Algebra 2 · Grades 10, 11

Factoring Polynomials: Cubes, Grouping and Quadratic Form

Quick answer

Factoring turns a polynomial into a product, and a product shows where the polynomial is zero. Beyond the quadratic patterns, three tools do most of the work. Sums and differences of cubes follow a³ ± b³ = (a ± b)(a² ∓ ab + b²). Grouping pairs terms that share a factor, as in x³ + 2x² − 9x − 18 = (x + 2)(x² − 9). And a polynomial in quadratic form, such as x⁴ − 5x² + 4, factors like a quadratic in x². Take out a greatest common factor first, and keep going until no factor splits further.

What you'll learn

  • Factor sums and differences of cubes
  • Factor four-term polynomials by grouping
  • Factor polynomials in quadratic form
  • Solve polynomial equations by factoring completely

Why factor

A factored polynomial shows its zeros at a glance. If (x+2)(x−3)(x+3)=0(x + 2)(x - 3)(x + 3) = 0, one of the factors must be 00, so x=−2x = -2, 33 or −3-3. The same polynomial multiplied out, x3+2x2−9x−18x^3 + 2x^2 - 9x - 18, hides them. Quadratics factor with the patterns from Algebra 1. Higher-degree polynomials need a few more tools, and the first step is always the same: take out the greatest common factor.

y = x³ + 2x² − 9x − 18 and its three zeros A cubic curve rising from the lower left, crossing the x-axis at −3, turning down a little above the axis, crossing again at −2, dipping to about −24 near x = 1.2, and rising through the x-axis at 3. Dots mark the three x-intercepts, one for each factor. -424-20-1010xy −3 −2 3
  • y = (x + 2)(x − 3)(x + 3)
y = x³ + 2x² − 9x − 18 and its three zeros

Sums and differences of cubes

Two cubes, added or subtracted, always factor:

a3−b3=(a−b)(a2+ab+b2)a3+b3=(a+b)(a2−ab+b2)a^3 - b^3 = (a - b)\left(a^2 + ab + b^2\right) \qquad\qquad a^3 + b^3 = (a + b)\left(a^2 - ab + b^2\right)

The signs follow a pattern. The binomial has the same sign as the original, the middle term of the trinomial has the opposite sign, and its last term is always positive. The trinomial never factors further over the real numbers.

Why the cube patterns work

Multiply the right side out:

(a−b)(a2+ab+b2)=a3+a2b+ab2−a2b−ab2−b3=a3−b3(a - b)\left(a^2 + ab + b^2\right) = a^3 + a^2b + ab^2 - a^2b - ab^2 - b^3 = a^3 - b^3

Every middle term appears twice with opposite signs and cancels, leaving only the two cubes. The factor theorem predicts the binomial too: x=bx = b makes x3−b3x^3 - b^3 zero, so x−bx - b must be a factor. Every factoring pattern is a multiplication, read backward, and multiplying out is the way to check one.

Factoring by grouping

A polynomial with four terms can often be split into two pairs that share a factor. Factor each pair, and if the same binomial appears twice, factor it out.

x3+2x2−9x−18=x2(x+2)−9(x+2)=(x+2)(x2−9)=(x+2)(x−3)(x+3)x^3 + 2x^2 - 9x - 18 = x^2(x + 2) - 9(x + 2) = (x + 2)\left(x^2 - 9\right) = (x + 2)(x - 3)(x + 3)

Quadratic form

A polynomial like x4−5x2+4x^4 - 5x^2 + 4 is a quadratic in disguise. With u=x2u = x^2 it reads u2−5u+4=(u−1)(u−4)u^2 - 5u + 4 = (u - 1)(u - 4). Put x2x^2 back and keep factoring:

x4−5x2+4=(x2−1)(x2−4)=(x−1)(x+1)(x−2)(x+2)x^4 - 5x^2 + 4 = \left(x^2 - 1\right)\left(x^2 - 4\right) = (x - 1)(x + 1)(x - 2)(x + 2)

Worked examples

Common mistakes

Practice problems

  1. Factor x3−27x^3 - 27.

    Answer

    (x−3)(x2+3x+9)(x - 3)\left(x^2 + 3x + 9\right)

    Full solution

    A difference of cubes with a=xa = x and b=3b = 3.

  2. Factor 8x3+18x^3 + 1.

    Answer

    (2x+1)(4x2−2x+1)(2x + 1)\left(4x^2 - 2x + 1\right)

    Full solution

    A sum of cubes with a=2xa = 2x and b=1b = 1.

