Equation -- Vieta's Formula

Relevant wiki: Vieta's Formula Problem Solving - Basic

Let the three integer roots of $x^3+kx^2-1329x-2007 = 0$ be $a$ , $b$ , and $c$ . Then by Vieta's formula, we have:

$\begin{cases} a+b+c = -k \\ ab+bc+ca = -1329 \\ abc = 2007 \end{cases}$

Since $2007 = 3^2\times 223$ , where both 3 and 223 are primes, the absolute values of the roots must be either $(|a|, |b|, |c|) = (3,3,223)$ or $(1, 9, 223)$ . Since $ab+bc+ca$ is negative, some of the roots must be negative. And as $abc$ is positive two of the roots must be negative. By try and error we note that $(-3, -3, 223)$ are the three roots, because $(-3)(-3) + (-3)(223)+(233)(-3) = -1329$ . Therefore, $k = - (a+b+c) = - (-3-3+223) = \boxed{-217}$ .

$Vieta’s$ $Formula$

$a$ , $b$ are the roots

(x-a)(x-b) = x2 - (a+b)x + ab

• x 1 + x 2 = -B/A

• x 1x 2 = C/A

$a$ , $b$ , $c$ are the roots

(x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+ac+bc)x - abc

\(Ax^3 + Bx^2 + Cx + D = 0

- x_1 + x_2 + x_3 = -B/A

- x_1 x_2 + x_1 x_3 + x_2 x_3 = C/A

- x_1x_2x_3 = -D/A\)

Main Solution

According to Vieta’s Formula $\begin{cases} a + b + c = -k \\ ab + ac + bc = -1329 \\ abc = 2007 \end{cases}$

Since 3^2 x 223 Equation 3 could be derived into $\begin{cases} a = +1, b = +9, c = +223 \\ a = +1, b = +3, c = +669 \\ a = +1, b = +1, c = +2007 \\ a = +3, b = +3, c = +223 \end{cases}$

Sub these roots into equation and get $a$ = - 3, $b$ = - 3, $c$ = 223

Therefore k = -217 is the correct answer