Topic 1.3 · Unit 1 · CA CCSS A-SSE.2–3

Factoring polynomials

Factoring is expanding run backwards, and there is a fixed order to try things in. Working through that order rather than guessing is what makes this reliable, and it is the skill the whole quadratics unit depends on.

The method

Always in this order:

  1. Greatest common factor first, every time. Pull out what every term shares. \( 6x^2 + 9x = 3x(2x + 3) \). Skipping this makes everything after it harder and sometimes impossible.
  2. Count the terms. Two terms → look for a difference of squares: \( a^2 - b^2 = (a+b)(a-b) \). Note \( a^2 + b^2 \) does not factor.
  3. Three terms, leading coefficient 1: for \( x^2 + bx + c \), find two numbers that multiply to \( c \) and add to \( b \). Those two numbers go straight into \( (x + \_)(x + \_) \).
  4. Three terms, leading coefficient not 1: for \( ax^2 + bx + c \), find two numbers multiplying to \( a \times c \) and adding to \( b \), split the middle term into those two, then factor in pairs.
  5. Check by expanding. Factoring is the one topic where checking is instant and catches nearly every error.

Signs, which is where most marks go: if \( c \) is positive, both numbers share the sign of \( b \). If \( c \) is negative, the numbers have opposite signs and the larger one carries the sign of \( b \).

Worked examples

Example 1: GCF then a pattern. Factor \( 2x^3 - 18x \).

GCF is \( 2x \): \( 2x(x^2 - 9) \). The bracket is now two terms and both are perfect squares, so it is a difference of squares: \( 2x(x+3)(x-3) \).

Pulling the GCF first is what revealed the pattern, without it, \( 2x^3 - 18x \) looks like nothing.

Example 2: trinomial with leading coefficient 1. Factor \( x^2 - 5x - 24 \).

Need two numbers multiplying to \( -24 \) and adding to \( -5 \). Since the product is negative the signs differ; since the sum is negative the larger number is negative. Pairs for 24: 1 and 24, 2 and 12, 3 and 8, 4 and 6. The pair 3 and 8 differs by 5, so take \( +3 \) and \( -8 \).

\( (x + 3)(x - 8) \). Check: \( -8x + 3x = -5x \) ✓ and \( 3 \times -8 = -24 \) ✓.

Example 3: leading coefficient not 1. Factor \( 3x^2 + 11x + 6 \).

\( a \times c = 18 \), and we need two numbers multiplying to 18 and adding to 11: that is 9 and 2. Split the middle term:

\( 3x^2 + 9x + 2x + 6 \). Factor in pairs: \( 3x(x + 3) + 2(x + 3) \). Both share \( (x+3) \), so it factors as \( (x + 3)(3x + 2) \).

Check by expanding: \( 3x^2 + 2x + 9x + 6 = 3x^2 + 11x + 6 \) ✓.

Practice · 15 problems

1–5 GCF and patterns, 6–11 trinomials, 12–15 harder cases and diagnosis.

  1. Factor \( 4x + 12 \).
    Show the full solution

    GCF 4. \( 4(x + 3) \)

  2. Factor \( 6x^2 - 15x \).
    Show the full solution

    GCF \( 3x \). \( 3x(2x - 5) \)

  3. Factor \( x^2 - 49 \).
    Show the full solution

    Difference of squares. \( (x+7)(x-7) \)

  4. Factor \( 9x^2 - 25 \).
    Show the full solution

    Both perfect squares: \( (3x)^2 - 5^2 \). \( (3x+5)(3x-5) \)

  5. Can \( x^2 + 16 \) be factored over the real numbers?
    Show the full solution

    No. A sum of squares does not factor; only the difference does. Does not factor

  6. Factor \( x^2 + 7x + 12 \).
    Show the full solution

    Two numbers multiplying to 12, adding to 7: 3 and 4. \( (x+3)(x+4) \)

  7. Factor \( x^2 - 9x + 20 \).
    Show the full solution

    Product positive and sum negative, so both negative: \( -4 \) and \( -5 \). \( (x-4)(x-5) \)

  8. Factor \( x^2 + 2x - 15 \).
    Show the full solution

    Product negative so signs differ; \( +5 \) and \( -3 \) add to \( +2 \). \( (x+5)(x-3) \)

  9. Factor \( x^2 - 3x - 28 \).
    Show the full solution

    Signs differ, larger negative: \( -7 \) and \( +4 \). \( (x-7)(x+4) \)

  10. Factor \( x^2 + 10x + 25 \).
    Show the full solution

    5 and 5. \( (x+5)^2 \)

  11. Factor \( 2x^2 + 14x + 24 \).
    Show the full solution

    GCF 2 first: \( 2(x^2 + 7x + 12) \), then the trinomial. \( 2(x+3)(x+4) \)

  12. Factor \( 2x^2 + 7x + 3 \).
    Show the full solution

    \( ac = 6 \); need two numbers multiplying to 6, adding to 7: 6 and 1. Split: \( 2x^2 + 6x + x + 3 = 2x(x+3) + 1(x+3) \). \( (x+3)(2x+1) \)

  13. Factor \( 3x^2 - 10x + 8 \).
    Show the full solution

    \( ac = 24 \); two numbers multiplying to 24 and adding to \( -10 \): \( -6 \) and \( -4 \). Split: \( 3x^2 - 6x - 4x + 8 = 3x(x-2) - 4(x-2) \). \( (x-2)(3x-4) \)

  14. A student factors \( x^2 - 16 \) as \( (x-4)^2 \). Find the error.
    Show the full solution

    \( (x-4)^2 \) expands to \( x^2 - 8x + 16 \), which has a middle term and the wrong sign on 16. A difference of squares factors into a sum times a difference: \( (x+4)(x-4) \). \( (x+4)(x-4) \)

  15. A student factors \( 2x^2 + 10x \) as \( (2x)(x + 5) \) and stops. Is that fully factored?
    Show the full solution

    It is correct but the GCF was not fully taken; it is conventionally written \( 2x(x+5) \), which is the same thing. The expression is fully factored either way, since \( x + 5 \) cannot factor further. Correct; write it as \( 2x(x+5) \)

← 1.2 Polynomial arithmetic

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