Unit 6 · Lesson 5 of 6

A general factoring strategy

About 25 minutes · 5 steps · one idea: a checklist that tells you which method to use

You'll be able to: factor any polynomial completely by following the same four checks every time.

  1. 1 Warm-up
  2. 2 The idea
  3. 3 Examples
  4. 4 Practice
  5. 5 Wrap-up

Step 1 of 5Warm-up

Two quick questions: one from last lesson, one from much earlier. Answer from memory first.

Factor \(x^3 - 8\).

\(x^3 = (x)^3\) and \(8 = 2^3\). Difference of cubes with \(a = x\), \(b = 2\).

SOAP: \((a - b)(a^2 + ab + b^2)\). Answer: \((x - 2)(x^2 + 2x + 4)\).

Solve \(-3 \lt 2x + 1 \le 7\). Give your answer in interval notation.

Subtract 1 from all three parts: \(-4 \lt 2x \le 6\).

Divide all three parts by 2: \(-2 \lt x \le 3\).

Answer: \((-2, 3]\). The round bracket leaves \(-2\) out; the square bracket keeps \(3\).

Step 2 of 5The idea

You now know several factoring methods. On a test, nobody tells you which one to use. This checklist decides for you.

Factor completely means: keep going until no factor can be factored any more. A number out front is fine.

1. GCF first. Take out the GCF. If the first term is negative, take out a negative.

2. Count the terms of what is left.

• 2 terms: difference of squares? sum or difference of cubes? (A sum of squares is prime.)

• 3 terms: perfect square trinomial? If not, \(x^2 + bx + c\) (two numbers) or \(ax^2 + bx + c\) (ac method).

• 4 terms: factor by grouping.

3. Look at each factor again. Can any factor be factored? If yes, go back to step 2 for that factor.

4. Check by multiplying back out.

Why this shows up on the placement test: test questions often say "factor completely," and the answer choices usually include a half-finished version that looks almost right.

Step 3 of 5Worked examples

Example 1: Factor completely \(3x^3 - 12x\)

  1. GCF: both terms have \(3\) and \(x\). \(3x^3 - 12x = 3x(x^2 - 4)\).
  2. Count the terms inside: 2 terms, \(x^2 - 4\). Both are squares with a minus: difference of squares.
  3. \(x^2 - 4 = (x + 2)(x - 2)\), so we have \(3x(x + 2)(x - 2)\).
  4. Look again: \(x + 2\) and \(x - 2\) can't be factored. Done. Check: \((x+2)(x-2) = x^2 - 4\), and \(3x(x^2 - 4) = 3x^3 - 12x\).

Example 2: Factor completely \(x^3 + 2x^2 - 9x - 18\)

  1. GCF: none (the terms share no common factor).
  2. 4 terms, so group: \(x^2(x + 2) - 9(x + 2)\). Take out \(-9\) so the parentheses match.
  3. Take out \((x + 2)\): \((x + 2)(x^2 - 9)\).
  4. Look again: \(x^2 - 9\) is a difference of squares, \((x + 3)(x - 3)\).
  5. Answer: \((x + 2)(x + 3)(x - 3)\).

Step 4 of 5Practice

Factor each one completely. Go through the checklist every time, even when it feels obvious. If you miss twice, open the worked solution. Then mark how it went.

Factor completely: \(5x^2 - 45\).

GCF: \(5(x^2 - 9)\).

2 terms inside, both squares, minus between: \(x^2 - 9 = (x + 3)(x - 3)\).

Answer: \(5(x + 3)(x - 3)\). Stopping at \(5(x^2 - 9)\) is not complete.

Factor completely: \(2x^2 + 12x + 18\).

GCF: \(2(x^2 + 6x + 9)\).

3 terms inside: \(x^2\) and \(9 = 3^2\) are squares, and \(2 \cdot x \cdot 3 = 6x\). Perfect square trinomial: \((x + 3)^2\).

Answer: \(2(x + 3)^2\).

Factor completely: \(x^3 - 6x^2 + 8x\).

GCF: every term has an \(x\). \(x(x^2 - 6x + 8)\).

3 terms inside, starting with \(x^2\): two numbers that multiply to 8 and add to \(-6\) are \(-2\) and \(-4\).

Answer: \(x(x - 2)(x - 4)\).

Factor completely: \(6x^2 - 7x - 3\).

GCF: none. 3 terms, not a perfect square (6 is not a square), and \(a \ne 1\): use the ac method.

\(ac = -18\). Multiply to \(-18\), add to \(-7\): \(-9\) and \(2\). Split: \(6x^2 - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3)\).

Answer: \((2x - 3)(3x + 1)\).

Factor completely: \(x^3 - 5x^2 - 4x + 20\).

GCF: none. 4 terms, so group: \(x^2(x - 5) - 4(x - 5) = (x - 5)(x^2 - 4)\).

Look again: \(x^2 - 4 = (x + 2)(x - 2)\).

Answer: \((x - 5)(x + 2)(x - 2)\). The answer \((x - 5)(x^2 - 4)\) is not complete.

Factor completely: \(x^4 - 81\).

2 terms: \(x^4 = (x^2)^2\) and \(81 = 9^2\). Difference of squares: \((x^2 + 9)(x^2 - 9)\).

Look again: \(x^2 + 9\) is a sum of squares (prime), but \(x^2 - 9 = (x + 3)(x - 3)\).

Answer: \((x^2 + 9)(x + 3)(x - 3)\).

Which is \(2x^3 - 18x\) factored completely?

  1. \(2x(x^2 - 9)\)
  2. \((2x^2 - 6x)(x + 3)\)
  3. \(2x(x + 3)(x - 3)\)
  4. \(x(2x + 6)(x - 3)\)

All four equal \(2x^3 - 18x\). Only one is finished.

(a) \(x^2 - 9\) is a difference of squares. Not complete.

(b) \(2x^2 - 6x\) still has a GCF of \(2x\). Not complete.

(c) GCF \(2x\), then \(x^2 - 9 = (x + 3)(x - 3)\). Nothing factors further. This one.

(d) \(2x + 6\) still has a GCF of 2. Not complete.

Factor completely: \(-3x^2 + 3x + 18\).

The first term is negative, so take out \(-3\): \(-3(x^2 - x - 6)\). (Every sign inside flips.)

Inside: two numbers that multiply to \(-6\) and add to \(-1\) are \(-3\) and \(2\).

Answer: \(-3(x - 3)(x + 2)\). Check: \((x-3)(x+2) = x^2 - x - 6\), and \(-3(x^2 - x - 6) = -3x^2 + 3x + 18\).

Step 5 of 5Wrap-up

You can now factor a polynomial completely with one checklist: GCF first, count the terms to pick the method, look at every factor again, and check by multiplying.

Read or watch next: OpenStax Intermediate Algebra 2e, 6.4 General Strategy for Factoring Polynomials (free textbook, with a strategy chart and mixed practice).

Also good: Khan Academy: Strategy in factoring quadratics (short videos). Quick rules: Factoring cheat sheet.

Something unclear? Ask your teacher bot a follow-up question, for example: "How do I know when a polynomial is factored completely?" It knows this course and can explain things another way.