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ACT Math, ACT Prep, Factorization & Simplification

ACT Math Factorization & Simplification Complete Study Guide

ACT Math: Factorization & Simplification – Complete Study Guide | The School of Mathematics
ACT Math — Full Lesson

Factorization & Simplification

A complete, test-focused breakdown of every Factorization & Simplification concept on the ACT — from pulling out a GCF to factoring trinomials, recognizing special patterns, simplifying complex algebraic expressions, and applying these skills to solve equations.

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What Is Factoring & Why It Matters

Factoring is the process of rewriting an expression as a product of simpler expressions (its factors). It is the algebraic equivalent of reverse-multiplication — you are un-distributing or un-multiplying.

Expanding (forward) (x + 3)(x + 4) = x² + 7x + 12

Multiplying factors → expanded form
Factoring (reverse) x² + 7x + 12 = (x + 3)(x + 4)

Expanded form → product of factors
Why the ACT Cares

Factoring unlocks four major ACT skills:

1. Solving quadratic and polynomial equations (set each factor = 0).
2. Simplifying rational expressions (cancel common factors).
3. Evaluating expressions by recognizing a pattern and substituting.
4. Identifying zeros, roots, and x-intercepts of functions.

The Master Factoring Checklist

Every time you see a polynomial to factor on the ACT, run through these steps in order:

1. GCF?
2. Binomial?
3. Trinomial?
4. Four Terms?
5. Factor Further?

Always extract a GCF first — it simplifies every subsequent step. Then identify the structure of what remains.

Factoring Out the GCF

The first step in every factoring problem is to look for a Greatest Common Factor shared by all terms. Pull it out front using the distributive property in reverse.

GCF(all terms) × (remaining polynomial)
// Think: what divides evenly into every term — coefficients AND variables

Finding the GCF of a Polynomial

  • Numeric GCF: Largest integer dividing all coefficients.
  • Variable GCF: Lowest power of each variable that appears in every term.
  • Combined GCF: Multiply numeric GCF × variable GCF.
📐 Worked Example — Numeric & Variable GCF
Factor: 12x³y² − 8x²y³ + 20x²y
01Numeric GCF of 12, 8, 20 → GCF = 4
02Variable GCF: x appears as x³, x², x² → lowest power = x²
03y appears as y², y³, y¹ → lowest power = y
04Combined GCF = 4x²y
05Divide each term: 12x³y²÷4x²y = 3xy ; −8x²y³÷4x²y = −2y² ; 20x²y÷4x²y = 5
4x²y(3xy − 2y² + 5)
📐 Worked Example — Binomial GCF
Factor: 3x(x + 5) − 7(x + 5)
01The entire binomial (x + 5) appears in both terms — it IS the GCF
02Factor it out: (x + 5)(3x − 7)
(x + 5)(3x − 7)
⚠️

Don't skip the GCF step! Missing a GCF at the start leads to much harder factoring later. If the leading coefficient in a trinomial is large, always check for a numeric GCF first — it may reduce a complex problem to a simple one.

Difference of Squares

The difference of squares is the most frequently tested factoring pattern on the ACT. It applies whenever you have two perfect-square terms being subtracted.

a² − b² = (a + b)(a − b)
// Note: a² + b² does NOT factor over the reals — sum of squares is PRIME
Recognition Checklist

✓ Exactly two terms
✓ Minus sign between them
✓ Both terms are perfect squares (coefficients and variable powers are both perfect squares)

📐 Worked Example — Basic
Factor: x² − 49
01x² = (x)² and 49 = (7)² → both perfect squares, minus sign ✓
02a = x, b = 7
(x + 7)(x − 7)
📐 Worked Example — With Coefficients
Factor: 25x² − 36y²
0125x² = (5x)² and 36y² = (6y)² ✓
02a = 5x, b = 6y
(5x + 6y)(5x − 6y)
📐 Worked Example — Double Difference of Squares
Factor completely: x⁴ − 81
01x⁴ = (x²)² and 81 = (9)² → first application: (x² + 9)(x² − 9)
02x² + 9 → sum of squares → cannot be factored further (prime over reals)
03x² − 9 = (x + 3)(x − 3) → difference of squares again
(x² + 9)(x + 3)(x − 3)
💡

Always factor completely. After one round, check whether any resulting factor can be factored again. x⁴ − 81 requires two rounds. The ACT will include the intermediate (incomplete) factorization as a trap answer choice.

