Solve for $a, b, c$ given that $a \le b \le c$, and

$a+b+c=-1$

$ab+bc+ac=-4$

$abc=-2$
1. **Identify the polynomial:**
   Given the conditions \(a \le b \le c\), \(a + b + c = -1\), \(ab + bc + ac = -4\), and \(abc = -2\), we recognize that \(a, b, c\) are the roots of a cubic polynomial. The polynomial can be written as:
   \[
   P(x) = x^3 + px^2 + qx + r
   \]
   Using Vieta's formulas, we know:
   \[
   p = -(a + b + c) = 1, \quad q = ab + bc + ac = -4, \quad r = -abc = 2
   \]
   Therefore, the polynomial is:
   \[
   P(x) = x^3 + x^2 - 4x + 2
   \]

2. **Find the roots of the polynomial:**
   To find the roots of \(P(x) = x^3 + x^2 - 4x + 2\), we can test possible rational roots using the Rational Root Theorem. The possible rational roots are the factors of the constant term (2) divided by the factors of the leading coefficient (1), which are \(\pm 1, \pm 2\).

   Testing \(x = 1\):
   \[
   P(1) = 1^3 + 1^2 - 4 \cdot 1 + 2 = 1 + 1 - 4 + 2 = 0
   \]
   So, \(x = 1\) is a root.

3. **Factor the polynomial:**
   Since \(x = 1\) is a root, we can factor \(P(x)\) as:
   \[
   P(x) = (x - 1)Q(x)
   \]
   where \(Q(x)\) is a quadratic polynomial. To find \(Q(x)\), we perform polynomial division of \(P(x)\) by \(x - 1\).

4. **Polynomial division:**
   Dividing \(x^3 + x^2 - 4x + 2\) by \(x - 1\):
   \[
   \begin{array}{r|rrr}
     1 & 1 & 1 & -4 & 2 \\
       &   & 1 & 2 & -2 \\
     \hline
       & 1 & 2 & -2 & 0 \\
   \end{array}
   \]
   The quotient is \(x^2 + 2x - 2\), so:
   \[
   P(x) = (x - 1)(x^2 + 2x - 2)
   \]

5. **Solve the quadratic equation:**
   To find the remaining roots, solve the quadratic equation \(x^2 + 2x - 2 = 0\) using the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   where \(a = 1\), \(b = 2\), and \(c = -2\):
   \[
   x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 8}}{2} = \frac{-2 \pm \sqrt{12}}{2} = \frac{-2 \pm 2\sqrt{3}}{2} = -1 \pm \sqrt{3}
   \]
   Thus, the roots are \(x = -1 + \sqrt{3}\) and \(x = -1 - \sqrt{3}\).

6. **Order the roots:**
   The roots of the polynomial are \(1\), \(-1 + \sqrt{3}\), and \(-1 - \sqrt{3}\). Given \(a \le b \le c\), we have:
   \[
   a = -1 - \sqrt{3}, \quad b = -1 + \sqrt{3}, \quad c = 1
   \]

The final answer is \(\boxed{a = -1 - \sqrt{3}, \, b = -1 + \sqrt{3}, \, c = 1}\)