 27x^2 + 270x + 972 = 2548$, or $x^3 + 27x^2 + 270x = 1576$. Now, the divisors of 1576 are 1, 2, 4, 8, &c. the two first of which are too small; but if we try $x=4$, that number is found to satisfy the terms of the equation. It remains, therefore, to divide by $x-4$, in order to find the two other roots; which division gives the quotient $x^2 + 31x + 394$; making therefore $x^2 + 31x = -394$, we shall find $x = -\frac{31}{2} \pm \sqrt{(\frac{961}{4} - \frac{1576}{4})}$; that is, two imaginary roots. Hence the numbers sought are 4, and $(4+18) = 22$. 729. Question 3. Required two numbers whose difference is 720, and such, that if the less be multiplied by the square root of the greater, the product may be 20736.