Perfect Square Trinomials

A perfect square trinomial is the expansion of a squared binomial. Recognizing it lets you factor in one step instead of working through the full trinomial method.

a² + 2ab + b² = (a + b)²
a² − 2ab + b² = (a − b)²
Recognition Pattern

First term is a perfect square → a²
Last term is a perfect square → b²
Middle term = exactly ±2ab (twice the product of the square roots of the first and last terms)

📐 Worked Example — Verify Then Factor
Factor: 4x² − 12x + 9
01First term: 4x² = (2x)² ✓ → a = 2x
02Last term: 9 = (3)² ✓ → b = 3
03Middle term check: 2ab = 2(2x)(3) = 12x. We have −12x → matches a² − 2ab + b² ✓
(2x − 3)²
📐 Worked Example — Recognizing a Non-Perfect-Square
Is x² + 8x + 12 a perfect square trinomial?
01First term: x² = (x)² ✓ → a = x
02Last term: 12 is NOT a perfect square ✗
03Not a perfect square trinomial → use the standard trinomial method instead
Not a PST → factors as (x + 2)(x + 6)
⚠️

Verify the middle term. Students often see two perfect square terms and assume it's a PST — but the middle term must be exactly 2ab. Always check all three conditions before concluding.

Factoring Trinomials When a = 1

The most common factoring task on the ACT: given x² + bx + c, find two binomials (x + p)(x + q) such that p + q = b and p × q = c.

x² + bx + c = (x + p)(x + q)
where: p + q = b   AND   p × q = c

The "Product-Sum" Method

Find two numbers that multiply to c and add to b. List factor pairs of c and check which pair sums to b.

📐 Worked Example — Both Factors Positive
Factor: x² + 9x + 20
01Need: p × q = 20 and p + q = 9
02Factor pairs of 20: (1,20) sum 21 ✗ ; (2,10) sum 12 ✗ ; (4,5) sum 9 ✓
03p = 4, q = 5
(x + 4)(x + 5)
📐 Worked Example — Mixed Signs
Factor: x² − 3x − 28
01Need: p × q = −28 and p + q = −3
02Product is negative → one factor positive, one negative
03Pairs of 28: (4,7) → try (−7, +4): product = −28 ✓, sum = −3 ✓
(x − 7)(x + 4)
📐 Worked Example — Both Factors Negative
Factor: x² − 11x + 30
01Need: p × q = 30 (positive) and p + q = −11 (negative)
02Product positive, sum negative → both factors are negative
03Negative pairs of 30: (−1,−30) sum −31 ✗ ; (−2,−15) sum −17 ✗ ; (−5,−6) sum −11 ✓
(x − 5)(x − 6)

Sign Rules Summary

c (last term)b (middle term)Both factors are…Example
PositivePositiveBoth positivex²+5x+6 → (+2)(+3)
PositiveNegativeBoth negativex²−5x+6 → (−2)(−3)
NegativePositiveLarger factor positivex²+x−6 → (+3)(−2)
NegativeNegativeLarger factor negativex²−x−6 → (−3)(+2)
💡

Verify by expanding. After factoring, quickly FOIL your answer: (x + p)(x + q) = x² + (p+q)x + pq. If you get back the original trinomial, you're correct. This 5-second check saves you from trap answers on the ACT.

Factoring Trinomials When a ≠ 1

When the leading coefficient isn't 1, you have ax² + bx + c with a ≠ 1. Two reliable methods: the AC Method (factoring by splitting the middle term) and Trial and Error.

Method 1 — The AC Method

Find ac
Find p, q with pq=ac and p+q=b
Split bx into px + qx
Factor by grouping
📐 Worked Example — AC Method
Factor: 6x² + 11x + 3
01a=6, b=11, c=3 → ac = 6×3 = 18
02Find p, q: pq = 18 and p+q = 11 → (2)(9) = 18 and 2+9 = 11 ✓
03Rewrite middle term: 6x² + 2x + 9x + 3
04Group: (6x² + 2x) + (9x + 3) = 2x(3x + 1) + 3(3x + 1)
05Factor out (3x + 1): (3x + 1)(2x + 3)
(3x + 1)(2x + 3)

Method 2 — Trial and Error

Write (dx + e)(fx + g) where d×f = a and e×g = c. Test combinations until the outer+inner products sum to b.