  3. Factor x3+3x2+2x+6x^3 + 3x^2 + 2x + 6 by grouping.

    Answer

    (x+3)(x2+2)(x + 3)\left(x^2 + 2\right)

    Full solution

    x2(x+3)+2(x+3)=(x+3)(x2+2)x^2(x + 3) + 2(x + 3) = (x + 3)\left(x^2 + 2\right). The second factor is a sum and does not factor over the reals.

  4. Factor 2x3−3x2−8x+122x^3 - 3x^2 - 8x + 12 completely.

    Answer

    (2x−3)(x−2)(x+2)(2x - 3)(x - 2)(x + 2)

    Full solution

    x2(2x−3)−4(2x−3)=(2x−3)(x2−4)x^2(2x - 3) - 4(2x - 3) = (2x - 3)\left(x^2 - 4\right), and x2−4x^2 - 4 is a difference of squares.

  5. Factor x4−13x2+36x^4 - 13x^2 + 36 completely.

    Answer

    (x−2)(x+2)(x−3)(x+3)(x - 2)(x + 2)(x - 3)(x + 3)

    Full solution

    With u=x2u = x^2: u2−13u+36=(u−4)(u−9)u^2 - 13u + 36 = (u - 4)(u - 9). Then x2−4x^2 - 4 and x2−9x^2 - 9 are differences of squares.

  6. Factor x4−16x^4 - 16 completely over the real numbers.

    Answer

    (x−2)(x+2)(x2+4)(x - 2)(x + 2)\left(x^2 + 4\right)

    Full solution

    (x2−4)(x2+4)\left(x^2 - 4\right)\left(x^2 + 4\right), and only the first factor splits further.

  7. Factor 3x3−243x^3 - 24 completely.

    Answer

    3(x−2)(x2+2x+4)3(x - 2)\left(x^2 + 2x + 4\right)

    Full solution

    Take out the 33 to get 3(x3−8)3\left(x^3 - 8\right), then use the difference of cubes.

  8. Solve x3−5x2−4x+20=0x^3 - 5x^2 - 4x + 20 = 0.

    Answer

    x=5x = 5, 22 or −2-2

    Full solution

    Group: x2(x−5)−4(x−5)=(x−5)(x−2)(x+2)=0x^2(x - 5) - 4(x - 5) = (x - 5)(x - 2)(x + 2) = 0.

  9. Solve x4−3x2−4=0x^4 - 3x^2 - 4 = 0, including complex solutions.

    Answer

    x=±2x = \pm 2 and x=±ix = \pm i

    Full solution

    (x2−4)(x2+1)=0\left(x^2 - 4\right)\left(x^2 + 1\right) = 0. So x2=4x^2 = 4, giving ±2\pm 2, or x2=−1x^2 = -1, giving ±i\pm i.

  10. A student factors x3+8x^3 + 8 as (x+2)(x2+2x+4)(x + 2)\left(x^2 + 2x + 4\right). What went wrong?

    Hint

    Multiply the student’s factors back out.

    Answer

    The middle sign of the trinomial should be negative: (x+2)(x2−2x+4)(x + 2)\left(x^2 - 2x + 4\right).

    Full solution

    The student’s product is x3+4x2+8x+8x^3 + 4x^2 + 8x + 8: the middle terms do not cancel. With x2−2x+4x^2 - 2x + 4 they do, and the product is x3+8x^3 + 8.

Frequently asked questions

How do you factor a difference of cubes?

a³ − b³ = (a − b)(a² + ab + b²). For example, x³ − 8 = (x − 2)(x² + 2x + 4).

How do you factor a sum of cubes?

a³ + b³ = (a + b)(a² − ab + b²). For example, x³ + 27 = (x + 3)(x² − 3x + 9).

When should I factor by grouping?

When a polynomial has four terms and the first pair and the last pair share a common factor after you factor each pair.

What is quadratic form?

A polynomial that looks like a quadratic in some expression, such as x⁴ − 5x² + 4, which is u² − 5u + 4 with u = x².

Does x² + 9 factor?

Not over the real numbers. A sum of two squares has no real factors, though over the complex numbers it is (x + 3i)(x − 3i).

What to learn next

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

  • CCSS.MATH.CONTENT.HSA.SSE.A.2Seeing Structure in ExpressionsUse the structure of an expression to identify ways to rewrite it.
  • CCSS.MATH.CONTENT.HSA.APR.B.3Arithmetic with Polynomials and Rational ExpressionsIdentify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial.
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