Example

Factoring 6pq29pq6p6pq^2 - 9pq - 6p

Factor 6pq29pq6p6pq^2 - 9pq - 6p completely.

Step 1 — Check for a GCF: The three terms share a greatest common factor of 3p3p. Factor it out: 3p(2q23q2)3p(2q^2 - 3q - 2)

Step 2 — Factor the trinomial: The expression inside the parentheses is a trinomial with a leading coefficient other than 11 (a1a \neq 1). Factor it using the trial and error or "ac" method. The binomial factors are (2q+1)(2q + 1) and (q2)(q - 2): 3p(2q+1)(q2)3p(2q + 1)(q - 2)

Step 3 — Check: Verify by multiplying: 3p(2q+1)(q2)=3p(2q24q+q2)=3p(2q23q2)=6pq29pq6p3p(2q + 1)(q - 2) = 3p(2q^2 - 4q + q - 2) = 3p(2q^2 - 3q - 2) = 6pq^2 - 9pq - 6p

The completely factored form is 3p(2q+1)(q2)3p(2q + 1)(q - 2).

0

1

Updated 2026-04-30

Contributors are:

Who are from:

Tags

OpenStax

Intermediate Algebra @ OpenStax

Ch.6 Factoring - Intermediate Algebra @ OpenStax

Algebra

Related