📐 Worked Example — Trial and Error
Factor: 2x² + 7x + 6
01a = 2 → factor pairs for first terms: (2x)(x)
02c = 6, positive → factor pairs: (1,6), (2,3), (3,2), (6,1)
03Try (2x + 3)(x + 2): outer = 4x, inner = 3x, sum = 7x ✓
(2x + 3)(x + 2)
📐 Worked Example — Negative Middle Term
Factor: 3x² − 10x + 8
01ac = 3×8 = 24. Need pq = 24 and p+q = −10 → both negative
02Negative pairs of 24: (−4)(−6) = 24 and −4 + (−6) = −10 ✓
03Rewrite: 3x² − 4x − 6x + 8
04Group: x(3x − 4) − 2(3x − 4) = (3x − 4)(x − 2)
(3x − 4)(x − 2)
💡

Always pull GCF first. If 2x² + 7x + 6 had been 4x² + 14x + 12, pulling out the GCF of 2 first gives 2(2x² + 7x + 6) — then factor the simpler trinomial. Missing the GCF forces you to work with larger numbers unnecessarily.

Factoring by Grouping

When a polynomial has four terms, factoring by grouping is usually the method. Pair the first two and last two terms, factor each pair separately, then factor out the common binomial.

ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y)
📐 Worked Example — Standard Grouping
Factor: x³ + 3x² + 5x + 15
01Group: (x³ + 3x²) + (5x + 15)
02Factor each group: x²(x + 3) + 5(x + 3)
03Common binomial factor (x + 3): (x² + 5)(x + 3)
(x² + 5)(x + 3)
📐 Worked Example — Regrouping When First Attempt Fails
Factor: 2xy − 6y + 3x − 9
01Group 1: (2xy − 6y) + (3x − 9) = 2y(x − 3) + 3(x − 3)
02Common binomial factor: (x − 3)
03Factor out: (2y + 3)(x − 3)
(2y + 3)(x − 3)
💡

If grouping doesn't work: Try a different pairing (first + third, second + fourth) or rearrange the terms. Sometimes you need to factor out −1 from one group to make the binomials match: factor (a − b) from one group and −(b − a) = (a − b) from the other.

Sum & Difference of Cubes

These patterns apply to binomials where both terms are perfect cubes. They appear less frequently than difference of squares but do show up on harder ACT questions.

a³ + b³ = (a + b)(a² − ab + b²)   ← Sum of cubes
a³ − b³ = (a − b)(a² + ab + b²)   ← Difference of cubes
Memory Device — SOAP

The sign pattern for cube factoring: S O A P

Same sign as the original operation (+ or −)
Opposite sign in the trinomial's second term
Always Positive for the last term of the trinomial

📐 Worked Example — Difference of Cubes
Factor: x³ − 27
01x³ = (x)³ and 27 = (3)³ → a = x, b = 3
02Difference of cubes: (a − b)(a² + ab + b²)
03Substitute: (x − 3)(x² + 3x + 9)
04Check: x² + 3x + 9 has discriminant 9 − 36 = −27 < 0 → cannot factor further ✓
(x − 3)(x² + 3x + 9)
📐 Worked Example — Sum of Cubes with Coefficient
Factor: 8x³ + 125
018x³ = (2x)³ and 125 = (5)³ → a = 2x, b = 5
02Sum of cubes: (a + b)(a² − ab + b²)
03Substitute: (2x + 5)((2x)² − (2x)(5) + (5)²)
04Simplify: (2x + 5)(4x² − 10x + 25)
(2x + 5)(4x² − 10x + 25)

Perfect Cubes to Memorize

1³=1   2³=8   3³=27   4³=64   5³=125
6³=216   7³=343   8³=512   9³=729   10³=1000

Simplifying Algebraic Expressions

Simplification means rewriting an expression in its most compact equivalent form. On the ACT this includes combining like terms, applying exponent rules, and using the distributive property carefully.

Combining Like Terms

Terms are like terms if they have identical variable parts (same variable, same exponent). Add or subtract their coefficients only.

📐 Worked Example — Combining Like Terms
Simplify: 5x² − 3xy + 2x² + 7xy − x²
01Group x² terms: 5x² + 2x² − x² = 6x²
02Group xy terms: −3xy + 7xy = 4xy
6x² + 4xy

Distributive Property & Expanding

Distribute carefully — especially with negative signs, which must multiply every term inside the parentheses.

