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Algebra 2 Polynomial functions

Factoring higher-degree polynomials

20 practice questions 0 video lessons Theory + worked examples

Factoring Higher-Degree Polynomials

Texas Algebra II (TEKS) • Standard 2A.7(D) • Polynomial Functions

Factoring Higher-Degree Polynomials is a topic in Polynomial Functions in the Texas Essential Knowledge and Skills (Algebra II, §111.40). It is aligned to Standard 2A.7(D), which requires students to determine the linear factors of a polynomial of degree three and four using algebraic methods.

Factoring rewrites a polynomial as a product using the GCF, difference of squares, grouping, and quadratic-form substitution.

Texas Algebra II (TEKS) › Polynomial Functions › Factoring Higher-Degree Polynomials  —  Standard 2A.7(D)

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Theory

Factoring rewrites a polynomial as a product. Work through a toolkit in order:
  • GCF first — always remove the greatest common factor.
  • Difference of squares: \(a^2-b^2=(a-b)(a+b)\).
  • Grouping for four terms.
  • Quadratic form: substitute for \(x^2\) in \(x^4+bx^2+c\).
Keep factoring until every factor is prime (fully factored).
Factoring toolkit Factoring toolkit Factoring toolkit 1. GCF first, always 2. difference of squares: a²-b²=(a-b)(a+b) 3. trinomial: x²+bx+c 4. grouping (four terms)
Try the tools in order, GCF first.
Factor by grouping Factor by grouping Factor by grouping x³ + 2x² + 3x + 6 = x²(x+2) + 3(x+2) = (x+2)(x²+3)
Factoring by grouping.

Key patterns:

\[a^2-b^2=(a-b)(a+b),\qquad x^2+(p+q)x+pq=(x+p)(x+q)\]
difference of squares and trinomial factoring patterns
Check by expanding — the product should return the original.

How to factor

  1. Remove the GCF.
  2. Count terms: two \(\to\) difference of squares; three \(\to\) trinomial; four \(\to\) grouping.
  3. Apply the matching pattern.
  4. Factor again until prime.
Example 1 — Greatest common factor
Factor \(6x^3-9x^2\).
Solution

Pull out the GCF \(3x^2\).

\(6x^3-9x^2\)\(=\)\(3x^2(2x-3)\)
factors as 3 x squared times 2 x minus 3
Example 2 — Difference of squares
Factor \(x^4-16\).
Solution

Apply the pattern twice.

\(x^4-16\)\(=\)\((x^2-4)(x^2+4)\)
\(=\)\((x-2)(x+2)(x^2+4)\)
factors as x minus 2 times x plus 2 times x squared plus 4
Example 3 — Factor by grouping
Factor \(x^3+2x^2+3x+6\).
Solution

Group in pairs and factor each.

\(x^2(x+2)+3(x+2)\)
\(=\)\((x+2)(x^2+3)\)
factors as x plus 2 times x squared plus 3
Example 4 — Trinomial in quadratic form
Factor \(x^4-5x^2+4\).
Solution

Treat \(x^2\) as the variable.

\((x^2-1)(x^2-4)\)
\(=\)\((x-1)(x+1)(x-2)(x+2)\)
factors into x minus 1, x plus 1, x minus 2, x plus 2

Common pitfalls

Always take the GCF first — it simplifies everything after.
Difference of squares can repeat: \(x^4-16\) factors twice.
\(a^2+b^2\) does not factor over the reals.

Frequently asked questions

What should you factor out first?

The greatest common factor (GCF).

How do you factor four terms?

Group them in pairs and factor each pair, then factor out the common binomial.

Does \(a^2+b^2\) factor?

Not over the real numbers; only \(a^2-b^2\) does.

When is a polynomial fully factored?

When every factor is prime and cannot be factored further.