Last updated on July 4th, 2025
The process of finding the factors of cubic polynomials is known as factoring cubic polynomials. Cubic polynomials are algebraic expressions with degree 3, and the standard form is ax3 + bx2 + cx + d, where a, b, c, and d are real numbers. In this article, we will discuss more on factoring cubic polynomials.
A polynomial that has exactly three terms with a degree of three is known as cubic trinomial. ‘Cubic’ refers to the highest degree of three, and the word ‘trinomial’ suggests the expression has three terms. The standard form of a cubic trinomial is:
ax3 + bx2 + cx, where the degree of the polynomial is three because the highest power of the polynomial is three. Where a, b, and c are coefficients, which can be any complex or real numbers.
x is the variable. 3x3 + 2x2 + 4x is an example of a cubic trinomial.
Factorizing a cubic trinomial means breaking down a cubic trinomial into its factors. This process is important for solving cubic equations. The following steps are used to break down a polynomial of the form ax³ + bx² + cx + d.
Step 1: Check for common factors.
Check for the common factors in the given polynomial. If there is a common factor, write it down so that we will get a quadratic equation, which will help us to solve the equation easily.
Example: x3 + 4x2 + 3x
In the given polynomial, we have the common term x. So we have to factor out x first.
x3 + 4x2 + 3x = x (x2 + 4x + 3).
Now we have a quadratic equation as x2 + 4x + 3
Step 2: Factor the quadratic.
After taking out the common factor, we are left with a quadratic equation. Now, we have to factorize that quadratic equation.
x2 + 4x + 3
In this equation, we have to look at two numbers; the sum of the numbers is 4, and the product of the numbers is 3. Therefore, we get an equation as,
x2 + 4x + 3 = (x + 1)(x + 3)
Step 3: Write the complete factorization.
Now we have to put the factorization of the quadratic equation to step 1,
x (x2 + 4x + 3) = x (x + 1)(x + 3)
Step 4: Verify the factorization.
Multiply the factors; if this matches the original polynomial expression, then the factorization is correct.
First, multiply the factorization of the quadratic polynomial,
(x + 1)(x + 3) = x2 + 3x + x + 3
= x2 + 4x + 3
Then multiply it by the x term, which we take as common in the first step.
x(x2 + 4x + 3) = x3+ 4x2 + 3x.
This matches the original polynomial, so the factorization is correct.
According to the rational root theorem, the possible roots of a cubic polynomial f(x) = ax3 + bx2 + cx + d can be determined by:
Possible rational roots = ±(factors of d)/(factors of a).
The following steps will help us understand rational root theorem better:
Step 1: Find the possible roots.
Using the rational root theorem, find the possible rational roots with the constant term d and the leading coefficient a. For example, x3 - 2x2 - 5x + 6.
To find the possible roots, we have to divide the factors of d (constant term) by the factors of a (leading coefficient).
Here, the coefficient d = 6
Leading term a = 1
Factors of 6: ±1, ±2, ±3, ±6
Factors of 1: ±1
So, the possible roots are = ±1, ±2, ±3, ±6.
Step 2: Test the possible roots.
Now we have to check all the possible roots of the polynomial. If the value is 0, that is considered as a root.
Try x = 1 in the polynomial
f(x) = x3 - 2x2 - 5x + 6
f(1) = (1)3 - 2(1)2 - 5(1) + 6
= 1 - 2 - 5 + 6 = 0
So, x = 1 is a root, and (x -1) is a factor.
Step 3: Use polynomial division.
We divide the polynomial by the factor that we got by using the long division or the synthetic division method.
(x3 - 2x2 - 5x + 6) ÷ (x - 1)
After dividing this, the equation will become x2 - x - 6.
Step 4: Factor the quadratic equation.
We got a quadratic equation after dividing the polynomial by the factor. Now we have to factorize that quadratic equation.
x2 - x - 6 is the quadratic equation.
In this, we have to look for two numbers that are, the sum of the numbers will be -1 and the product of the numbers is -6.
The numbers are -3 and 2
x2 - x - 6 = (x - 3)(x + 2)
Step 5: Write the final factorization.
After factorizing the equation, we have to combine all the factors to get the final answer.
f(x) = (x - 1)(x - 3)(x + 2)
x3 - 2x2 - 5x + 6 = (x - 1)(x - 3)(x + 2)
We can use special math rules called identities when a cubic polynomial has only two terms. Mentioned below are two main cases:
Case 1: No constant terms.
If there are no constant terms and the equation consists of only variable terms, the expression might look like,
ax3 + bx2 or ax3 + cx
In this case, we can take the common x terms,
ax3 + bx2 = x2 (ax + b)
ax3 + cx = x(ax2 + c)
Case 2: Equations that include constants
If one term is just a number like ax3 + d, then we will use a special identity to factor:
If both the terms are perfect cubes, like 8x3 + 27, use:
a3 + b3 = (a + b)(a2 - ab + b2)
If both the terms are cubes but with a minus, like x3 - 8, use:
a3 - b3 = (a - b)(a2 + ab + b2)
These help us to break the expression into smaller parts.
Factoring cubic polynomials is used in many real-life applications, especially in fields like engineering, economics, physics, and computer science. Some of the applications are mentioned below:
Mistakes are common while dealing with problems that involve the factorization of cubic polynomials. However, some mistakes can be avoided if we have prior knowledge about them. Given below are some of the common mistakes and the ways to avoid them.
Factor the cubic polynomial: x to the power 3 + 6x to the power2 + 11x + 6.
x3 + 6x2 + 11x + 6 = (x + 1)(x + 2)(x + 3)
Group the first two and last two terms: (x3 + 6x2) + (11x + 6)
Grouping doesn’t guarantee a common factor, so try rational roots instead.
Try x = -1:
(-1)3 + 6(-1)2 + 11(-1) + 6 = -1 + 6 - 11 + 6 = 0
So, (x + 1) is a factor.
Now divide:
(x3 + 6x2 + 11x + 6) ÷ (x + 1) = x2 + 5x + 6
Factor the quadratic equation, we get: (x + 2)(x + 3)
Therefore, (x + 1)(x + 2)(x + 3) are the factors.
Factor: x cube - 4x square - 7x + 10
(x - 1)(x - 5)(x + 2)
Using rational root, try x = 1
x3 - 4x2 - 7x + 10 = (1)3 - 4(1)2 - 7(1) + 10 = 0
So, (x -1) is a factor.
Divide: (x3 - 4x2 - 7x + 10) ÷ (x - 1)
Now we get the factors: (x - 5)(x + 2)
Final answer: (x - 1)(x - 5)(x + 2)
Factor x cube - 8
(x - 2)(x2. + 2x + 4)
This is a difference between cubes. Use the identity:
a3 - b3 = (a - b)(a2. + ab + b2)
x3 - 23 = (x - 2)(x2. + 2x + 4)
Factor: 2x cube + 4x square + 2x
2x(x + 1)2
First, take the common factor:
2x3 + 4x2 + 2x = 2x(x2 + 2x + 1)
The quadratic expression factors as:
x2 + 2x + 1 = (x + 1)2.
Combine both the answers to get the final answer: 2x(x + 1)2
Factor: x cube + 3x square + 2x
x(x + 1)(x + 2)
Take out the common factor,
x3 + 3x2 + 2x = x(x2 + 3x + 2)
Factorize the quadratic equation, x2 + 3x + 2 = (x + 1)(x + 2)
Final factor: x(x + 1)(x + 2)
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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