I’m prepping for a test and factorising quadratics is making my brain do tiny cartwheels. I can handle the ones where the x^2 coefficient is 1 (like x^2 + 5x + 6), but when there’s a number glued to x^2 I start second-guessing every sign and pair of numbers.
Example: 6x^2 – x – 12. I tried the AC method: a*c = -72. So I hunted for two numbers that multiply to -72 and add to -1. I scribbled pairs like 9 and -8, 8 and -9, 12 and -6, etc. I picked -9 and 8 and split the middle term: 6x^2 – 9x + 8x – 12. Then I tried grouping, but I keep messing the binomials so they don’t match. One attempt looked like 3x(2x – 3) + 4(2x – 3) and then I panicked that I messed a sign earlier and erased it. Other times I get something like 3x(2x – 3) + 4(3x + 4), which obviously goes nowhere. What’s the clean, reliable way to do this without turning my scratch paper into a novella?
Another one that trips me up: 4x^2 + 13x + 3. I tried setting it up as (4x + ?)(x + ?). I played with 1 and 3 in a few spots, but my middle term kept coming out 15x or 10x instead of 13x. Is there a trick to choosing which factor pair goes with which binomial so I don’t brute-force every combination?
Bonus: how can I quickly tell when a quadratic won’t factor nicely over integers so I know to stop and use another method during the test?
Any help appreciated!
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3 Responses
Totally get the cartwheel feeling-when a ≠ 1, I use one tidy pattern every time: the AC/cross method, with a sign rule. Think of factoring (ax^2 + bx + c) like fitting two Lego bricks: (px + r)(qx + s) where p·q = a and r·s = c, and the “snap test” is ps + qr = b. Signs: if c > 0, the two signs match (both + if b > 0, both − if b < 0); if c < 0, the signs differ. Now your examples. 6x^2 − x − 12: a·c = −72. Pick −9 and 8 (sum −1). Split and group: 6x^2 − 9x + 8x − 12 = 3x(2x − 3) + 4(2x − 3) = (2x − 3)(3x + 4). You actually had this right-when the grouped binomials don’t match at first, factor a negative to make them match. Second: 4x^2 + 13x + 3: a·c = 12, use 12 and 1. 4x^2 + 12x + x + 3 = 4x(x + 3) + 1(x + 3) = (x + 3)(4x + 1). A quick “cross” version avoids the rewrite: try p·q = 4, r·s = 3; test (4x + 1)(x + 3): cross-sum 4·3 + 1·1 = 12 + 1 = 13-ding! Finally, the fast “should I even try?” test: compute the discriminant D = b^2 − 4ac. If D is a perfect square (after pulling out any common factor first), it factors nicely over the integers; if not, don’t waste time-use completing the square or the quadratic formula. For your two: D = 289 = 17^2 and D = 121 = 11^2, so both factor. It’s like a combination lock: pick the right factor pairs, cross-check the middle number, and click-open!
Use the AC split-and-group trick-like finding two Lego bricks that click to ac and line up to b: for 6x^2 − x − 12, ac = −72, pick −9 and 8, so 6x^2 − 9x + 8x − 12 = (2x − 3)(3x + 4), and for 4x^2 + 13x + 3, ac = 12, pick 12 and 1, so 4x^2 + 12x + x + 3 = (x + 3)(4x + 1). Quick “should I stop?” test: b^2 − 4ac must be a perfect square (after pulling out any common factor) for the quadratic to factor nicely over the integers.
When a ≠ 1, I lean on the AC (product–sum) trick so I don’t turn my scratch paper into a novella: find m and n with m*n = a*c and m+n = b, split bx into mx+nx and factor by grouping; quick sanity check-if b^2 − 4ac isn’t a perfect square, it won’t factor nicely over integers.
Example runs: 6x^2 − x − 12 → ac = −72, pick 8 and −9 ⇒ 6x^2 − 9x + 8x − 12 = (3x + 4)(2x − 3); 4x^2 + 13x + 3 → ac = 12, pick 12 and 1 ⇒ 4x^2 + 12x + x + 3 = (4x + 1)(x + 3).