I’m preparing for a test on factorising quadratics, and I keep getting stuck when the coefficient of x^2 isn’t 1 and the signs mix. For example, with 6x^2 + 7x − 5, I know ac = −30, and I can find 10 and −3 to split the middle term. I try grouping like (6x^2 + 10x) + (−3x − 5), but I’m not confident I’m doing it the best way, and sometimes I pick a pair that doesn’t lead to clean factors. I also get confused about when I should factor out a GCF first, especially with things like −8x^2 + 14x − 3 or 12x^2 − 8x − 3.
Could someone lay out a reliable step-by-step checklist for factorising ax^2 + bx + c over the integers? Specifically: when to pull a GCF, how to choose the right pair for ac, and how to handle negatives so the grouping works without sign mistakes. Also, is there a quick way to tell if a quadratic won’t factor over the integers so I don’t waste time? I tried using the discriminant (b^2 − 4ac) to see if it’s a perfect square, but I’m not sure if that’s the right idea here.
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3 Responses
Here’s my trusty “quadratic recipe” (served with a sprinkle of sign-sense): 1) First pull out the GCF of all coefficients; if the leading term is negative, also factor out −1 to make life sunnier. 2) Quick irreducibility check: after the GCF step, compute the discriminant D = b^2 − 4ac; if D isn’t a nonnegative perfect square, it won’t factor over the integers-save your time and move on. 3) If it passes, do the ac method: find integers p and q with pq = ac and p + q = b. Sign tip: if ac < 0, p and q have opposite signs; if ac > 0, they share the sign of b. 4) Split the middle term: ax^2 + bx + c = ax^2 + px + qx + c. 5) Group in pairs and factor each; if the common binomial doesn’t match, factor a negative from the second pair so it does. 6) Factor the common binomial, and you’re done. Worked example: 6x^2 + 7x − 5 → ac = −30; choose 10 and −3. Then 6x^2 + 10x − 3x − 5 = 2x(3x + 5) − 1(3x + 5) = (2x − 1)(3x + 5). Negative leading? Tame it first: −8x^2 + 14x − 3 = −(8x^2 − 14x + 3); now ac = 24, pick −12 and −2, so −[8x^2 − 12x − 2x + 3] = −[4x(2x − 3) − 1(2x − 3)] = −(4x − 1)(2x − 3). And a quick “nope” example: 12x^2 − 8x − 3 has D = 64 + 144 = 208, not a perfect square, so it won’t factor over the integers.
Here’s the checklist I use when the signs get grumpy and a ≠ 1: first pull out the greatest common factor (including a −1 if the leading coefficient is negative so you work with a > 0). After that, you’ve got a “primitive” quadratic; now do a quick discriminant check Δ = b^2 − 4ac-if Δ isn’t a perfect square, it won’t factor over the integers (so you can stop and save your sanity). If it might factor, I like the “cross” method because it avoids the guessy part of grouping: pick factor pairs u·w = a and v·z = c, choose signs so v·z = c, and check whether u·z + v·w = b; when it matches, you’ve got (u x + v)(w x + z). Sign tip: if c > 0 then v and z have the same sign (the same as b); if c < 0 they have opposite signs, and the larger-in-absolute-value one shares the sign of b. Simple worked example: 6x^2 + 7x − 5. Factors of a are (1,6) or (2,3); factors of c are (5, −1) or (−5, 1). Try u = 3, w = 2 with v = 5, z = −1: u z + v w = 3(−1) + 5·2 = −3 + 10 = 7, so 6x^2 + 7x − 5 = (3x + 5)(2x − 1). Using the same checklist: −8x^2 + 14x − 3 → factor out −1 first: −(8x^2 − 14x + 3); Δ = 196 − 96 = 100 = 10^2, so it factors, and the cross check gives (4x − 1)(2x − 3), hence −(4x − 1)(2x − 3). Meanwhile 12x^2 − 8x − 3 has Δ = 64 + 144 = 208, not a square, so don’t waste time-irreducible over the integers. I still scribble a tiny table of factor pairs when I’m unsure; it keeps me from overthinking the signs (which I definitely do on cranky days!).
Ooh, this is my favorite kind of pattern-hunt! Here’s a reliable flow I use: first, always pull out the greatest common factor (including a negative if a < 0 so the leading term is positive); that makes everything cleaner. Then compute ac and look for two integers m, n with m + n = b and mn = ac; sign rules help a lot: if ac > 0, m and n have the same sign as b; if ac < 0, one is positive and one is negative, and the larger absolute value matches the sign of b. Split bx into mx + nx and factor by grouping, but choose m and n with divisibility in mind so grouping “clicks”: try to make one of m share a factor with a and the other share a factor with c (that’s the secret sauce that makes binomials match). Example: 6x^2 + 7x − 5 → ac = −30, pick 10 and −3, so 6x^2 + 10x − 3x − 5 = 2x(3x + 5) − 1(3x + 5) = (3x + 5)(2x − 1). For negatives, I always normalize first: −8x^2 + 14x − 3 = −(8x^2 − 14x + 3) and 8x^2 − 14x + 3 = (4x − 1)(2x − 3), so the original is −(4x − 1)(2x − 3). A quick irreducibility check after pulling the GCF is the discriminant: if Δ = b^2 − 4ac isn’t a perfect square, it won’t factor over the integers; for 12x^2 − 8x − 3, Δ = 64 + 144 = 208, not a square, so stop. If you’re prone to sign slips, the 2×2 “box” version of grouping is super steady: place ax^2 and c on the diagonal, put mx and nx in the other boxes, and factor rows/columns by GCF-no sign guesswork. When I was learning this, I kept a little “ac table” and circled factor pairs with arrows to b; once I started forcing m to “share” factors with a and c, grouping suddenly felt like solving a clean puzzle instead of wrestling with signs.