Solve the cubic equation x^3 + 3x^2 + 2x - 504 = 0.

Step-by-step solution

1. 1

   Factor the variable part and read the structure. The first three terms share an $x$:

   $x^3+3x^2+2x=x\left(x^2+3x+2\right)=x(x+1)(x+2)$

   so the equation is

   $x(x+1)(x+2)=504$

   That is: find three consecutive integers whose product is $504$. Recognising this shape turns a cubic into a one-line search.

2. 2

   Search near the cube root. Three consecutive integers near $n$ multiply to roughly $n^3$, and $\sqrt[3]{504}\approx7.96$, so try the triple centred near $8$:

   $7\times8\times9=504\ \checkmark$

   Hence $x=7$ is a root. Confirming directly: $343+147+14-504=0$ ✓.

3. 3

   Depress the cubic by synthetic division. Dividing by $x-7$ with coefficients $1,\ 3,\ 2,\ -504$:

   $1\quad\to\quad 3+7=10\quad\to\quad 2+70=72\quad\to\quad -504+504=0$

   The zero remainder confirms the root, and the quotient is $x^2+10x+72$:

   $x^3+3x^2+2x-504=(x-7)\left(x^2+10x+72\right)$

4. 4

   Solve the quadratic factor. Its discriminant is

   $10^2-4(1)(72)=100-288=-188<0$

   so the remaining two roots are complex — meaning $x=7$ is the only real solution.

5. 5

   Simplify the complex roots. Since $\sqrt{-188}=i\sqrt{4\cdot47}=2i\sqrt{47}$:

   $x=\frac{-10\pm2i\sqrt{47}}{2}=-5\pm i\sqrt{47}$

6. 6

   Verify all three roots. Substituting $x=7$ gives exactly $0$ ✓. Substituting $-5\pm i\sqrt{47}$ gives $0$ to within $1.1\times10^{-14}$ in complex arithmetic ✓. As a structural check, the three roots sum to $7+(-5)+(-5)=-3=-b/a$ ✓ and multiply to $7\left(25+47\right)=504=-d/a$ ✓.