Quadratic Expressions & Factoring

The reverse gear meets its first real hill

M1 ended with a promise: expanding has a reverse gear. You used it there to undo a monomial — spot the shared $2x$ in $6x^2 + 10x$ and pull it out front. Now look at $x^2 + 5x + 6$. Check its terms for a common factor: the coefficients $1$, $5$, $6$ share nothing, and the $6$ carries no $x$. The GCF move comes up empty — yet this expression is a product, $(x+2)(x+3)$, wearing its expanded costume. The SAT loves this disguise, and the whole story of quadratic equations (that’s M3, one unit ahead) turns on being able to strip it off. So: given the offspring, how do you find the parents?

Interrogate the product

You know from M1 exactly what expanding two parentheses does. Run it on $(x+m)(x+n)$ with the letters left in:

Read it like a detective. Whatever $m$ and $n$ were, the expanded costume reports on them: the constant term is their product, and the middle coefficient is their sum. The two numbers aren’t gone — they’re sitting in plain sight, encoded. So factoring $x^2 + 5x + 6$ is a hunt with two clues:

Watch the hunt happen at the board first — one wrong pair crossed out and all:

Now hunt yourself — the slider walks every pair of $c$, and you watch the sum needle:

Predict before you slide: load the $x^2 - 8x + 12$ preset. The product is positive $12$, the sum is negative $8$ — so should the pair be two positives, two negatives, or one of each? Decide, then sweep and see where the needle lands. Then load $x^2 + x - 12$: the product went negative, and now the pair must straddle zero — feel how much smaller the sums get when the two numbers fight instead of team up. That’s the sign compass, and your hands just learned it: $c$ positive → both numbers wear $b$‘s sign; $c$ negative → opposite signs, and the bigger number takes $b$‘s.

One more, important precisely because it fails: load $x^2 + 3x + 1$ and sweep the whole track. Nothing multiplies to $1$ and adds to $3$. That’s not you failing — that trinomial genuinely doesn’t factor over the integers, and on the SAT, recognizing a dead end fast is worth as much as factoring. (M3 gives you the tool that cracks even those.)

When the lead isn’t 1: send the hunt through a·c

$2x^2 + 7x + 3$ breaks the pattern above — the lead coefficient $2$ contaminates the simple sum–product reading. The fix is one extra stop: run the same hunt on $a \cdot c = 2 \cdot 3 = 6$, still adding to $7$. That’s $1$ and $6$. They don’t drop into parentheses directly; instead they split the middle term, and M1’s grouping instinct finishes it:

$2x^2 + 7x + 3 = 2x^2 + x + 6x + 3 = x(2x+1) + 3(2x+1) = (2x+1)(x+3)$

Read the third step twice — both halves handed you the same parenthesis, $(2x+1)$. That’s not luck; it’s the guarantee the hunt bought you. If your two groups ever disagree, the pair is wrong (or a sign slipped) — the method audits itself.

The trinomial with a missing term

Factor $x^2 - 9$. Don’t reach for a new rule — run the hunt you already own: two numbers that multiply to $-9$ and add to… well, there’s no $x$-term, so they add to $0$. That forces the pair $3$ and $-3$:

$x^2 - 9 = (x-3)(x+3)$

Expand it back and watch why the middle term vanished: $-3x + 3x = 0$. The pattern is worth knowing by name — a difference of squares, $A^2 - B^2 = (A-B)(A+B)$ — because the SAT dresses it up in coefficients: $4x^2 - 25 = (2x-5)(2x+5)$, both pieces perfect squares, a minus between. But it’s not a separate trick. It’s the hunt, in the special case where the pair is forced to be $\pm$ twins.

Its evil twin is the trap: $x^2 + 9$, a sum of squares, does not factor at all — try the hunt: multiply to $+9$, add to $0$ is impossible, since two numbers with a positive product are on the same side of zero and can’t cancel. If you catch yourself writing $(x+3)^2$ for either of these, expand it: $(x+3)^2 = x^2 + 6x + 9$. There’s a $6x$ in there that neither $x^2 - 9$ nor $x^2 + 9$ ever had. M1’s equivalence test settles it in one line.

Completing the square — the name is literal

Some quadratics refuse every hunt ($x^2 + 6x + 2$: multiply to $2$, add to $6$ — nothing). For those, algebra has a move so geometric its name is a construction manual. Take $x^2 + 6x$ and draw it: an $x$-by-$x$ square, plus a $6$-wide strip of $x$-height standing beside it. Now cut the strip in half — two slabs of $3x$ — and wrap one around the corner. What you’ve built is almost a bigger square, $(x+3)$ on each side… except the corner has a hole in it. A $3 \times 3$ hole. Exactly $9$.

Watch the square get built — and the classic trap get crossed out:

Drag the fold slowly and watch the hole open. That hole is the whole method: the pieces you HAVE ($x^2 + 6x$) are a completed square minus its corner patch:

$x^2 + 6x = (x+3)^2 - 9$

So $x^2 + 6x + 2$ — the one the hunt couldn’t touch — surrenders to arithmetic: it’s $(x+3)^2 - 9 + 2 = (x+3)^2 - 7$. The recipe your hands just learned: half of $b$ (that’s why the strip splits — only halves wrap evenly), square it (the corner patch), add it and subtract it (patch the hole, pay for the patch).

Predict before you drag: load the $x^2 + 10x + 25$ preset. The strip is $10$ wide — how wide is each wrapped half, and how big will the corner hole be? Now drag — and notice the $+25$ you brought along pays for the hole exactly. That’s what “perfect square trinomial” means: $x^2 + 10x + 25 = (x+5)^2$, no leftovers. Then try $x^2 + 5x + 3$: an odd strip still folds, the halves are just $\tfrac{5}{2}$ wide — fractions are the method working, not the method breaking.

One warning, because this is where points die: the sign of half rides along. For $x^2 - 8x + 5$, half of $-8$ is $-4$, so the square is $(x-4)^2$ — and the patch is $(-4)^2 = 16$, positive. Square the half after keeping its sign, and never forget the “subtract” half of add-and-subtract: $(x-4)^2$ alone claims a $+16$ you were never given.

The one thing to remember

Every factoring move in this unit is the expansion formula read backwards. The trinomial reports the pair’s sum and product — hunt it. No pair? Check for two squares and a minus. Still nothing? Complete the square: half of $b$, squared, added and subtracted — and if you forget why, fold the strip and look at the hole.

The three patterns

The sign compass for the hunt: $c > 0$ → both numbers wear $b$‘s sign. $c < 0$ → opposite signs, the bigger number takes $b$‘s. No pair works → it doesn’t factor over the integers.

Completing the square ($x^2 + bx + c$): half of $b$ (keep the sign) → square it → add AND subtract it → bundle: $\left(x + \tfrac{b}{2}\right)^2 + \left(c - \tfrac{b^2}{4}\right)$.

Perfect square trinomial: $x^2 \pm 2vx + v^2 = (x \pm v)^2$ — the constant is a square AND the middle is twice its root. Both checks, always.