find the eigen value for λ^3-11λ^2+40λ^2+36λ=0

To find the eigenvalues, we need to solve the cubic equation given:

\[
\lambda^3 - 11\lambda^2 + 40\lambda - 36 = 0
\]

### Step-by-Step Solution

1. **Correct the Equation:**  
   Notice that the term \(40\lambda^2\) seems to be a typographical error since it conflicts with \(-11\lambda^2\). Assuming the correct term is \(40\lambda\) instead of \(40\lambda^2\), the equation should be:

   \[
   \lambda^3 - 11\lambda^2 + 40\lambda - 36 = 0
   \]

2. **Look for Rational Roots:**  
   We use the Rational Root Theorem to find possible rational roots of the cubic polynomial. The possible rational roots are the factors of the constant term \(-36\) divided by the factors of the leading coefficient \(1\):

   Factors of \(-36\): \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36 \)

   Since the leading coefficient is \(1\), the possible rational roots are:

   \[
   \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36
   \]

3. **Test Possible Roots:**

   We test each of these possible roots by substituting them into the polynomial.

   - **Testing \( \lambda = 2 \):**

   \[
   2^3 - 11 \cdot 2^2 + 40 \cdot 2 - 36 = 8 - 44 + 80 - 36 = 0
   \]

   So, \( \lambda = 2 \) is a root.

4. **Factor the Polynomial:**

   Since \( \lambda = 2 \) is a root, we can factor out \( (\lambda - 2) \) from the polynomial using synthetic division or polynomial long division.

   **Synthetic Division:**

   ```
   2 |  1  -11   40  -36
     |       2   -18   44
     ------------------
        1   -9   22    0
   ```

   The quotient from synthetic division is \( \lambda^2 - 9\lambda + 18 \). Therefore, we can write:

   \[
   \lambda^3 - 11\lambda^2 + 40\lambda - 36 = (\lambda - 2)(\lambda^2 - 9\lambda + 18)
   \]

5. **Solve the Quadratic Factor:**

   The remaining quadratic equation is:

   \[
   \lambda^2 - 9\lambda + 18 = 0
   \]

   To solve it, we use the quadratic formula:

   \[
   \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]

   where \(a = 1\), \(b = -9\), and \(c = 18\).

   \[
   \lambda = \frac{-(-9) \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot 18}}{2 \cdot 1}
   \]

   \[
   \lambda = \frac{9 \pm \sqrt{81 - 72}}{2}
   \]

   \[
   \lambda = \frac{9 \pm \sqrt{9}}{2}
   \]

   \[
   \lambda = \frac{9 \pm 3}{2}
   \]

   So, the roots are:

   \[
   \lambda_1 = \frac{9 + 3}{2} = 6, \quad \lambda_2 = \frac{9 - 3}{2} = 3
   \]

### Conclusion

The eigenvalues of the polynomial equation \( \lambda^3 - 11\lambda^2 + 40\lambda - 36 = 0 \) are:

\[
\lambda_1 = 2, \quad \lambda_2 = 6, \quad \lambda_3 = 3
\]

These are the three roots of the cubic equation.

Would you like to go through any particular part in more detail or have any questions?

### Related Questions

1. How do you perform synthetic division for polynomials?
2. What is the Rational Root Theorem, and how is it applied?
3. How can you factor a polynomial once a root is found?
4. What are the steps to solve a quadratic equation using the quadratic formula?
5. Can a cubic polynomial have fewer than three real roots? If so, under what conditions?

### Tip

When solving polynomials, always double-check the coefficients and terms for any typographical errors before proceeding with the solution.

Learn how to find the eigenvalues of the cubic equation λ^3 - 11λ^2 + 40λ - 36. This problem involves solving a cubic equation using the Rational Root Theorem and identifying eigenvalues. Suitable for students in Grades 10-12.

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