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

Step-by-step solution

1. 1

   Hunt for a rational root. By the Rational Root Theorem, any rational root divides the constant $2$, so the candidates are $\pm1,\pm2$. Testing $x=1$:

   $1-3+2=0\ \checkmark$

   so $x=1$ is a root and $(x-1)$ is a factor. Note the missing $x^2$ term means its coefficient is $0$ — that placeholder must be carried into the division.

2. 2

   Divide by (x - 1) using synthetic division. With coefficients $1,\ 0,\ -3,\ 2$ and divisor root $1$:

   $1\ \to\ 0+1=1\ \to\ -3+1=-2\ \to\ 2-2=0$

   The final $0$ is the remainder, confirming the root, and the quotient is $x^2+x-2$:

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

3. 3

   Factor the quadratic. Two numbers multiplying to $-2$ and adding to $+1$ are $+2$ and $-1$:

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

4. 4

   Collect the repeated factor. The factor $(x-1)$ appears twice:

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

   That is the signal of a double root — worth stating explicitly rather than listing $x=1$ twice as if it were two different answers.

5. 5

   Read off the solutions.

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

   A cubic has three roots counted with multiplicity, so this accounts for all of them.

6. 6

   Interpret the double root graphically. At a root of even multiplicity the curve touches the axis without crossing: $x=1$ is a local minimum of $f(x)=x^3-3x+2$ (indeed $f'(x)=3x^2-3=0$ at $x=1$, so $f$ and $f'$ vanish together there). At $x=-2$ the graph crosses normally.

7. 7

   Verify the factorisation and the roots. Comparing $x^3-3x+2$ with $(x-1)^2(x+2)$ at every integer from $-40$ to $39$ gives exact agreement ✓, and substituting $x=1$ and $x=-2$ each returns exactly $0$ ✓. Vieta also checks out: the roots sum to $1+1-2=0=-b/a$ ✓.