Solving cubic equations by factorization

According to the fundamental theorem of algebra, a cubic equation always has 3 roots (taking into account multiplicity), and at least one of the roots is real, since if the root is a complex number, then its complex conjugate will also be its root. Thus, a cubic polynomial a(x) = a₀*x³ + a₁*x² + a₂*x + a₃, a₀ ≠ 0, can always be factored into two factors, one of which is linear and the other is quadratic: a(x) = (x - x₀) (a₀x² + bx + c).
In turn, a polynomial of the second degree a₀x² + bx + c can have 2 different real roots, 1 real root or 2 complex conjugate roots.
Accordingly, we obtain such cases of factorization of the polynomial a(x):

1) a(x) = a₀ (x - x₀) (x - x₁) (x - x₂)
2) a(x) = a₀ (x - x₀) (x - x₁)²
3) a(x) = a₀ (x - x₀)³
4) a(x) = (x - x₀) (a₀x² + bx + c), if b² - 4a₀c < 0.
Thus, equating each factor in the expansion to zero, we will find all the roots of the cubic equation in each case. Let's consider the solving of cubic equations by factorization using examples.

Examples of solving cubic equations by factorization

Example 1.
Solve the equation x³ - 3x² - 4x + 6 = 0.
Solution
The divisors of the free term are the numbers: ±1, ±2, ±3, ±6. This means that the roots of the equation must be sought among them. By simple substitution we verify that the root of the equation is the number 1. Therefore, the original equation is equivalent to (x - 1)*(a₀x² + bx + c) = 0.
To find the polynomial, a₀x² + bx + c, you need to divide the left side of the original equation by x - 1. To divide a polynomial by a binomial, we will use Horner's scheme.


Answer: -1- √7, 1 ,-1+√7.

Example 2.
Solve the equation -2x³ + 3x² - 4x - 9 = 0.
Solution.
The divisors of the free term are the numbers: ±1, ±3, ±9. The divisors of the higher coefficient are the numbers: ±1, ±2. This means that the roots of the original equation can be among the numbers: ±1, ±3, ±9, .
A simple substitution shows that -1 is the root of the equation. Using Horner's scheme, we divide the left side of the original equation by x + 1.


Thus, -2x³ + 3x² - 4x - 9 = (x + 1)(-2x² + 5x - 9). Therefore, the original equation is equivalent to (x + 1) (-2x² + 5x - 9)=0. Solving the quadratic equation -2x² + 5x - 9 = 0, we obtain, что его дискриминант < 0, therefore, it has no real roots.
Answer: -1.

Example 3. Solve the equation 2x³ - x² - 8x + 4 = 0.
Solution.
The divisors of the free term are the numbers: ±1, ±2, ±4. The divisors of the higher coefficient are the numbers: ±1, ±2.
This means that the roots of the original equation can be among the numbers: ±1, ±2, ±4.
A simple substitution shows that 2 is the root of the equation. Using Horner's scheme, we divide the left side of the original equation by x - 2.


Thus, 2x³ - x² - 8x + 4 = (x - 2)(2x² + 3x - 2). Therefore, the original equation is equivalent to (x - 2) (2x² + 3x - 2) = 0. Solving the quadratic equation 2x² + 3x - 2 = 0, we get,


Answer: -2, , 2.

Another way to factorize a polynomial of the third degree is the method of indefinite coefficients. It is quite cumbersome, but sometimes it is very useful when solving various kinds of problems, and not only in the case of factorization. The factorization of any polynomial of the third degree can be represented as follows a(x) = (x-x₀)*(a₀x² + bx + c). Opening the brackets, we get a(x) = a₀x³ + x²(b - a₀x₀) + x*(c - bx₀) - cx₀. Now equating the coefficients for the same powers of x and the free terms in the original polynomial and in the polynomial a(x), we get a system of four equations and four unknowns a₀, b, c and x₀. Let's consider the application of the method of undefined coefficients using examples.

Example 4.
Solve the equation x³ + 2x² - 5x - 6 = 0.
Solution.
Since any polynomial of degree 3 can be represented as a₀x³ + x²(b - a₀x₀) + x*(c - bx₀) - cx₀, then equating the coefficients for the same degrees of x, we obtain the following system of equations:
Or We find from the first equation x₀ = b - 2 and substitute it into the remaining two. We get


Now let's find the variable c from the first equation and substitute it into the second.


By opening the brackets in the second equation and solving it, we find b:


If b=4, then c=3, x₀ = 2. Therefore, x³ + 2x² - 5x - 6 = (x - 2)(x² - 4x + 3)=(x - 2)(x + 1)(x + 3).
If b = 1, then c = -6, x₀ = -1. Therefore, x³ + 2x² - 5x - 6 = (x + 1)(x² + x - 6)=(x + 1)(x + 3)(x - 2).
If b = -1, then c = -2, x₀ = -3. Therefore, x³ + 2x² - 5x - 6=(x + 3)(x² - x - 2) = (x + 3)(x - 2)(x + 1).
Thus, the original equation is equivalent to the equation (x + 3)(x - 2)(x + 1) = 0.
Equating each of the factors to zero, we obtain the roots of the equation x = -3, x = 2, x = -1.
Answer: -3, -1, 2.

Example 5.
Solve the equation 2x³ + x² - 5x + 2 = 0.
Solution.
By equating the corresponding coefficients for the same powers of x, we obtain the following system of equations:

Or We find from the first equation x₀ =  and substitute it into the remaining two. We get
Now let's find the variable c from the first equation and substitute it into the second.


Multiplying the left and right sides of the second equation by 4 and opening the brackets, we find b:


If b=2, then c=-4, x₀ = . Следовательно, 2x³ + x² - 5x + 2 = (x - )(2x² + 2x - 4) = 2(x - )(x - 1)(x + 2).
If b = 3, then c = -2, x₀ = 1. Therefore, 2x³ + x² - 5x + 2 = (x - 1)(2x² + 3x - 2)=2(x - 1)(x - )(x + 2).
If b = -3, then c = 1, x₀ = -2. Therefore, 2x³ + x² - 5x + 2 = (x + 2)(2x² - 3x + 1) = 2(x + 2)(x - )(x - 1).
Therefore, the original equation is equivalent to the equation 2(x + 2)(x - )(x - 1) = 0.
Equating each of the factors to zero, we obtain the roots of the equation x = -2, x = , x = 1.
Answer: -2, , 1.