The first move every time: find the piece every term carries, and set it outside.
Point 4 of 5 in this lesson: 8.2.4 A whole bracket can be the common factor
In Lesson 8.1 you learned that factoring is multiplication run in reverse: handed the area of a rectangle, you recover its sides. Now we meet the very first move you should try on any factoring problem — pulling out the common factor. If every term in a polynomial carries the same piece, you set that piece outside a set of parentheses and write what is left inside. We keep Stage 8's steady color habit: a factor — a side length, the piece you pull out — is teal; an expanded term, a product, a piece of the area — is amber; a sign trap or a "not fully factored" warning — is red; and the blue frame is the structural rectangle that holds it all together.
By the end you will be able to spot the shared piece every term carries, build the greatest common factor from the numbers and the letters, pull out a leading minus sign without losing a term, treat a whole bracket like (x + y) as one bundle, and — every single time — multiply your answer back out to prove it is right.
Look at ma + mb. Both terms are built from the same piece, m: the first is m times a, the second is m times b. When a piece sits inside every term, it is a common factor, and the distributive law — read backward — lets you set it out front:
m·a + m·b = m(a + b)
The picture makes it obvious. Lay the two terms side by side as one rectangle of shared height m: the left piece is m tall and a wide, the right piece is m tall and b wide. Measured piece by piece, the area is ma + mb. But the whole strip is simply m tall and a + b wide — so its area is also m(a + b). One rectangle, two ways to measure it; the two measurements must be equal.
If a piece appears in every term, it is a common factor. Set it outside the parentheses, and write what is left of each term inside. That is the distributive law, run in reverse.
Set the shared height m and the two widths a and b. Watch the strip split into the two amber pieces m·a and m·b — then see the teal height m come out front as m(a + b).
When the terms carry numbers and letters, you do not want just any common factor — you want the greatest one, the GCF, so nothing more can be pulled out afterward. Build it in two halves:
GCF = (the gcd of the coefficients) × (each shared letter raised to its lowest power)
Take 6x2 + 9x. The coefficients are 6 and 9, whose gcd is 3. Both terms carry x; the lowest power present is x1 (from the second term). So the GCF is 3x, and dividing each term by it leaves 2x + 3:
6x2 + 9x = 3x(2x + 3)
The lowest-power rule is the part students forget. With 12a3b − 18a2b2, the gcd of 12 and 18 is 6; the lowest power of a is a2, and the lowest power of b is b1. So the GCF is 6a2b:
12a3b − 18a2b2 = 6a2b(2a − 3b)
A three-term example works exactly the same way. For 4x3 + 8x2 + 12x, the gcd of 4, 8, 12 is 4, and the lowest power of x across all three terms is x1, giving GCF 4x:
4x3 + 8x2 + 12x = 4x(x2 + 2x + 3)
Factor 12a3b − 18a2b2.
Numbers: gcd(12, 18) = 6. Letter a: lowest power is a2. Letter b: lowest power is b1. So GCF = 6a2b.
Divide each term: 12a3b ÷ 6a2b = 2a, and 18a2b2 ÷ 6a2b = 3b. So the answer is 6a2b(2a − 3b).
Build a binomial c₁·xe₁ + c₂·xe₂ with the steppers. The machine takes the gcd of the two coefficients and the lowest power of x, then shows the GCF and the fully factored form.
A factor does not have to be a number or a letter — it can be a sign. Pulling out −1 flips the sign of every term that was inside. So −a − b = −(a + b), because −1 times a is −a and −1 times b is −b. The minus reaches all the way through the parentheses.
This matters most when the leading term is negative. A tidy habit: if the first term is negative, factor the sign out along with the GCF. Take −2x2 + 4x. Pull out −2x and every inside sign flips:
−2x2 + 4x = −2x(x − 2)
Check it by expanding: −2x·x = −2x2 and −2x·(−2) = +4x. ✓ The second sign flipped from −2 inside to +4x outside, exactly as it should.
| Inside, before | Pull out −1 | Inside, after |
|---|---|---|
| −a − b | −( … ) | a + b |
| −a + b | −( … ) | a − b |
When you pull out a negative, every sign inside flips, not just the first one. −2x2 + 4x is −2x(x − 2), not −2x(x + 2). Forget to flip the +4x and your answer expands to the wrong thing. Always multiply back to catch it.
