HighSchoolQ

➕ Algebra 1 · Factoring

Factoring Trinomials: notes and common mistakes

Study notes

Key ideas

  • Factoring a trinomial is un-multiplying two binomials. For x2 + bx + c, find integers p and q with pq = c and p + q = b. Then x2 + bx + c = (x + p)(x + q).
  • Signs (for nonzero b and c): when c is positive, p and q have the same sign (the sign of b). When c is negative, they have opposite signs, and the one larger in size has the sign of b.
  • Always take out the GCF first (a negative GCF if the leading coefficient is negative), and keep it in the final answer.
  • When a ≠ 1, use the ac method: find two integers with product ac and sum b, split the middle term, then factor by grouping.
  • If no pair of integers works, the trinomial is prime (cannot be factored using integers). Always check by expanding.
  • Perfect square trinomials, u2 ± 2uv + v2 = (u ± v)2, are the case where the two numbers are equal. A factored form reveals the zeros: a product is 0 only when a factor is 0.

Steps

  1. Write the trinomial in standard form (descending powers). Take out the GCF, using a negative GCF if the leading coefficient is negative.
  2. If the coefficient of x2 is 1: list integer pairs with product c, pick the pair with sum b, and write (x + p)(x + q).
  3. If the coefficient of x2 is not 1: find two integers with product ac and sum b. Split the middle term, factor each pair, and pull out the common binomial.
  4. If no pair works, say the trinomial is prime.
  5. Expand to check. Make sure the GCF is in your final answer and that no factor can be factored further.

Formulas

  • Multiplying two binomials: (x + p)(x + q) = x2 + (p + q)x + pq
  • Factoring $x^2 + bx + c$: x2 + bx + c = (x + p)(x + q) when pq = c and p + q = b
  • ac method: Find integers m and n with mn = ac and m + n = b. Then ax2 + bx + c = ax2 + mx + nx + c, and you factor by grouping.
  • Perfect square trinomials: u2 + 2uv + v2 = (u + v)2 and u2 − 2uv + v2 = (u − v)2

Watch out

  • Right product, wrong signs: expand to catch it. (x + 2)(x + 5) is x2 + 7x + 10, not x2 − 7x + 10.
  • Do not drop the GCF: 6x2 + 12x − 18 = 6(x + 3)(x − 1).
  • When a ≠ 1, look for the product ac, not c.
  • Both end terms being perfect squares does not make a perfect square trinomial. Check that the middle term is twice the product of the square roots.
  • “Prime” only means “cannot be factored using integers.” List every integer pair, including negative ones, before you decide.

Why it works

  • Why do the numbers multiply to c and add to b? Because (x + p)(x + q) = x2 + (p + q)x + pq: expanding adds the two cross terms and multiplies the two constants.
  • Why does the ac method work? In (rx + s)(tx + u) the cross products ru and st add to b and multiply to (rt)(su) = ac. So the two numbers with product ac and sum b are exactly those cross terms. Splitting the middle term into them lets you factor by grouping.
  • Why factor at all? A product is 0 only when one of its factors is 0, so a factored quadratic shows the zeros (the x-values where it equals 0) at a glance.

Common mistakes

✗ Mistakex2 − 7x + 10 = (x + 2)(x + 5) (right product, wrong signs).
✓ InsteadBoth pairs multiply to 10, but the middle term −7x needs −2 and −5: (x − 2)(x − 5). Expanding (x + 2)(x + 5) gives x2 + 7x + 10, which catches the mistake.
✗ Mistake6x2 + 12x − 18 = (x + 3)(x − 1) (dropping the GCF).
✓ InsteadThe GCF 6 is part of the answer: 6(x + 3)(x − 1). Without it, you only have x2 + 2x − 3, one sixth of the original.
✗ MistakeTo factor 4x2 + 8x + 3, look for two numbers with product 3 and sum 8.
✓ InsteadWhen a ≠ 1, use the product ac = 12 and sum 8: the pair is 2 and 6. Then 4x2 + 8x + 3 = (2x + 1)(2x + 3).
✗ Mistakex2 + 12x + 25 is a perfect square trinomial because x2 and 25 are perfect squares.
✓ InsteadThe middle term must be twice the product of the square roots: 2 · x · 5 = 10x, not 12x. So it is not a perfect square trinomial. No integer pair has product 25 and sum 12, so it is prime.
✗ MistakeIf the positive pairs do not work, the trinomial must be prime. For example, x2 − 13x + 36 looks prime.
✓ InsteadCheck negative pairs too. When c is positive and b is negative, both numbers are negative: (−4)(−9) = 36 and −4 + (−9) = −13, so x2 − 13x + 36 = (x − 4)(x − 9).

Teach it back

Explain to a friend how to factor 2x2 − 3x − 20 completely and how to check your answer. Say why you look for two numbers with product −40.

Explaining it out loud shows what you really know. Did you…

  • I checked for a GCF first. There is none: 2, −3 and −20 share only the factor 1.
  • I multiplied a · c = 2 × (−20) = −40 and found two integers with product −40 and sum −3. They are 5 and −8.
  • I explained why: in (rx + s)(tx + u) the two cross products ru and st add to b and multiply to (rt)(su) = ac, so the split must use two numbers with product ac = −40 and sum b = −3.
  • I split the middle term and grouped: 2x2 + 5x − 8x − 20 = x(2x + 5) − 4(2x + 5) = (2x + 5)(x − 4).
  • I expanded to check: (2x + 5)(x − 4) = 2x2 − 8x + 5x − 20 = 2x2 − 3x − 20.

Your 3-week memory plan

  1. TodayRead the lesson, then try the flashcards.
  2. TomorrowRecall it from memory first, then check.
  3. In 3 daysTake the true/false and practice questions.
  4. In 1 weekMix it with older topics and teach it back.
  5. In 3 weeksA last quick review before it fades.

Downloads a calendar file (.ics). Works with Google, Apple and Outlook calendars. Nothing is sent to us.