7.2 Factoring Methods for Quadratics, Cubics, and Polynomials

A Content Area and a Prerequisite

Factoring: Factoring methods applied to quadratics, cubics, and polynomials.

The Advanced Algebra and Functions test lists factoring as one of its eleven content areas, worth 1 to 2 of the 20 questions. Its real importance is larger: solving quadratics, finding polynomial roots, and simplifying rational expressions all run through factoring. Weak factoring shows up as lost points across four content areas, not one.

The Ordered Procedure

Work the steps in this order every time. The order is not arbitrary — each step makes the next easier.

1. Greatest common factor (GCF) — always first.
2. Count the terms:
   • 2 terms → difference of squares, or sum/difference of cubes
   • 3 terms → trinomial methods
   • 4 terms → grouping
3. Check whether any factor factors again.

Step 1: Greatest Common Factor

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

Take the largest numerical factor and the lowest power of each shared variable.

$6x^2y + 9xy^2 - 3xy = 3xy(2x + 3y - 1)$

The trailing $-1$ matters. When a term is entirely consumed by the GCF, it leaves a 1 behind, not nothing — dropping it is a frequent error.

Extracting the GCF first often exposes a familiar form:

$2x^2 - 18 = 2(x^2 - 9) = 2(x+3)(x-3)$

Without the GCF step, $2x^2 - 18$ does not look like a difference of squares.

Step 2A: Two Terms

Difference of Squares

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

$x^2 - 49 = (x+7)(x-7)$ $9x^2 - 25 = (3x+5)(3x-5)$ $x^4 - 16 = (x^2+4)(x^2-4) = (x^2+4)(x+2)(x-2)$

That last example shows why step 3 exists: the first factoring was incomplete.

A sum of squares does not factor over the real numbers. $x^2 + 49$ is prime. Offering $(x+7)(x+7)$ is wrong — that expands to $x^2 + 14x + 49$.

Sum and Difference of Cubes

$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$

$x^3 - 27 = (x-3)(x^2 + 3x + 9)$

The sign pattern: the binomial takes the same sign as the original; the middle term of the trinomial takes the opposite sign; the last term is always positive.

Step 2B: Three Terms

Leading Coefficient 1

For $x^2 + bx + c$, find two numbers whose product is $c$ and whose sum is $b$.

$x^2 + 11x + 24$: product 24, sum 11 → 3 and 8 → $(x+3)(x+8)$ $x^2 - 5x - 24$: product −24, sum −5 → −8 and 3 → $(x-8)(x+3)$

Sign guide:

Perfect Square Trinomials

$a^2 + 2ab + b^2 = (a+b)^2 \qquad a^2 - 2ab + b^2 = (a-b)^2$

$x^2 - 12x + 36 = (x-6)^2$

Recognize these by checking whether the first and last terms are perfect squares and the middle term is twice the product of their roots.

Leading Coefficient Not 1 — the AC Method

For $ax^2 + bx + c$:

1. Compute $a \cdot c$.
2. Find two numbers with product $ac$ and sum $b$.
3. Split the middle term using them.
4. Factor by grouping.

Factor $6x^2 + 11x - 10$.

1. $ac = 6(-10) = -60$
2. Product −60, sum 11 → 15 and −4
3. $6x^2 + 15x - 4x - 10$
4. Group: $3x(2x + 5) - 2(2x + 5) = (2x+5)(3x-2)$

Check by expanding: $6x^2 - 4x + 15x - 10 = 6x^2 + 11x - 10$ ✓

The method is mechanical and reliable, which makes it preferable to guess-and-check when $a$ has several factor pairs.

Step 2C: Four Terms — Grouping

Pair the terms and extract a GCF from each pair. If the remaining binomials match, factor it out.

$x^3 + 4x^2 - 3x - 12$ $= x^2(x + 4) - 3(x + 4)$ $= (x+4)(x^2 - 3)$

The sign discipline matters: factoring $-3$ (not $+3$) from the second pair is what makes both binomials read $(x+4)$.

$2x^3 - 6x^2 + 5x - 15 = 2x^2(x-3) + 5(x-3) = (x-3)(2x^2+5)$

If the binomials do not match, try reordering the terms before concluding the expression is prime.

Factoring Cubics

The AAF specification names cubics explicitly. Three approaches cover nearly all cases:

1. GCF, if available: $2x^3 - 8x = 2x(x^2-4) = 2x(x+2)(x-2)$
2. Sum/difference of cubes, if it is a two-term cube.
3. Grouping, for four terms.

Verification

Every factoring can be checked by expanding, and under untimed conditions there is no reason not to. A faster check for a specific answer: substitute a convenient value into both the original and the factored form.

$6x^2 + 11x - 10$ at $x = 1$: $6 + 11 - 10 = 7$ $(2x+5)(3x-2)$ at $x = 1$: $(7)(1) = 7$ ✓

On a multiple-choice test this converts factoring from a production task into a verification task — expand or test each choice and keep the one that reproduces the original.