Cubic Equation Solving
Example 3 (Difficulty level - 5):
Let's solve the following third degree equation:
3x3 + 4x2 - 4x - 3 = 0
Solution:
Let's factor the left part of original cubic equation, using the cubes difference formula:
a3 – b3 = (a - b)(a2 + ab + b2)
3x3 + 4x2 - 4x - 3 =
3(x3 - 1) + 4x2 - 4x = 3(x - 1)(x2 + x + 1) + 4x(x - 1) = (x - 1)(3x2 + 3x + 3 + 4x) = (x - 1)(3x2 + 7x + 3)
So the original cubic equation is equivalent to the following set of the equations:
x - 1 = 0
3x2 + 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 ± √
b2 - 4 × a × c
/2 × a x
1,2 =
- 7 ± √
72 - 4 × 3 × 3
/2 × 3
