Factoring by Trial and Error: 6x^2 - 17x + 12

Problem Setup

• Goal: Factor the quadratic 6x^2 - 17x + 12 by trial and error.
• Reminder: Quadratic binomials come from the product of two binomials via FOIL (First, Outer, Inner, Last).
• If we factor as
  $(ax + b)(cx + d)$
  then the expansion is
  $acx^2 + (ad + bc)x + bd.$
• For our target 6x^2 - 17x + 12, we need:
  • $ac = 6$ (the coefficient of x^2)
  • $bd = 12$ (the constant term)
  • $ad + bc = -17$ (the middle term coefficient)
• Plan: enumerate possible factor pairs for 6 and 12, assign signs, and check the resulting middle term via FOIL.

Key Concepts and Sign Reasoning

• The middle term comes from the sum of the outer and inner products: ad + bc.
• Since the constant term 12 is positive and the middle term is negative (-17x), we should have b and d with the same sign as each other (both negative), assuming a and c are positive (as they multiply to 6).
• Therefore, consider negative factor pairs for the last term: bd = 12 with b < 0 and d < 0.
• Signs:
  • To get a negative middle term, we typically need at least one of ad or bc to be negative; with a and c positive, negative b and d make both ad and bc negative, giving a negative middle term.
• Trial-and-error approach: try different combinations of factor pairs for the first term and the last term, compute the outside and inside products, and compare to -17x.

Factor Pair Enumeration (FOIL framework)

• Possible factor pairs for the leading coefficient 6 (from ac = 6):
  • Case A: (a, c) = (1, 6)
  • Case B: (a, c) = (2, 3)
• Possible factor pairs for the constant term 12 (from bd = 12):
  • (b, d) = (1, 12), (2, 6), (3, 4)
• With negative signs (to get the negative middle term), consider:
  • (b, d) = (-1, -12), (-2, -6), (-3, -4)

Trial Cases and Results

• Case A: (a, c) = (1, 6)
  • Subcase A1: (b, d) = (-1, -12)
  • ad = ad = 1(-12) = -12
  • bc = bc = (-1)6 = -6
  • ad + bc = -12 + (-6) = -18
  • Middle term would be
    $-18x$ (not -17x).
  • Subcase A2: (b, d) = (-2, -6)
  • ad = 1*(-6) = -6
  • bc = (-2)*6 = -12
  • ad + bc = -6 + (-12) = -18
  • Middle term would be
    $-18x$ (not -17x).
  • Subcase A3: (b, d) = (-3, -4)
  • ad = 1*(-4) = -4
  • bc = (-3)*6 = -18
  • ad + bc = -4 + (-18) = -22
  • Middle term would be
    $-22x$ (not -17x).
• Case B: (a, c) = (2, 3)
  • Subcase B1: (b, d) = (-1, -12)
  • ad = 2*(-12) = -24
  • bc = (-1)*3 = -3
  • ad + bc = -24 + (-3) = -27
  • Middle term would be
    $-27x$ (not -17x).
  • Subcase B2: (b, d) = (-2, -6)
  • ad = 2*(-6) = -12
  • bc = (-2)*3 = -6
  • ad + bc = -12 + (-6) = -18
  • Middle term would be
    $-18x$ (not -17x).
  • Subcase B3: (b, d) = (-3, -4)
  • ad = 2*(-4) = -8
  • bc = (-3)*3 = -9
  • ad + bc = -8 + (-9) = -17
  • Middle term matches: $-17x$

Correct Factorization

• The combination that matches is:
  • $(a, b) = (2, -3)$ and $(c, d) = (3, -4)$
  • So the factorization is
    $(2x - 3)(3x - 4).$
• Verification by FOIL:
  • First: $2x imes 3x = 6x^2$
  • Outer: $2x imes (-4) = -8x$
  • Inner: $(-3) imes 3x = -9x$
  • Last: $(-3) imes (-4) = 12$
  • Sum of outer and inner: $-8x + (-9x) = -17x$
  • Final expression: $6x^2 - 17x + 12$
• This matches the original quadratic: $6x^2 - 17x + 12.$

Final Note on Method and Sign Logic

• The trial-and-error method worked here by systematically testing factor pairs for the leading and trailing terms and enforcing the sign requirement so that the middle term becomes negative.
• The quote from the transcript: "See, this is why it's called trial and error. We try until we find the right combination." reflects the iterative approach.
• The identified factor pairs provide a clean factorization that can be checked with FOIL.

Summary and Practical Takeaways

• To factor quadratics of the form $ax^2 + bx + c$ when a and c are composite (like 6 and 12), enumerate factor pairs for a and c and for c (bd).
• Use the sign rule: if the middle term b is negative and the constant term c is positive, you will typically use negative values for both b and d so that bd is positive and ad + bc yields a negative middle term.
• Always verify candidate factorizations by FOIL:
  • Expand to ensure the coefficients match: $acx^2 + (ad + bc)x + bd$.
• The concrete result for this example is:
  • $6x^2 - 17x + 12 = (2x - 3)(3x - 4).$

Connections and Real-World Relevance

• Factoring quadratics is fundamental in solving quadratic equations and appears in physics, engineering, and economics when modeling parabolic relationships.
• The trial-and-error approach demonstrates an essential problem-solving strategy: use structure (FOIL, product-sum relationships) to prune possibilities and verify quickly with expansion.

Quick Reference Formulas

• Factored form:
  $(ax + b)(cx + d) = acx^2 + (ad + bc)x + bd.$
• Conditions for our example:
  • ac = 6,d = 12,
    ad + bc = -17.
• Final verified factorization:
  • $(2x - 3)(3x - 4) = 6x^2 - 17x + 12.$