Mathwords logoReference LibraryMathwords

Linear Factorization

Linear Factorization

A factored form of a polynomial in which each factor is a linear polynomial.

 

Example: A linear factorization of 2x3 – 6x2 + 4x is 2x(x – 1)(x – 2).

Key Formula

p(x)=an(x−r1)(x−r2)⋯(x−rn)p(x) = a_n(x - r_1)(x - r_2) \cdots (x - r_n)
Where:
  • p(x)p(x) = The original polynomial of degree n
  • ana_n = The leading coefficient of the polynomial
  • r1,r2,…,rnr_1, r_2, \ldots, r_n = The roots (zeros) of the polynomial, which may be real or complex
  • nn = The degree of the polynomial

Worked Example

Problem: Find the linear factorization of p(x) = 3x³ − 12x² + 9x.
Step 1: Factor out the greatest common factor (GCF) from all terms.
3x3−12x2+9x=3x(x2−4x+3)3x^3 - 12x^2 + 9x = 3x(x^2 - 4x + 3)
Step 2: Factor the remaining quadratic. Find two numbers that multiply to 3 and add to −4. Those numbers are −1 and −3.
x2−4x+3=(x−1)(x−3)x^2 - 4x + 3 = (x - 1)(x - 3)
Step 3: Write the complete linear factorization by combining the GCF with the factored quadratic.
3x3−12x2+9x=3x(x−1)(x−3)3x^3 - 12x^2 + 9x = 3x(x - 1)(x - 3)
Step 4: Verify: Each factor — 3, x, (x − 1), and (x − 3) — is either a constant or a linear polynomial. The roots are x = 0, x = 1, and x = 3, matching a degree-3 polynomial with three roots.
Answer: The linear factorization is 3x(x − 1)(x − 3).

Another Example

Problem: Find the linear factorization of p(x) = x⁴ − 5x² + 4.
Step 1: Recognize this as a quadratic in disguise. Let u = x², so the expression becomes u² − 5u + 4.
u2−5u+4=(u−1)(u−4)u^2 - 5u + 4 = (u - 1)(u - 4)
Step 2: Substitute back u = x².
x4−5x2+4=(x2−1)(x2−4)x^4 - 5x^2 + 4 = (x^2 - 1)(x^2 - 4)
Step 3: Factor each quadratic further using the difference of squares pattern.
(x2−1)(x2−4)=(x−1)(x+1)(x−2)(x+2)(x^2 - 1)(x^2 - 4) = (x - 1)(x + 1)(x - 2)(x + 2)
Answer: The linear factorization is (x − 1)(x + 1)(x − 2)(x + 2). The leading coefficient is 1, and the four roots are x = ±1 and x = ±2.

Frequently Asked Questions

Does every polynomial have a linear factorization?
Yes, over the complex numbers. The Linear Factorization Theorem (a consequence of the Fundamental Theorem of Algebra) guarantees that every polynomial of degree n ≥ 1 can be written as a product of exactly n linear factors. Over the real numbers alone, some factors may be irreducible quadratics (e.g., x² + 1 has no real roots), so a full linear factorization requires complex numbers.
How is linear factorization different from regular factoring?
Regular factoring may stop at irreducible quadratic or higher-degree factors. Linear factorization means you break the polynomial down completely so that every factor has degree 1. For example, x⁴ + 4 factors over the reals as (x² + 2x + 2)(x² − 2x + 2), but those quadratics are not linear. A full linear factorization would require complex roots.

Linear factorization vs. Irreducible factorization over the reals

Linear factorization breaks a polynomial into all degree-1 factors (which may involve complex numbers). Irreducible factorization over the reals allows degree-2 factors that have no real roots, such as x² + 1. A polynomial like x² + 1 is already in irreducible real form but factors as (x − i)(x + i) in its linear factorization.

Why It Matters

Linear factorization reveals all the zeros of a polynomial at a glance, making it essential for solving polynomial equations and graphing. It is the backbone of the Fundamental Theorem of Algebra, which you encounter repeatedly in algebra and precalculus. In applications, factored form makes it easy to analyze where a function crosses or touches the x-axis.

Common Mistakes

Mistake: Forgetting to factor out the leading coefficient, so the product of factors doesn't match the original polynomial.
Correction: Always pull out the leading coefficient first. For example, 2x² − 8 should be written as 2(x − 2)(x + 2), not just (x − 2)(x + 2), which only equals x² − 4.
Mistake: Stopping the factorization too early by leaving a reducible quadratic (like x² − 9) instead of breaking it into linear factors.
Correction: Check every quadratic factor to see if it can be factored further. Use the discriminant b² − 4ac: if it is non-negative, the quadratic splits into two real linear factors.

Related Terms