4.1 GCF, Grouping, and Special Products

4.1 GCF, Grouping, and Special Products

College Board’s Next-Generation ACCUPLACER Advanced Algebra and Functions (AAF) test weights factoring at 5–10% of the 20-item computer-adaptive exam — typically 1–2 items. Table 11 names the skill as factoring methods applied to quadratics, cubics, and polynomials. That is not a request to guess two numbers for a trinomial and stop. An AAF item can ask you to extract a greatest common factor, group four terms, recognize a difference of squares or a sum of cubes, or finish a perfect-square trinomial. The adaptive engine can also stack those moves: pull out a GCF, then keep factoring.

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations

Always start with the greatest common factor

The greatest common factor (GCF) of a polynomial is the largest monomial that divides every term. For the integer-coefficient polynomials AAF uses, that means three pieces at once:

1. The GCF of the coefficients — the largest positive integer that divides each coefficient.
2. The lowest power of each variable that appears in every term.
3. A sign convention: factor a negative leading coefficient when it cleans up later signs, but do not invent a minus that is not in the data.

Worked example: 6x³y − 9x²y²

Factor 6x^3y − 9x^2y^2 completely.

• Coefficients: GCF of 6 and 9 is 3.
• Variable x: the powers are 3 and 2, so take x².
• Variable y: the powers are 1 and 2, so take y.

The monomial GCF is 3x^2y. Divide each original term by that monomial:

6x^3y / 3x^2y = 2x

9x^2y^2 / 3x^2y = 3y

The second original term is negative, so the leftover binomial is 2x − 3y, not 2x + 3y.

Result: 6x^3y − 9x^2y^2 = 3x^2y(2x − 3y).

Check by distributing: 3x^2y · 2x = 6x^3y and 3x^2y · (−3y) = −9x^2y^2. Both original terms return. The binomial 2x − 3y has no remaining common factor and is not a special product, so the factorization is complete.

Two coefficient errors show up constantly. Taking only the numerical GCF leaves variables behind: 3(2x^3y − 3x^2y^2) is factored, but not by the greatest common factor. Taking too high a power of y is worse: 3x^2y^2 cannot divide the first term, which has only y^1.

Trap: stopping after the GCF when more factoring remains

The GCF is step one, not the finish line. If the leftover polynomial is a special product or a factorable trinomial, you must continue. Completely factored means every remaining polynomial factor is prime over the integers: it cannot be written as a product of non-constant integer-coefficient polynomials of lower degree.

Classic incomplete stop:

4x^2 − 16 = 4(x^2 − 4) ← not finished

x^2 − 4 is a difference of squares, so the complete factorization is 4(x − 2)(x + 2).

Another incomplete stop:

2x^3 − 18x = 2x(x^2 − 9) = 2x(x − 3)(x + 3).

If a stem asks you to factor completely and you report 2x(x^2 − 9), the leftover difference of squares is still sitting there. Both answers look factored, which is why this trap is so effective on a multiple-choice CAT.

The same trap appears after a cubes formula. x^3 − 8 is already a difference of cubes, but 2x^3 − 16 is not finished at 2(x^3 − 8). Pull the 2, then apply the cube template. Section 4.3 works that example in full; the lesson here is mechanical: GCF, then look again.

Factoring by grouping (four terms)

When a polynomial has four terms and no monomial GCF empties the expression, try grouping. Split the four terms into two pairs, factor each pair, and hope a common binomial appears.

Template:

ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y)

Worked grouping: x³ + 5x² + 2x + 10

Group the first two terms and the last two:

x^3 + 5x^2 + 2x + 10 = x^2(x + 5) + 2(x + 5)

The common binomial is x + 5:

= (x^2 + 2)(x + 5)

x^2 + 2 is a sum of squares, not a difference, so it does not factor over the reals. The factorization is complete over the integers.

If the third term is negative, keep the minus with the second group. The cubic x^3 + 3x^2 − 4x − 12 in Section 4.3 pairs as x^2(x + 3) − 4(x + 3) = (x^2 − 4)(x + 3), and then a difference of squares finishes the job. Grouping is how four-term cubics become products of binomials.

