Solve the cubic equation x³ + 2x² + 7x + 4 = 0.

Step-by-step solution

1. 1

   Test every rational candidate. By the rational root theorem the only possibilities are divisors of $4$:

   $f(1) = 14, \quad f(-1) = -2, \quad f(2) = 34, \quad f(-2) = -10, \quad f(4) = 128, \quad f(-4) = -56$

   None is zero, so the cubic is irreducible over $\mathbb{Q}$ and no exact factorisation by inspection is available.

2. 2

   Bracket the single real root. A cubic always has at least one real root. The sign change

   $f(-1) = -2 < 0, \qquad f(0) = 4 > 0$

   pins a root in $(-1, 0)$. Moreover $f'(x) = 3x^2 + 4x + 7$ has discriminant $16 - 84 < 0$, so $f' > 0$ everywhere: $f$ is strictly increasing and there is exactly one real root.

3. 3

   Refine with Newton's method. Iterating $x_{n+1} = x_n - \dfrac{x_n^3 + 2x_n^2 + 7x_n + 4}{3x_n^2 + 4x_n + 7}$ from $x_0 = -0.7$ converges quickly to

   $x_1 \approx -0.6535951654$

   At 40-digit working precision the residual is below $10^{-39}$. (A common slip is to stop at a coarse estimate such as $-0.6954$; there $f = -0.2368$, not $0$.)

4. 4

   Divide out the real root. Synthetic division by $\left(x - x_1\right)$ leaves the quadratic

   $x^2 + 1.3464048346\,x + 6.1199963094 = 0$

   with zero remainder, where the coefficients are $2 + x_1$ and $7 + x_1(2 + x_1)$.

5. 5

   Solve the quadratic for the complex pair. Its discriminant is negative:

   $\Delta = 1.3464048346^2 - 4(6.1199963094) = -22.6671792591$

   $x = \frac{-1.3464048346 \pm i\sqrt{22.6671792591}}{2} \approx -0.6732024173 \pm 2.3805030592\,i$

6. 6

   Cross-check with Vieta's formulas. The three roots must sum to $-2$ and multiply to $-4$:

   $-0.6535951654 + 2(-0.6732024173) = -2.0000000000 \ \checkmark$

   $(-0.6535951654)\left(0.6732024173^2 + 2.3805030592^2\right) = -4.0000000000 \ \checkmark$

   Both identities close, confirming the numerics independently of the iteration.