4.3 Factoring Cubics and Higher Polynomials

4.3 Factoring Cubics and Higher Polynomials

Table 11’s factoring skill explicitly includes cubics and polynomials, not only quadratic trinomials. On AAF, a cubic is almost always an integer-coefficient polynomial that factors cleanly: a GCF comes out, grouping produces a common binomial, or a possible rational root splits off a linear factor and leaves a quadratic you already know how to handle. You are not being asked to grind Cardano’s formula. You are being asked to factor completely and, when the stem is an equation, to apply the zero product property.

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

GCF first, even on cubics

A cubic with a monomial GCF is a quadratic in disguise once the GCF is out.

3x^3 + 6x^2 − 24x = 3x(x^2 + 2x − 8) = 3x(x + 4)(x − 2)

The leftover trinomial is the a = 1 skill from Section 4.2. Forgetting 3x as a factor is the same GCF trap as in Section 4.1, now wearing a degree-3 costume. Completely factored form still includes the 3 and the x; (x + 4)(x − 2) alone is a different polynomial.

Worked example: 2x³ − 16

Factor 2x^3 − 16 completely over the integers.

Step 1 — GCF. Both terms are divisible by 2:

2x^3 − 16 = 2(x^3 − 8)

Step 2 — leftover binomial. x^3 − 8 = x^3 − 2^3 is a difference of cubes, not a difference of squares. Difference-of-cubes template: a^3 − b^3 = (a − b)(a^2 + ab + b^2) with a = x and b = 2.

x^3 − 8 = (x − 2)(x^2 + 2x + 4)

Step 3 — is the quadratic finished? Discriminant of x^2 + 2x + 4 is 4 − 16 = −12 < 0. It has no real linear factors, so it is prime over the integers.

Result: 2x^3 − 16 = 2(x − 2)(x^2 + 2x + 4).

Incomplete answers to watch:

• 2(x^3 − 8) stopped after the GCF.
• 2(x − 2)^3 confuses a difference of cubes with a cubed binomial: (x − 2)^3 = x^3 − 6x^2 + 12x − 8, not x^3 − 8.
• (2x − 2)(x^2 + 2x + 4) hides the constant 2 inside a binomial and, worse, 2x − 2 = 2(x − 1) uses the wrong linear factor.

Distribute to check: (x − 2)(x^2 + 2x + 4) = x^3 + 2x^2 + 4x − 2x^2 − 4x − 8 = x^3 − 8, then multiply by 2 to recover 2x^3 − 16.

Grouping four-term cubics, then factor completely

Many AAF cubics arrive as four terms. Group, factor the common binomial, then look at each leftover factor with the Section 4.1 special-product list.

Worked example: x³ + 3x² − 4x − 12

Group the first pair and the second pair, keeping the minus with the second group:

x^3 + 3x^2 − 4x − 12 = (x^3 + 3x^2) + (−4x − 12)

= x^2(x + 3) − 4(x + 3)

The shared binomial is x + 3:

= (x^2 − 4)(x + 3)

Do not stop. x^2 − 4 is a difference of squares:

x^2 − 4 = (x − 2)(x + 2)

Complete factorization: x^3 + 3x^2 − 4x − 12 = (x − 2)(x + 2)(x + 3).

Multiply stepwise to check:

(x − 2)(x + 2) = x^2 − 4, then (x^2 − 4)(x + 3) = x^3 + 3x^2 − 4x − 12.

The original cubic returns.

If you grouped as (x^3 − 4x) + (3x^2 − 12) instead, you still succeed: x(x^2 − 4) + 3(x^2 − 4) = (x + 3)(x^2 − 4), then the same difference of squares. Two legal groupings, one complete factorization.

The AAF trap is reporting (x^2 − 4)(x + 3) as the final answer on a factor-completely stem. That product is correct but not complete. Every leftover special product is still your job.

Another four-term cubic in the same family: x^3 − x^2 − 4x + 4 = x^2(x − 1) − 4(x − 1) = (x^2 − 4)(x − 1) = (x − 2)(x + 2)(x − 1). Same two-step pattern: grouping, then difference of squares.

Possible rational roots as a setup

Sometimes a cubic does not group. Then treat possible rational roots as a short integer checklist, not as a graduate-level theorem.

For a polynomial with integer coefficients, any rational root written in lowest terms p/q must have p dividing the constant term and q dividing the leading coefficient. On AAF, cubics are often monic (leading coefficient 1), so possible rational roots are simply the integer factors of the constant term.

