Solve the equation x^3 - 3x^2 + 4 = 0.

Step-by-step solution

1. 1

   Test the rational candidates. By the rational root theorem the possibilities are $\pm1, \pm2, \pm4$. Checking: $f(1) = 1 - 3 + 4 = 2$, $f(-1) = -1 - 3 + 4 = 0$, $f(2) = 8 - 12 + 4 = 0$. Two candidates work straight away, which already hints at a repeated factor.

2. 2

   Divide by (x − 2) using synthetic division. Write the coefficients including the missing linear term: $1,\ -3,\ 0,\ 4$. Bringing down $1$ and multiplying by $2$ at each step gives the row $1,\ -1,\ -2$ with remainder $0$, so

   $x^3 - 3x^2 + 4 = (x-2)\left(x^2 - x - 2\right)$

   Inserting the $0$ placeholder is essential — omitting it shifts every subsequent coefficient.

3. 3

   Factor the quadratic. Two numbers with product $-2$ and sum $-1$ are $-2$ and $1$:

   $x^2 - x - 2 = (x-2)(x+1)$

4. 4

   Write the complete factorisation.

   $x^3 - 3x^2 + 4 = (x-2)^2(x+1)$

   The factor $(x-2)$ appeared twice, confirming the earlier hint.

5. 5

   State the roots and their multiplicities.

   $x = 2 \ \text{(double root)}, \qquad x = -1$

6. 6

   Verify. Expanding $(x-2)^2(x+1) = \left(x^2 - 4x + 4\right)(x+1) = x^3 - 3x^2 + 4$ — the $x$ terms $-4x$ and $+4x$ cancel, which is why the original has no linear term. Since $x = 2$ has even multiplicity the curve touches the axis there without crossing, while it crosses at $x = -1$.