If grouping fails on the pairing you chose, try a different pairing (swap the middle two terms, or pair first-with-third). If no pairing produces a shared binomial, the polynomial may be prime over the integers, or it may need a rational-root search (Section 4.3).

Signs inside a group matter. x^3 − 2x^2 − 3x + 6 groups as x^2(x − 2) − 3(x − 2) only after you factor −3 from the second pair −3x + 6. Factoring +3 from that pair produces +3(−x + 2), which is not the same binomial as x − 2. Matching binomials is the whole point of grouping; mismatched signs mean you factored the second pair with the wrong sign.

Special products: memorize the templates

AAF does not give you a formula sheet. Special-product templates must be automatic, because they are faster than the general ac method and they catch complete-factoring traps.

Sum of squares a^2 + b^2 does not factor over the real numbers. x^2 + 9 is prime over the reals. Do not invent (x + 3)(x − 3) — that product is x^2 − 9.

SOAP is the memory aid for cubes: Same, Opposite, Always Positive. For a^3 + b^3, the binomial uses the same sign (a + b) and the trinomial starts with the opposite middle sign (−ab) and always ends with + b^2. For a^3 − b^3, the binomial uses the minus (a − b) and the trinomial’s middle sign flips to +ab.

Worked example: x³ + 8 (sum of cubes)

x^3 + 8 = x^3 + 2^3. Here a = x and b = 2.

Sum-of-cubes template: (a + b)(a^2 − ab + b^2)

= (x + 2)(x^2 − x·2 + 2^2) = (x + 2)(x^2 − 2x + 4)

The quadratic x^2 − 2x + 4 has discriminant 4 − 16 = −12 < 0, so it has no real linear factors. Over the integers (and over the reals) the factorization stops at (x + 2)(x^2 − 2x + 4).

Check by distributing:

(x + 2)(x^2 − 2x + 4) = x^3 − 2x^2 + 4x + 2x^2 − 4x + 8 = x^3 + 8.

The middle terms cancel. That cancellation is why the cube formulas work.

Wrong answers on this item usually swap the middle sign (x^2 + 2x + 4) or cube the binomial: (x + 2)^3 = x^3 + 6x^2 + 12x + 8, which is a different polynomial. The cube of a binomial is not a sum of cubes.

Worked example: 4x² − 12x + 9 (perfect-square trinomial)

A trinomial Ax^2 + Bx + C is a perfect square when:

• A is a perfect square (4 = 2^2),
• C is a perfect square (9 = 3^2),
• and |B| equals twice the product of those square roots (2 · 2 · 3 = 12).

Here the middle sign is minus, so the template is a^2 − 2ab + b^2 = (a − b)^2 with a = 2x and b = 3.

Result: 4x^2 − 12x + 9 = (2x − 3)^2.

FOIL check: (2x − 3)(2x − 3) = 4x^2 − 6x − 6x + 9 = 4x^2 − 12x + 9.

If the middle coefficient had been +12, the square would be (2x + 3)^2. If the middle had been −12 but the constant −9, it would not be a perfect square — the constant term of a real square is never negative.

Difference of squares, including hidden ones

Two terms, both perfect squares, minus in between: factor as conjugates.

9x^2 − 25 = (3x − 5)(3x + 5)

x^4 − 16 is a difference of squares twice: (x^2 − 4)(x^2 + 4) = (x − 2)(x + 2)(x^2 + 4). The factor x^2 + 4 is a sum of squares and stays.

After a GCF, look again: 18y^2 − 8 = 2(9y^2 − 4) = 2(3y − 2)(3y + 2).

AAF decision order

On a 20-item CAT you do not have time to try every method at random. Use this order every time:

1. GCF of all terms.
2. Count the remaining terms: two (special binomial), three (trinomial — Section 4.2), or four (grouping).
3. For two terms: difference of squares, or sum/difference of cubes. Never factor a sum of squares over the reals.
4. For three terms: test the perfect-square pattern, then the general ac method.
5. Factor completely — if a leftover factor is itself a special product, continue.
6. Check by multiplying (FOIL or distribution). The product must match the original polynomial, including signs.

That checklist is Table 11 factoring at the methods level. Quadratic trinomials with a ≠ 1 get their own section next; cubics that need a rational-root setup come after that.