13.3 Factoring Techniques

13.3 Factoring Techniques

Quick Answer: Factoring rewrites a polynomial as a product of simpler polynomials. Always pull out the GCF first. Factor trinomials by finding numbers that multiply to the constant product and add to the middle coefficient; use difference of squares for $a^2 - b^2$. Factor completely, then use the zero product property to solve equations set equal to zero.

Factoring is the reverse of the multiplication skills from Section 13.1. On PERT Math it appears both as “which expression is equivalent?” simplify items and as the gateway to solving quadratic equations without the quadratic formula. Treat every factoring problem as: undo FOIL / undoistribute until nothing factors further over the integers.

Strategy Order (Use This Every Time)

1. GCF of all terms (numbers and variables).
2. Count terms:
   • 2 terms: difference of squares (or later sum/difference of cubes if your course includes them).
   • 3 terms: trinomial patterns ($x^2+bx+c$ or $ax^2+bx+c$).
   • 4 terms: factor by grouping.
3. Factor completely—look again inside each factor for another GCF or special pattern.
4. Optional: check by multiplying factors back to the original polynomial.

Greatest Common Factor (GCF)

The GCF is the largest monomial that divides every term. Factor it out using the distributive property in reverse: $ab + ac = a(b + c)$.

Worked Example 1: Numeric and Variable GCF

Problem: Factor $12x^3 - 18x^2 + 6x$.

GCF of coefficients: $6$ (largest integer dividing 12, 18, and 6).

GCF of variables: $x$ (smallest power of $x$ present).

GCF: $6x$.

$12x^3 - 18x^2 + 6x = 6x(2x^2 - 3x + 1)$

Check: $6x \cdot 2x^2 = 12x^3$, $6x \cdot (-3x) = -18x^2$, $6x \cdot 1 = 6x$. Good.

Note: You can still factor the trinomial $2x^2 - 3x + 1 = (2x - 1)(x - 1)$ later when factoring completely.

Worked Example 2: Factoring Out −1

Problem: Factor $-x^2 + 5x - 6$.

Factoring out $-1$ makes the leading coefficient positive, which simplifies trinomial search:

$-x^2 + 5x - 6 = -1(x^2 - 5x + 6) = -(x - 2)(x - 3)$

Either $-(x-2)(x-3)$ or $(-x+2)(x-3)$ style forms may appear; multiply to verify equivalence to the original.

Trinomials of the Form $x^2 + bx + c$

Find two integers $m$ and $n$ such that:

$m \cdot n = c \quad \text{and} \quad m + n = b$

Then $x^2 + bx + c = (x + m)(x + n)$.

Worked Example 3: Positive Factors

Problem: Factor $x^2 + 9x + 20$.

Pairs of factors of $20$: $(1,20)$, $(2,10)$, $(4,5)$ and negatives.

Which pair adds to $9$? $4 + 5 = 9$.

$x^2 + 9x + 20 = (x + 4)(x + 5)$

FOIL check: $x^2 + 5x + 4x + 20 = x^2 + 9x + 20$.

Worked Example 4: Negative Constant (Opposite Signs)

Problem: Factor $x^2 + 2x - 15$.

Need product $-15$ and sum $+2$. Opposite signs: try $(-3, 5)$ because $-3 + 5 = 2$ and $(-3)(5) = -15$.

$x^2 + 2x - 15 = (x - 3)(x + 5)$

Sign rule of thumb:

Worked Example 5: Both Factors Negative

Problem: Factor $x^2 - 8x + 12$.

Product $+12$, sum $-8$ → both negative: $-2$ and $-6$.

$x^2 - 8x + 12 = (x - 2)(x - 6)$

Trinomials $ax^2 + bx + c$ with $a \neq 1$

Use the $ac$ method (factoring by grouping):

1. Compute $a \cdot c$.
2. Find integers $m$, $n$ with $m \cdot n = ac$ and $m + n = b$.
3. Rewrite $bx$ as $mx + nx$.
4. Factor by grouping the four terms.

Worked Example 6: ac Method

Problem: Factor $6x^2 + 11x + 3$.

$a = 6$, $c = 3$, $ac = 18$. Pairs of 18 that add to $11$: $2$ and $9$.

Rewrite: $6x^2 + 2x + 9x + 3$

Group: $(6x^2 + 2x) + (9x + 3) = 2x(3x + 1) + 3(3x + 1)$

Factor common binomial: $(2x + 3)(3x + 1)$

FOIL check: $2x \cdot 3x = 6x^2$, $2x \cdot 1 = 2x$, $3 \cdot 3x = 9x$, $3 \cdot 1 = 3$ → $6x^2 + 11x + 3$. Perfect.

