Find two numbers that add to the middle and multiply to the end.
Point 1 of 5 in this lesson: 8.4.1 The shape of a quadratic trinomial
In Lesson 8.3 you ran the special-product formulas backward — a difference of squares, a perfect square. But most quadratics are not special. A trinomial like x2 + 5x + 6 fits no formula, yet it still factors into two neat pieces: (x + 2)(x + 3). The trick is a tiny puzzle — find two numbers that add to the middle coefficient and multiply to the constant — and a picture, the cross, that checks your guess in one glance. We keep the steady color habit of Stage 8: a FACTOR or side length is teal, an expanded term of the product is amber, a sign trap is red, and the structural frame (the rectangle, the axes) is blue.
By the end you will be able to factor any x2 + px + q by hunting for its magic pair, draw the cross diagram to confirm the middle term, read the signs off the constant before you even guess, and stretch the method to the harder case where the leading coefficient is not 1.
When you multiply two simple binomials you get a trinomial. Watch the shape carefully, because factoring is just reading this multiplication backward:
(x + m)(x + n) = x2 + (m + n)x + mn
The middle coefficient is the sum m + n; the constant is the product mn. So if someone hands you x2 + px + q and asks for the two factors, you are really being asked: what two numbers add to p and multiply to q? Find them and the factors are (x + m)(x + n).
The area picture makes the shape physical. Lay x2 + 5x + 6 out as tiles: one big x·x square, five x-strips, and six unit squares. They snap together into a single rectangle whose sides are x + 2 and x + 3. The constant 6 is the little 2×3 block of units in the corner; the 5 strips split as 2 + 3 along the two edges.
Every factorable x2 + px + q comes from (x + m)(x + n), where m + n = p (the middle) and m · n = q (the end). Factoring is the search for that pair.
So factoring x2 + px + q boils down to one search: find two integers m and n with m + n = p and m · n = q. The smart way to search is to list the factor pairs of the product q and check which pair adds to p. Always start from the product — it has only a handful of pairs — then test the sum.
Take x2 + 5x + 6. The product is 6, whose positive factor pairs are 1·6 and 2·3. Their sums are 7 and 5. We want 5, so the pair is 2 and 3: x2 + 5x + 6 = (x + 2)(x + 3). For x2 + 7x + 12, the pairs of 12 are 1·12, 2·6, 3·4 with sums 13, 8, 7 — so the pair is 3 and 4, giving (x + 3)(x + 4).
Negatives come along for the ride. For x2 − 5x + 6 we need a product of +6 but a sum of −5; that forces both numbers negative: −2 and −3, so (x − 2)(x − 3). For x2 + x − 6 the product is −6 (opposite signs) and the sum is +1: the pair is +3 and −2, giving (x + 3)(x − 2).
Factor x2 + 8x + 15.
Product is 15: pairs 1·15 (sum 16) and 3·5 (sum 8). The sum we want is 8, so the pair is 3 and 5.
Answer: (x + 3)(x + 5). Check by expanding: x2 + 5x + 3x + 15 = x2 + 8x + 15 ✓.
Set the middle coefficient p and the constant q. The machine lists every integer factor pair of q, marks the one that adds to p, and writes the factored form.
The cross (or "X") diagram is the fastest way to confirm a factoring — and the tool that gives the method its name. Write the two factors as two short columns, one above the other, splitting each into its x-part and its number. Then multiply along the two diagonals of the X and add the results: that sum must rebuild the middle term.
For x2 + 5x + 6 = (x + 2)(x + 3), lay the factors in a 2×2 grid. Down each column the two x's multiply to x2 and the two numbers 2 · 3 to 6. Across the diagonals, x·3 = 3x and 2·x = 2x, and 3x + 2x = 5x — the middle term. All three pieces of the trinomial appear, so the factoring is correct.
You can also lay the same check as a small table — handy for neatness:
| x | +3 | |
| x | x2 | 3x |
| +2 | 2x | 6 |
The four cells sum to x2 + 3x + 2x + 6 = x2 + 5x + 6. ✓
Choose the two numbers m and n in (x + m)(x + n). The cross draws itself and reports the trinomial it builds — watch the diagonals add to the middle term.
