Example

Factoring 3x2+5x+23x^2 + 5x + 2

Factor 3x2+5x+23x^2 + 5x + 2 using the trial and error method for trinomials whose leading coefficient is not 1.

Step 1 — Find factor pairs of the first term. The only way to factor 3x23x^2 into a product of two first-degree terms is x3xx \cdot 3x. Place these as the first terms of the binomials: (x)(3x)(x\quad)(3x\quad).

Step 2 — Find factor pairs of the last term. Since all terms of the trinomial are positive, only positive factors of 2 are needed. The only factor pair is 1 and 2.

Step 3 — Test arrangements. The two factors 1 and 2 can be placed in either order, creating two trial factorizations:

  • (x+1)(3x+2)(x + 1)(3x + 2): inner product =13x=3x= 1 \cdot 3x = 3x, outer product =x2=2x= x \cdot 2 = 2x, sum =5x= 5x
  • (x+2)(3x+1)(x + 2)(3x + 1): inner product =23x=6x= 2 \cdot 3x = 6x, outer product =x1=x= x \cdot 1 = x, sum =7x= 7x

The first arrangement produces a middle term of 5x5x, which matches.

Step 4 — Check by multiplying: (x+1)(3x+2)=3x2+2x+3x+2=3x2+5x+2(x + 1)(3x + 2) = 3x^2 + 2x + 3x + 2 = 3x^2 + 5x + 2

The factored form is (x+1)(3x+2)(x + 1)(3x + 2). This example illustrates that even when the first term and the last term each have only one factor pair, the placement of the last-term factors still matters — swapping 1 and 2 between the two binomials changes the inner and outer products and produces a different middle term.

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Updated 2026-04-21

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