Worked Example 7: Leading Coefficient and Negative Middle

Problem: Factor $4x^2 - 8x - 5$.

$ac = 4(-5) = -20$. Need product $-20$, sum $-8$: try $2$ and $-10$ ($2 + (-10) = -8$).

$4x^2 + 2x - 10x - 5 = 2x(2x + 1) - 5(2x + 1) = (2x - 5)(2x + 1)$

Check: $(2x - 5)(2x + 1) = 4x^2 + 2x - 10x - 5 = 4x^2 - 8x - 5$.

Difference of Squares

$a^2 - b^2 = (a + b)(a - b)$

Both parts must be perfect squares (or perfect-square monomials), and the operation between them must be subtraction.

Worked Example 8: Basic Difference of Squares

Problem: Factor $x^2 - 49$.

$x^2 - 49 = x^2 - 7^2 = (x + 7)(x - 7)$

Worked Example 9: Coefficients and Higher Powers

Problem: Factor $9x^2 - 16$.

$9x^2 - 16 = (3x)^2 - 4^2 = (3x + 4)(3x - 4)$

Problem: Factor $x^4 - 81$.

First: $x^4 - 81 = (x^2)^2 - 9^2 = (x^2 + 9)(x^2 - 9)$.

Continue on $x^2 - 9$: $(x^2 + 9)(x + 3)(x - 3)$.

Over the reals/integers in PERT scope, $x^2 + 9$ does not factor further as a difference of squares (it is a sum of squares).

Trap: Trying to factor $x^2 + 49$ as $(x+7)(x-7)$—that product is $x^2 - 49$, not $x^2 + 49$.

Factoring Completely

“Factor completely” means apply GCF, special patterns, and trinomial methods until each polynomial factor is irreducible over the integers.

Worked Example 10: Multi-Step Complete Factorization

Problem: Factor completely: $2x^3 - 50x$.

Step 1 — GCF: $2x(x^2 - 25)$

Step 2 — Difference of squares: $2x(x - 5)(x + 5)$

Stopping at $2x(x^2 - 25)$ is incomplete if the question says “factor completely.”

Worked Example 11: GCF Then Trinomial

Problem: Factor completely: $3x^2 + 18x + 24$.

GCF $3$: $3(x^2 + 6x + 8)$

Trinomial: $x^2 + 6x + 8 = (x + 2)(x + 4)$

Complete: $3(x + 2)(x + 4)$

Factoring to Solve Equations

If a product equals zero, at least one factor equals zero (zero product property):

$A \cdot B = 0 \implies A = 0 \text{ or } B = 0$

Always move all terms to one side first so the equation reads $\text{polynomial} = 0$.

Worked Example 12: Solve by Factoring

Problem: Solve $x^2 - 5x = 14$.

Step 1 — Set equal to zero: $x^2 - 5x - 14 = 0$

Step 2 — Factor: Find numbers with product $-14$ and sum $-5$: $-7$ and $2$.

$(x - 7)(x + 2) = 0$

Step 3 — Zero product: $x - 7 = 0$ or $x + 2 = 0$ → $x = 7$ or $x = -2$.

Check: $7^2 - 5(7) = 49 - 35 = 14$. $(-2)^2 - 5(-2) = 4 + 10 = 14$. Both solutions work.

Worked Example 13: Include the GCF Factor

Problem: Solve $2x^2 = 8x$.

$2x^2 - 8x = 0 \Rightarrow 2x(x - 4) = 0 \Rightarrow x = 0 \text{ or } x = 4$

Trap: Dividing both sides by $2x$ at the start loses the solution $x = 0$. Factoring keeps every solution.

FOIL Reverse Mental Model

When you FOIL $(x + m)(x + n)$, you get $x^2 + (m+n)x + mn$. Factoring reads that identity backward: the middle coefficient is a sum, the constant is a product. Practice saying “what two numbers multiply to ___ and add to ___?” until it is automatic under adaptive-test pressure.

Quick Reference Table

PERT Habits That Raise Accuracy

1. Write “= 0” before solving; never solve by factoring an expression that is not set to zero.
2. Pull GCF first—it shrinks coefficients and often reveals a difference of squares.
3. List factor pairs systematically when stuck; do not guess randomly forever.
4. Multiply back when two answer choices look similar.
5. Watch for “factor completely” wording so you do not stop one step early.

With operations, division, and factoring in place, you are ready for quadratic equations, rational expressions, and application problems that assemble these skills into multi-step PERT Math items.