By Sky Smith

Updated: Feb. 27, 2025 7:24 pm EST

© Kamil Zajaczkowski/Shutterstock

Factoring cubic polynomials is a powerful tool that reveals a function’s zeros, indicating where the graph changes direction and simplifying deeper analysis. While quadratic factoring is straightforward, cubics often require a systematic approach. Below is a proven, expert‑approved method to factor any degree‑3 polynomial efficiently.

Step 1 – Grouping

Identify a pattern where the polynomial can be split into two groups that share a common factor. For example, consider F(x) = x³ – x² – 4x + 4. Group the terms:

 x²(x – 1) – 4(x – 1)

Pull out the shared binomial factor (x – 1):

(x² – 4)(x – 1)

Apply the difference‑of‑squares rule to the remaining quadratic:

(x – 2)(x + 2)(x – 1)

All factors are now prime.

Step 2 – Sum or Difference of Cubes

When a polynomial consists of two terms, each a perfect cube, use the standard identities:

• Sum: (x³ + y³) = (x + y)(x² – xy + y²)
• Difference: (x³ – y³) = (x – y)(x² + xy + y²)

Example: G(x) = 8x³ – 125 factors as

(2x – 5)(4x² + 10x + 25)

The quadratic is irreducible over the integers, so factoring stops here.

Step 3 – Extract a Greatest Common Factor

Check if a variable or constant multiplies all terms. For H(x) = x³ – 4x, factor out x:

H(x) = x(x² – 4)

Then apply the difference‑of‑squares trick:

H(x) = x(x – 2)(x + 2)

Step 4 – Use the Factor Theorem

When grouping, cubes, and GCFs are insufficient, find a rational root using the Factor Theorem. For P(x) = x³ – 4x² – 7x + 10, test integer candidates ±1, ±2, ±5, ±10. We find

P(5) = 0

Thus (x – 5) is a factor. Dividing by this binomial yields

P(x) = (x – 5)(x² + x – 2)

The quadratic factors further:

(x – 5)(x – 1)(x + 2)

References

• Lamar University: Factoring Polynomials