Example: x^3 − 2x^2 − 5x + 6. Possible rational roots: ±1, ±2, ±3, ±6.

Test x = 1: 1 − 2 − 5 + 6 = 0. So x − 1 is a factor.

The leftover quadratic is the Section 4.2 skill: x^3 − 2x^2 − 5x + 6 = (x − 1)(x^2 − x − 6), and x^2 − x − 6 = (x − 3)(x + 2) because −3 and 2 multiply to −6 and add to −1.

Complete: (x − 1)(x − 3)(x + 2).

You do not need synthetic division displayed as a tableau if you can factor the leftover quadratic by the ac method. The rational-root list is only a setup that produces the first linear factor. After that, you are back in GCF, grouping, and trinomials.

If none of the possible rational roots work, the cubic does not factor into integer-coefficient linear factors. AAF almost never puts you in that dead end on a factoring item; the CAT is testing whether you try the list, not whether you can prove irreducibility.

For a non-monic cubic such as 2x^3 − 3x^2 − 3x + 2, possible rational roots include fractions whose denominators divide 2: ±1, ±2, ±1/2. Testing x = 2 gives 16 − 12 − 6 + 2 = 0, so x − 2 is a factor and the leftover quadratic can be read off as 2x^2 + x − 1 = (2x − 1)(x + 1). Completely factored: (x − 2)(2x − 1)(x + 1). Keep the method AAF-sized: test a short list, split off one linear factor, finish with Section 4.2.

Higher than cubic, still the same toolkit

A difference of squares can hide inside a degree-4 polynomial:

x^4 − 81 = (x^2 − 9)(x^2 + 9) = (x − 3)(x + 3)(x^2 + 9)

x^4 − 16 from Section 4.1 is the same pattern. After grouping a degree-4 four-term polynomial, you may get a quadratic that is itself a difference of squares. Factor completely still means: keep going until every leftover polynomial is prime over the integers.

A GCF of x^2 on a quartic produces a quadratic: x^4 + 5x^3 − 24x^2 = x^2(x^2 + 5x − 24) = x^2(x + 8)(x − 3). The factor x^2 is a repeated linear factor. Completely factored form is often written x^2(x + 8)(x − 3), which already encodes the double root at x = 0 if you later solve the equation.

From a factorization to solutions: zero product property

Factoring is how you solve polynomial equations that split over the integers. The zero product property says: if a product of numbers is zero, then at least one factor is zero. So

(x − 2)(x + 3) = 0 implies x − 2 = 0 or x + 3 = 0,

hence x = 2 or x = −3.

From the complete factorization in the first worked cubic:

(x − 2)(x + 2)(x + 3) = 0 ⇒ x = 2, x = −2, or x = −3.

From 2(x − 2)(x^2 + 2x + 4) = 0, the constant 2 is never zero, x = 2 is a root, and x^2 + 2x + 4 = 0 has no real roots (discriminant −12). The only real solution of 2x^3 − 16 = 0 is x = 2.

Two cautions, so this section does not steal the polynomials chapter:

• Setting a product equal to 6, or equal to x, is not a zero-product situation. Move everything to one side first so the equation reads polynomial = 0, then factor. x(x − 2) = 8 is not solved by x = 8 or x − 2 = 8.
• A quadratic factor that is prime over the integers may still have real roots via the quadratic formula. On a factor-over-the-integers item you leave that quadratic as is. On a solve-over-the-reals item you would continue in Solving Polynomial Equations.

Substitute each candidate root back into the original polynomial as a last check. For x^3 + 3x^2 − 4x − 12, x = 2 gives 8 + 12 − 8 − 12 = 0; x = −2 gives −8 + 12 + 8 − 12 = 0; x = −3 gives −27 + 27 + 12 − 12 = 0. All three check.

AAF checklist for cubics and higher

1. GCF of all terms, including a power of x if every term has one.
2. Four terms → grouping; two terms → cubes or squares; three terms → Section 4.2.
3. If grouping fails, test possible rational roots (factors of the constant, over factors of the leading coefficient).
4. Factor completely, including a difference of squares that appears after grouping.
5. If the stem is an equation, write product = 0 and apply the zero product property.
6. Check by expanding, or by substituting each candidate root back into the original polynomial.

Factoring is 5–10% of AAF and only 1–2 CAT items, but those items are cheap points if the checklist is automatic and expensive if you stop one special product too soon. The decision tree in Section 4.1 still governs every degree.