Cubic Equation Solving
Example 3 (Difficulty level - 5):
Let's solve the following third degree equation:
3x^{3} + 4x^{2} - 4x - 3 = 0
Solution:
Let's factor the left part of original cubic equation, using the cubes difference formula:
a^{3} – b^{3} = (a - b)(a^{2} + ab + b^{2})
3x^{3} + 4x^{2} - 4x - 3 =
3(x^{3} - 1) + 4x^{2} - 4x = 3(x - 1)(x^{2} + x + 1) + 4x(x - 1) = (x - 1)(3x^{2} + 3x + 3 + 4x) = (x - 1)(3x^{2} + 7x + 3)
So the original cubic equation is equivalent to the following set of the equations:
x - 1 = 0
3x^{2} + 7x + 3 = 0
The first equation of the set is a linear equation and has the root x = -1.
The second equation is a quadratic equation solved by the quadratic formula, so the roots of the square equation are:
x
_{1,2} =
- b ± √
b^{2} - 4 × a × c
/2 × a
x
_{1,2} =
- 7 ± √
7^{2} - 4 × 3 × 3
/2 × 3
Answer: x
_{1} = -1, x
_{2} =
, x
_{3} =