Before you guess any numbers, read the signs off the trinomial — they cut your search in half. Everything hinges on the sign of the constant q:
| Sign of constant q | The two numbers… | …and the rest |
|---|---|---|
| q > 0 | share a sign | that shared sign is the sign of p |
| q < 0 | have opposite signs | the bigger-magnitude one takes the sign of p |
Why? Because m · n = q. If q is positive, the two numbers must have the same sign (two positives or two negatives), and their sum p then carries that sign. If q is negative, the numbers must have opposite signs; their sum leans toward whichever has the larger magnitude, so that one takes p's sign.
Two worked cases. For x2 − x − 6: the constant is −6, so opposite signs; we need product −6 and sum −1, so the bigger-magnitude number is negative — that's −3 and +2, giving (x − 3)(x + 2). For x2 − 7x + 12: the constant is +12, so same sign, and since p = −7 is negative, both are negative — −3 and −4, giving (x − 3)(x − 4).
The most common mistake is to nail the sum and forget the product's sign. In x2 + x − 6 the numbers +2 and −1 sum to +1 but multiply to −2, not −6 — wrong. The correct pair +3 and −2 hits both: sum +1, product −6. Always check the product, not just the sum.
Factor x2 + 2x − 15.
Constant −15 → opposite signs; sum +2 → bigger one is positive. Pairs of 15: 1·15, 3·5. We need a difference of 2, so +5 and −3.
Answer: (x + 5)(x − 3). Check: x2 − 3x + 5x − 15 = x2 + 2x − 15 ✓.
What about ax2 + bx + c when a ≠ 1? The cross still works — you just have two columns to choose. Now you split a across the top of the cross as well as c, and the diagonal products must add to the middle coefficient b.
Take 2x2 + 7x + 3. Split the leading 2 as 2x · x and the constant 3 as 1 · 3. Arrange the cross so the diagonals give 2x·3 = 6x and 1·x = x, summing to 7x — the middle term. So 2x2 + 7x + 3 = (2x + 1)(x + 3).
Two more, the same way. 3x2 − 10x + 8 splits as (3x − 4)(x − 2): diagonals 3x·(−2) = −6x and (−4)·x = −4x sum to −10x ✓. And 6x2 + 11x + 3 = (3x + 1)(2x + 3): diagonals 3x·3 = 9x and 1·2x = 2x sum to 11x ✓.
Factor 3x2 − 10x + 8.
Constant +8 with negative middle → both numbers negative. Split 3 as 3x · x and 8 as (−4)·(−2).
Diagonals: 3x·(−2) = −6x and (−4)·x = −4x, total −10x ✓.
Answer: (3x − 4)(x − 2).
Pick the four pieces of (ax + b)(cx + d). The widget expands them so you can see how a·c lands on x2, b·d on the constant, and the diagonals on the middle term.
A trinomial x2 + px + q factors as (x + m)(x + n) exactly when two integers m, n add to p and multiply to q — so search the factor pairs of q and pick the one summing to p. The cross diagram confirms a guess: the column products rebuild the two ends, and the two diagonal products must add to the middle term. Read the signs off q before guessing: if q > 0 the numbers share a sign (the sign of p); if q < 0 they have opposite signs and the bigger one takes p's sign. When the leading coefficient isn't 1, split a across the top of the cross too and make the diagonals add to b. And always — always — expand your answer to check it equals the original.
What if a polynomial has four terms and no formula fits? Next you'll learn factoring by grouping: pair the terms, factor each pair, and watch a shared bracket appear that you can pull out front.
Work each one out first, then open the answer to check your thinking.
Six questions to lock it in. Tap the answer you think is right.
This lesson develops CCSS A-SSE.B.3a — factoring a quadratic expression to reveal the zeros of the function it defines — and lays the groundwork for solving quadratics by factoring in A-REI.B.4b. It builds directly on the structure-spotting of A-SSE.A.2 and the special-product work of Lesson 8.3. The single most common misconception is matching only the product or only the sum: a student finds two numbers that multiply to q but forgets to check they add to p, or vice versa. The antidote built into this lesson is the cross diagram, which forces both diagonal products and their sum to be checked at once — and the unbreakable habit of expanding the answer back to confirm it equals the original. Sign errors are the second pitfall; the sign table in 8.4.4 (read the sign of q first) turns a guessing game into a short, directed search.