📐 Worked Example — Negative Distribution Trap
Simplify: 3(x² − 4x + 1) − 2(x² + 3x − 5)
01Distribute 3: 3x² − 12x + 3
02Distribute −2 (EVERY term): −2x² − 6x + 10
03Combine: (3x² − 2x²) + (−12x − 6x) + (3 + 10)
x² − 18x + 13

Exponent Rules for Simplification

RuleFormExample
Product rulexᵃ · xᵇ = xᵃ⁺ᵇx³ · x⁵ = x⁸
Quotient rulexᵃ / xᵇ = xᵃ⁻ᵇx⁷ / x² = x⁵
Power rule(xᵃ)ᵇ = xᵃᵇ(x²)⁴ = x⁸
Product to power(xy)ᵃ = xᵃyᵃ(2x)³ = 8x³
Quotient to power(x/y)ᵃ = xᵃ/yᵃ(x/3)² = x²/9
Negative exponentx⁻ᵃ = 1/xᵃx⁻³ = 1/x³
Zero exponentx⁰ = 1 (x≠0)(5x)⁰ = 1
Fractional exponentx^(m/n) = ⁿ√(xᵐ)x^(3/2) = (√x)³
📐 Worked Example — Multi-Rule Simplification
Simplify: (3x²y)³ / (9x³y²)
01Expand numerator: (3)³ · (x²)³ · y³ = 27x⁶y³
02Divide: 27x⁶y³ / 9x³y² = (27/9) · x⁶⁻³ · y³⁻²
03Simplify: 3 · x³ · y¹
3x³y

Simplifying Radical Expressions

√(ab) = √a · √b   ← Split under radicals
√(a/b) = √a / √b   ← Quotient under radical
√(a²) = |a|   ← Perfect square out of radical
📐 Worked Example — Simplifying Radicals
Simplify: √(75x⁴y³)
01Factor under radical: √(25 · 3 · x⁴ · y² · y)
02Pull out perfect squares: √25 · √(x⁴) · √(y²) · √(3y)
03Simplify: 5 · x² · y · √(3y)
5x²y√(3y)

Simplifying Rational Expressions

A rational expression is a fraction with polynomials in the numerator and/or denominator. Simplifying one means factoring both and canceling common factors — never canceling terms.

The 3-Step Process

Step 1: Factor the numerator completely.
Step 2: Factor the denominator completely.
Step 3: Cancel any factor that appears in both numerator and denominator. State restrictions (denominator ≠ 0).

📐 Worked Example — Standard Simplification
Simplify: (2x² + x − 6) / (x² − x − 2)
01Factor numerator: 2x² + x − 6 → ac = −12, pairs summing to 1: (4,−3) → (2x − 3)(x + 2)
02Factor denominator: x² − x − 2 → (x − 2)(x + 1)
03Write as fraction: [(2x − 3)(x + 2)] / [(x − 2)(x + 1)]
04No common factors → already fully simplified
(2x − 3)(x + 2) / [(x − 2)(x + 1)], where x ≠ 2, −1
📐 Worked Example — Canceling a Common Factor
Simplify: (x² − 9) / (x² + 4x + 3)
01Factor numerator: x² − 9 = (x + 3)(x − 3)
02Factor denominator: x² + 4x + 3 = (x + 3)(x + 1)
03Cancel (x + 3): what remains is (x − 3)/(x + 1)
(x − 3)/(x + 1), where x ≠ −3, −1
📐 Worked Example — Opposite Factors
Simplify: (3 − x) / (x² − 9)
01Factor denominator: x² − 9 = (x + 3)(x − 3)
02Numerator (3 − x) = −(x − 3) → note the opposite factor
03Rewrite: −(x − 3) / [(x + 3)(x − 3)]
04Cancel (x − 3): −1/(x + 3)
−1/(x + 3), where x ≠ ±3
⚠️

Terms vs. Factors — the #1 ACT error:
ILLEGAL: (x + 5)/(x + 7) → 5/7 by "canceling x" ✗
LEGAL: x(x + 5) / [x(x + 7)] → (x + 5)/(x + 7) by canceling the factor x ✓
You can only cancel a factor that multiplies the entire numerator or the entire denominator.

Solving Equations by Factoring

Factoring is the primary tool for solving quadratic and polynomial equations on the ACT. The foundation is the Zero Product Property.

Zero Product Property

If A × B = 0, then A = 0 OR B = 0 (or both).

