Example

Factoring 3y2+22y+73y^2 + 22y + 7

Factor 3y2+22y+73y^2 + 22y + 7 completely using the trial and error method.

Step 1 — Write in descending order. The trinomial 3y2+22y+73y^2 + 22y + 7 is already in descending order of degrees.

Step 2 — Find factor pairs of the first term. The only way to factor 3y23y^2 into first-degree terms is y3yy \cdot 3y. Place these in the binomials: (y)(3y)(y\quad)(3y\quad).

Step 3 — Find factor pairs of the third term. Since all terms are positive, only positive factors of 7 are needed. The only factor pair is 1 and 7.

Step 4 — Test all arrangements:

  • (y+1)(3y+7)(y + 1)(3y + 7): inner product =13y=3y= 1 \cdot 3y = 3y, outer product =y7=7y= y \cdot 7 = 7y, sum =10y= 10y
  • (y+7)(3y+1)(y + 7)(3y + 1): inner product =73y=21y= 7 \cdot 3y = 21y, outer product =y1=y= y \cdot 1 = y, sum =22y= 22y

The second arrangement produces the required middle term 22y22y.

Step 5 — Check by multiplying: (y+7)(3y+1)=3y2+y+21y+7=3y2+22y+7(y + 7)(3y + 1) = 3y^2 + y + 21y + 7 = 3y^2 + 22y + 7

The factored form is (y+7)(3y+1)(y + 7)(3y + 1). Like the example of 3x2+5x+23x^2 + 5x + 2, both the squared term and the constant have only one factor pair each, yet swapping the placement of 1 and 7 between the binomials produces very different middle terms (10y10y versus 22y22y). This reinforces the importance of testing every arrangement systematically.

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Updated 2026-04-21

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