Find all real and imaginary solutions to the equation x^3 - 6x^2 - 13x + 78 = 0.

Steps to Solve

1. Identify the polynomial The given equation is a cubic polynomial: $$ x^3 - 6x^2 - 13x + 78 = 0. $$

2. Use the Rational Root Theorem To find possible rational roots, consider the factors of 78. The possible values are ±1, ±2, ±3, ±6, ±13, ±26, ±39, ±78.

3. Test possible roots Substitute the possible rational roots into the polynomial to identify any that yield zero:

   • Testing ( x = 6 ): $$ 6^3 - 6(6)^2 - 13(6) + 78 = 216 - 216 - 78 + 78 = 0 $$ Therefore, ( x = 6 ) is a root.

4. Perform synthetic division Now divide the polynomial by ( (x - 6) ):

   • The coefficients are ( 1, -6, -13, 78 ).
   • Performing synthetic division gives us: $$ x^3 - 6x^2 - 13x + 78 = (x - 6)(x^2 + 0x - 13) $$

5. Solve the quadratic equation To find the other roots, solve: $$ x^2 - 13 = 0 $$ This can be rearranged as: $$ x^2 = 13 $$

6. Find the square roots Taking the square root of both sides yields: $$ x = \sqrt{13} \quad \text{and} \quad x = -\sqrt{13} $$

7. List all solutions The solutions to the equation are: $$ x = 6, \quad x = \sqrt{13}, \quad x = -\sqrt{13} $$