The common factor does not have to be a single symbol — it can be an entire bracket. Treat a group like (x + y) as one indivisible bundle, exactly as if it were a single letter. Then in a(x + y) + b(x + y), the bundle (x + y) sits in both terms, so pull it out front:
a(x + y) + b(x + y) = (x + y)(a + b)
It reads just like am + bm = m(a + b), only now the shared piece m happens to be the bundle (x + y). The signs travel with the leftover pieces, so 4(m − n) − x(m − n) keeps the shared (m − n) and gathers 4 − x inside:
4(m − n) − x(m − n) = (m − n)(4 − x)
Factor a(x + 1) + 5(x + 1).
Both terms carry the bundle (x + 1). Pull it out front; what is left of each term is a and 5. So the answer is (x + 1)(a + 5). Expand to check: (x + 1)·a + (x + 1)·5 = a(x + 1) + 5(x + 1). ✓
Factoring rewrites a sum as a product without changing its value — so the surest proof you did it right is to expand your answer and compare. Distribute the piece you pulled out across every term inside the parentheses; you should land back on exactly the polynomial you started with.
Take the answer from Section 8.2.1's family, 3x(2x + 3). Distribute: 3x·2x = 6x2 and 3x·3 = 9x, so the product is 6x2 + 9x — the original. ✓ That round trip catches the two classic mistakes at once: a leftover that is too small (dropping the +3) shows up as a missing term, and a missed sign shows up as a wrong sign.
The most common slip is pulling the factor out but leaving an incomplete quotient — writing 6x2 + 9x = 3x(2x) and dropping the +3. Expanding 3x·2x gives only 6x2, so the 9x vanished — the answer fails its own check. Every term inside must be the original term divided by the GCF.
Build a factored form k·x(a·x + b) with the steppers. The machine multiplies it back out term by term and shows the polynomial it came from — proof the factoring is exact.
The first move in factoring is always to look for the piece every term carries. If a piece appears in every term it is a common factor, and you set it outside the parentheses with the leftovers inside: ma + mb = m(a + b). To pull out the most you can, build the GCF from the gcd of the coefficients times the lowest power of each shared letter — so 6x2 + 9x = 3x(2x + 3) and 12a3b − 18a2b2 = 6a2b(2a − 3b). When the lead term is negative, factor the sign out too and flip every inside term: −2x2 + 4x = −2x(x − 2). The shared piece can even be a whole bracket: a(x + y) + b(x + y) = (x + y)(a + b). And whatever you pull out, multiply back to check.
Once the common factor is gone, what is left often hides a famous pattern. Next you will run the multiplication formulas backward — spotting a difference of squares a2 − b2 = (a + b)(a − b) and a perfect-square trinomial — to factor in a single glance.
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 A-SSE.A.2 (use the structure of an expression to identify ways to rewrite it) and A-SSE.B.3a (factor a quadratic — here, the prerequisite step of pulling out the greatest common factor), resting on the distributive property students met in 6.EE.A.3/4. Pulling out the GCF is the universal first move that makes every later method — special products, cross-multiplication, grouping — cleaner, so it is worth over-practicing. The two most common misconceptions are (1) pulling out a factor but leaving an incomplete quotient, e.g. writing 6x² + 9x → 3x(2x) and dropping the +3; and (2) forgetting that the GCF must include the lowest power of each shared variable, not the highest. A third, sign-related slip is failing to flip every inside term when a negative is factored out. The single antidote to all three is the habit baked into Section 8.2.5: multiply your answer back out and compare it to the original — a dropped term, a wrong power, or a missed sign all reveal themselves instantly.