This is why factoring to solve equations works: once the polynomial equals zero, each factor can independently be set to zero to find solutions.

Standard 4-Step Process

Move all terms to one side (= 0)
Factor completely
Set each factor = 0
Solve each equation
📐 Worked Example — Quadratic Equation
Solve: x² − 5x = 14
01Move all terms to left: x² − 5x − 14 = 0
02Factor: need pq = −14 and p+q = −5 → (−7)(+2) → (x − 7)(x + 2) = 0
03Set each factor to zero: x − 7 = 0 → x = 7   OR   x + 2 = 0 → x = −2
x = 7 or x = −2
📐 Worked Example — Don't Divide Away a Root!
Solve: 3x² = 12x
01Move all terms to left: 3x² − 12x = 0
02Factor GCF: 3x(x − 4) = 0
03Set each factor to zero: 3x = 0 → x = 0   OR   x − 4 = 0 → x = 4
x = 0 or x = 4
⚠️

Never divide both sides by a variable to "cancel" it. In 3x² = 12x, dividing both sides by x gives 3x = 12, so x = 4 only — you lose the solution x = 0! Always move everything to one side and factor instead.

📐 Worked Example — Higher-Degree Equation
Solve: x³ − 4x = 0
01Factor GCF first: x(x² − 4) = 0
02Factor further: x(x + 2)(x − 2) = 0
03Set each factor to zero: x=0, x+2=0 → x=−2, x−2=0 → x=2
x = −2, 0, or 2 (three solutions)

Using Factoring to Find Zeros of a Function

The zeros (or roots) of f(x) are the x-values where f(x) = 0. They are the x-intercepts of the graph. Factor f(x) and set each factor to zero.

📐 Worked Example — Zeros of a Function
Find all zeros of f(x) = 2x³ + 2x² − 12x
01Factor GCF: 2x(x² + x − 6)
02Factor trinomial: 2x(x + 3)(x − 2)
03Set each factor = 0: x = 0, x = −3, x = 2
Zeros: x = −3, 0, 2

Test Day Strategy

Complete Factoring Pattern Reference

Pattern 01
GCF
ka + kb + kc = k(a + b + c)
Always check first — every time
Pattern 02
Difference of Squares
a² − b² = (a+b)(a−b)
a² + b² does NOT factor
Pattern 03
Perfect Square Trinomial
a² ± 2ab + b² = (a ± b)²
Verify: middle term = ±2ab
Pattern 04
Trinomial (a = 1)
x² + bx + c = (x+p)(x+q)
pq=c, p+q=b
Product-Sum method
Pattern 05
Trinomial (a ≠ 1)
ax² + bx + c
AC method or trial & error
Find p,q where pq=ac, p+q=b
Pattern 06
Grouping (4 terms)
ax+ay+bx+by
= a(x+y)+b(x+y)
= (a+b)(x+y)
Pair, factor each pair, extract common binomial
Pattern 07
Sum of Cubes
a³ + b³ = (a+b)(a²−ab+b²)
SOAP: Same Opposite Always Positive
Pattern 08
Difference of Cubes
a³ − b³ = (a−b)(a²+ab+b²)
Trinomial factor is always irreducible

Top Traps to Avoid

  • Canceling terms, not factors: (x+3)/(x+5) ≠ 3/5.
  • Forgetting x = 0 as a solution when a variable is a factor.
  • Trying to factor a² + b² (sum of squares — it doesn't factor over reals).
  • Not factoring completely — one more round may be possible.
  • Skipping the GCF step and working with larger coefficients.
  • Dividing both sides by a variable (loses a solution).
  • Forgetting to state that the equation must equal zero before using Zero Product Property.

Speed Strategies

  • Always factor GCF first — it simplifies every subsequent step.
  • For x² + bx + c: use the sign rules table to know which sign combinations to test.
  • Verify any factoring by quickly FOILing your answer.
  • Plug in answer choices to check solutions — faster than re-solving on some problems.
  • Spot difference of squares instantly: two terms, perfect squares, minus sign.
  • For higher-degree polynomials, always look for GCF + pattern in what remains.
~8–12 questions per test Highest ROI algebra topic Appears on every ACT

Put It to the Test

You've covered every factoring pattern and simplification technique. Now build speed and accuracy with ACT-style timed questions. Work through all five quizzes — each one targets different patterns and difficulty levels.

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