D · S · T · C · G
Difference of squares | Sum/Square (perfect) | Trinomial | Cubes | GCF first always!
Always pull out GCF FIRST before doing anything else!
G · F · O · (Greatest Factor Out)
Find the Greatest common factor → Factor it Out → write what's left in parentheses.
📝 Example
Factor: \( 6x^3 + 9x^2 - 3x \)
GCF of 6, 9, 3 =
3 | GCF of \(x^3, x^2, x\) =
x
→ \( 3x(2x^2 + 3x - 1) \)
1
Factor completely: \( 12x^4 - 8x^3 + 4x^2 \)
easy
What is the GCF of 12, 8, 4? What is the GCF of \(x^4, x^3, x^2\)?
2
Factor out the GCF: \( 15a^2b^3 - 10ab^4 + 5a^3b^2 \)
easy
Two variables! Find GCF for both \(a\) and \(b\) separately.
Students forget to include the GCF in the final answer, or they don't factor out ALL the GCF at once.
D·O·T·S
Difference · Of · Two · Squares → \((a+b)(a-b)\)
Must be: subtraction, both terms are perfect squares.
📝 Example
Factor: \( 25x^2 - 49 \)
\( = (5x)^2 - 7^2 \) ← identify \(a = 5x,\ b = 7\)
\( = (5x + 7)(5x - 7) \) ✓
3
Factor: \( x^2 - 64 \)
easy
4
Factor completely: \( 2x^2 - 50 \)
easy
GCF FIRST!
Don't forget to pull out the GCF before applying DOTS!
5
Factor completely: \( x^4 - 81 \)
tricky!
Apply DOTS twice! \(x^4 = (x^2)^2\) and then one factor can be factored again!
\( x^2 + 64 \) cannot be factored (sum, not difference). SUM of squares is PRIME over reals!
S · O · A · P
Square first term | Outer–inner = \(2ab\) | Add or subtract | Perfect square last
→ Check: is the middle term exactly \(2 \cdot \sqrt{\text{first}} \cdot \sqrt{\text{last}}\) ?
📝 Example
Factor: \( x^2 + 10x + 25 \)
\(\sqrt{x^2} = x,\quad \sqrt{25} = 5,\quad 2(x)(5) = 10x\) ✓ (matches!)
\( = (x + 5)^2 \)
6
Factor: \( x^2 - 14x + 49 \)
easy
Check: \( 2 \cdot x \cdot 7 = 14x\) ✓ Minus sign → \((x - 7)^2\)
7
Factor: \( 4x^2 + 12x + 9 \)
medium
First term is \( (2x)^2 \), last term is \(3^2\). Check middle: \(2(2x)(3) = 12x\) ?
Students write \((x+5)(x+5)\) instead of \((x+5)^2\) — both are correct but know they are the same!
M · A · S · S
Find two numbers that Multiply to \(c\) · And · Sum to \(b\)
Sign rules: both + → both positive | both − → bigger negative, smaller positive
📝 Example
Factor: \( x^2 + 7x + 12 \)
Need: multiply to \(12\), add to \(7\)
Pairs: \(1 \times 12,\ 2 \times 6,\ \mathbf{3 \times 4}\) → \(3+4=7\) ✓
\( = (x+3)(x+4) \)
8
Factor: \( x^2 + 8x + 15 \)
easy
Multiply to 15 AND add to 8 → ?, ?
9
Factor: \( x^2 - 5x - 24 \)
medium
Multiply to −24 and add to −5. Signs are different when product is negative!
10
Factor: \( x^2 - 11x + 28 \)
easy
Product positive + sum negative → both numbers are negative!
When \(c\) is negative, the two numbers have OPPOSITE signs. Don't make both the same sign!
A · C · S · P · (AC Split)
A·C method: multiply \(a \times c\) → find factors that add to \(b\) → Split middle term → factor by grouping (Pair up)
📝 Example — AC Method
Factor: \( 6x^2 + 11x + 4 \)
\(a \times c = 6 \times 4 = 24\). Need factors of 24 that add to 11 → \(3 + 8 = 11\) ✓
Split: \( 6x^2 + 3x + 8x + 4 \)
Group: \( 3x(2x+1) + 4(2x+1) \)
\( = (3x+4)(2x+1) \)
11
Factor: \( 2x^2 + 7x + 3 \)
easy
\(a \times c = 2 \times 3 = 6\). What two numbers multiply to 6 and add to 7?
12
Factor: \( 3x^2 - 10x + 8 \)
medium
AC method
\(a \times c = 24\). Need multiply to 24, add to −10. Both negative!
13
Factor completely: \( 4x^2 - 8x - 5 \)
tricky!
\(a \times c = -20\). Opposite signs. Find pair for −20 that adds to −8.
Students forget to check if there's a GCF BEFORE using the AC method. Always check GCF first!
S · O · A · P
Same sign | Opposite sign | Always positive | Prime (can't factor more)
\(a^3 \pm b^3 = (a \underbrace{\pm}_{\text{S}} b)(a^2 \underbrace{\mp}_{\text{O}} ab + b^2\underbrace{\ \ +\ \ }_{\text{A}})\)
14
Factor: \( x^3 + 27 \)
easy
\( 27 = 3^3 \). So \(a = x,\ b = 3\). Use sum of cubes formula!
15
Factor: \( 8x^3 - 125 \)
medium
\( 8x^3 = (2x)^3 \) and \( 125 = 5^3 \). So \(a = 2x,\ b = 5\).
The trinomial part \((a^2 \mp ab + b^2)\) CANNOT be factored further. Students waste time trying to factor it!
G · P · F · R (Group · Pair · Factor · Repeat)
When you have 4 terms: group first 2 & last 2 → factor each pair → the parentheses must match → factor out the common binomial
📝 Example
Factor: \( x^3 + 2x^2 + 3x + 6 \)
Group: \( (x^3 + 2x^2) + (3x + 6) \)
Factor each: \( x^2(x + 2) + 3(x + 2) \)
Common binomial: \( (x+2)(x^2+3) \)
16
Factor by grouping: \( 2x^3 - 3x^2 + 4x - 6 \)
easy
17
Factor by grouping: \( ax - ay + bx - by \)
medium
variables!
Group \( (ax - ay) + (bx - by) \) → factor \(a\) from first, \(b\) from second
If binomials inside parentheses DON'T match after factoring each pair → try re-grouping differently!
G · D · T · C · (Strategy Checklist)
Always ask in order:
1. GCF? 2. Difference of squares? 3. Trinomial? 4. Cubes? 5. Grouping?
18
Factor completely: \( 3x^3 - 75x \)
easy
2 steps!
Step 1: Pull out GCF. Step 2: Look at what's left — what type is it?
19
Factor completely: \( 2x^3 + 8x^2 + 8x \)
medium
GCF first, then check — is what's left a perfect square trinomial?
20
Factor completely: \( x^4 - 13x^2 + 36 \)
tricky!
Substitution!
Let \( u = x^2 \). Then this becomes \( u^2 - 13u + 36 \). Factor as trinomial, then substitute back!
After: \( (x^2 - 4)(x^2 - 9) \) → can each be factored more?
After factoring once — always ask "can I factor this further?" Don't stop too early!
✅ Check your answers by expanding/multiplying back!
If you expand your answer and get the original expression → you're correct!
p. 1 · Factoring Self-Study