Factoring Trinomials Examples: 46 Problems with Answers

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Factoring Trinomials.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

Factoring a trinomial of the form ax2+bx+c into a product of two binomials by finding two numbers that multiply to ac and add to b.

You are reverse-engineering FOIL. If (x+p)(x+q)=x2+(p+q)x+pq, then you need two numbers p and q whose sum is b and whose product is c (when a=1). When a≠1, use the AC method: find two numbers that multiply to ac and add to b, then split the middle term and factor by grouping.

Read the full concept explanation →

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: Reverse FOIL by finding the pair whose product is ac and whose sum is b.

Common stuck point: The procedure for factoring trinomials is the easy part; the trap is using c instead of ac when a≠1. Asking "Can I find two numbers that multiply to c (or ac) and add to b?" first is what keeps a correct-looking calculation from being attached to the wrong concept.

Sense of Study hint: Ask: Can I find two numbers that multiply to c (or ac) and add to b?

Worked Examples

Example 1

easy
Factor x2+5x+6.

Answer

(x+2)(x+3)

First step

1
Step 1: Find two numbers that multiply to 6 and add to 5.

Full solution

  1. 2
    Step 2: 2×3=6 and 2+3=5. The numbers are 2 and 3.
  2. 3
    Step 3: Factor: (x+2)(x+3).
  3. 4
    Check: (x+2)(x+3)=x2+3x+2x+6=x2+5x+6 ✓
For x2+bx+c, find two numbers p,q where pq=c and p+q=b. Then factor as (x+p)(x+q). This reverses the FOIL multiplication process.

Example 2

hard
Factor 2x2+7x+3 using the AC method.

Example 3

medium
Factor 3x2+11x+6 using the AC method.

Example 4

medium
Factor 2x2−7x+3.

Example 5

medium
Factor completely: 2x2+10x+12.

Example 6

medium
Factor 6x2+11x+4.

Example 7

hard
Factor 4x2−12x+9.

Example 8

hard
Factor 6x2−7x−20.

Example 9

hard
Solve 2x2+5x−3=0 by factoring.

Example 10

hard
Factor x4−5x2+4 completely.

Example 11

challenge
Factor x2+6x+9−4y2 completely.

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy
Factor x2−7x+12.

Example 2

medium
Factor x2+2x−15.

Example 3

easy
Factor x2+5x+6.

Example 4

easy
Factor x2+7x+12.

Example 5

easy
Factor x2−7x+10.

Example 6

easy
Factor x2+2x−8.

Example 7

easy
Factor x2−x−6.

Example 8

easy
Factor x2+6x+9.

Example 9

easy
Factor x2−10x+25.

Example 10

easy
Factor x2+8x+15.

Example 11

medium
Factor 2x2+7x+3.

Example 12

medium
Factor 3x2−10x+8.

Example 13

medium
Factor 6x2+11x−10.

Example 14

medium
Factor completely: 2x2+10x+12.

Example 15

medium
Factor x2+xy−6y2.

Example 16

medium
Factor 4x2+12x+9.

Example 17

medium
Factor x2−13x+40.

Example 18

medium
Factor x2−11x+28.

Example 19

medium
Factor 5x2+13x−6.

Example 20

challenge
Factor completely: 6x3−5x2−6x.

Example 21

challenge
Factor x4−5x2+4 completely.

Example 22

challenge
Factor 6x2+x−12.

Example 23

easy
Factor x2+9x+20.

Example 24

easy
Factor x2−8x+15.

Example 25

easy
Factor x2+4x−12.

Example 26

easy
Factor x2+11x+18.

Example 27

easy
Factor x2−5x−14.

Example 28

medium
Factor x2+12x+36.

Example 29

medium
Factor x2−9x+14.

Example 30

medium
Factor x2+3x−10.

Example 31

medium
Solve x2−5x+6=0 by factoring.

Example 32

medium
Factor x2+13x+42.

Example 33

hard
Factor completely: 3x3−12x2+12x.

Example 34

hard
Factor x2−10xy+16y2.

Example 35

hard
Factor 4x2+4x−15.
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Background Knowledge

These ideas may be useful before you work through the harder examples.

factoringpolynomial multiplication