Algebra 1 - Factoring Trinomials ax^2 + bx + c Foldable - Free Printable

Let’s go through each quadratic expression and factor it step by step. We’ll check if it can be factored into two binomials, or if it’s prime (can’t be factored with integers).

---

1. 2x² + 8x + 2

Factor out the GCF first:
All terms divisible by 2 → 2(x² + 4x + 1)
Now try to factor x² + 4x + 1.
We need two numbers that multiply to 1 and add to 4 → only possibilities are 1 and 1 → 1+1=2 ≠ 4.
So x² + 4x + 1 is prime.
→ Final: 2(x² + 4x + 1) — but wait! The image shows this as “prime” in a box? Actually, no — looking at the image layout, the “prime” label is under 2b² + 3b - 6. Let’s double-check our work.

Wait — maybe I misread. Let me recheck 2x² + 8x + 2:

GCF = 2 → 2(x² + 4x + 1). Can we factor x² + 4x + 1? Discriminant: b² - 4ac = 16 - 4 = 12 → not a perfect square → so yes, irreducible over integers. So technically, after factoring out GCF, the trinomial is prime. But sometimes “prime” means the whole expression can’t be factored further *at all*. Since we factored out 2, it’s not prime — it’s partially factored.

But in many school contexts, if you can factor out a GCF and then the rest is prime, they still consider the original expression “factored” as GCF times prime trinomial. However, some teachers say “prime” only if NO factoring is possible at all.

Looking at the image again — for 2x² + 8x + 2, the student wrote:
> 2x² + x + 4x + 2
> x(2x + 1) + 2(2x + 1)
> (x + 2)(2x + 1)

That’s WRONG. Because 2x² + 8x + 2 does NOT equal (x+2)(2x+1). Let’s multiply: (x+2)(2x+1) = 2x² + x + 4x + 2 = 2x² + 5x + 2 — which is NOT the same as 2x² + 8x + 2.

So the student made an error here.

Correct approach:

2x² + 8x + 2 = 2(x² + 4x + 1) — and x² + 4x + 1 cannot be factored further using integers → so final answer should be 2(x² + 4x + 1), or if required to leave as is, note that it's not factorable beyond GCF.

But let’s compare to what’s shown in the image — perhaps the problem was meant to be 2x² + 5x + 2? That would factor to (2x+1)(x+2). Maybe typo in problem?

Wait — user uploaded image of completed worksheet. Our job is to verify correctness.

Actually, let’s systematically solve ALL problems correctly, ignoring the student’s answers unless needed for context.

Since ExplainAnswer is false, TeachTopic is false, GeneratePractice is false — we just need to provide correct final answers for each expression.

Let’s list all 12 expressions from the image grid:

Top row:
1. 2x² + 8x + 2
2. 3c² - 7c + 2
3. 4x² + 8x - 5

Second row:
4. 6g² - 13g + 6
5. 9p² + 6p - 8
6. 10x² + 19x - 15

Third row:
7. 2b² + 3b - 6
8. 3a² + 30a + 63
9. 6n² + 10n + 4

And one labeled “prime” — which is #7: 2b² + 3b - 6

Let’s factor each properly.

---

1. 2x² + 8x + 2

GCF = 2 → 2(x² + 4x + 1)
Discriminant of x² + 4x + 1: 16 - 4 = 12 → not perfect square → doesn't factor over integers.
→ Answer: 2(x² + 4x + 1)
*(Note: Some might write "prime" if they don’t allow GCF extraction, but standard practice is to factor out GCF first.)*

But wait — let’s test if original can be factored without GCF? Try AC method: a*c = 2*2=4. Need two numbers multiplying to 4, adding to 8 → none. So indeed, only GCF.

However, in the image, student tried to split middle term incorrectly. So correct answer is 2(x² + 4x + 1)

But let’s hold on — maybe the problem was mistyped? If it were 2x² + 5x + 2, then it factors to (2x+1)(x+2). Given that the student wrote that, perhaps there’s a typo in the problem statement? But we must go by what’s written: 2x² + 8x + 2.

I think safest is to factor out GCF and stop.

---

2. 3c² - 7c + 2

AC method: a*c = 3*2=6. Find two numbers that multiply to 6, add to -7 → -6 and -1.

Split middle term: 3c² -6c -c + 2
Group: (3c² -6c) + (-c + 2) = 3c(c - 2) -1(c - 2) = (3c -1)(c - 2)

Check: (3c-1)(c-2) = 3c² -6c -c +2 = 3c² -7c +2 ✓

→ Answer: (3c - 1)(c - 2)

---

3. 4x² + 8x - 5

AC = 4*(-5) = -20. Need two numbers multiplying to -20, adding to 8 → 10 and -2.

Split: 4x² +10x -2x -5
Group: (4x² +10x) + (-2x -5) = 2x(2x + 5) -1(2x + 5) = (2x -1)(2x + 5)

Check: (2x-1)(2x+5) = 4x² +10x -2x -5 = 4x² +8x -5 ✓

→ Answer: (2x - 1)(2x + 5)

---

4. 6g² - 13g + 6

AC = 6*6=36. Need two numbers multiplying to 36, adding to -13 → -9 and -4.

Split: 6g² -9g -4g +6
Group: (6g² -9g) + (-4g +6) = 3g(2g - 3) -2(2g - 3) = (3g - 2)(2g - 3)

Check: (3g-2)(2g-3) = 6g² -9g -4g +6 = 6g² -13g +6 ✓

→ Answer: (3g - 2)(2g - 3)

---

5. 9p² + 6p - 8

AC = 9*(-8) = -72. Need two numbers multiplying to -72, adding to 6 → 12 and -6.

Split: 9p² +12p -6p -8
Group: (9p² +12p) + (-6p -8) = 3p(3p + 4) -2(3p + 4) = (3p - 2)(3p + 4)

Check: (3p-2)(3p+4) = 9p² +12p -6p -8 = 9p² +6p -8 ✓

→ Answer: (3p - 2)(3p + 4)

---

6. 10x² + 19x - 15

AC = 10*(-15) = -150. Need two numbers multiplying to -150, adding to 19 → 25 and -6.

Split: 10x² +25x -6x -15
Group: (10x² +25x) + (-6x -15) = 5x(2x + 5) -3(2x + 5) = (5x - 3)(2x + 5)

Check: (5x-3)(2x+5) = 10x² +25x -6x -15 = 10x² +19x -15 ✓

→ Answer: (5x - 3)(2x + 5)

---

7. 2b² + 3b - 6

AC = 2*(-6) = -12. Need two numbers multiplying to -12, adding to 3 → possible pairs: (4,-3), (6,-2), etc. 4 + (-3)=1; 6 + (-2)=4; none give 3.

Try discriminant: b² - 4ac = 9 - 4*2*(-6) = 9 + 48 = 57 → not perfect square → doesn’t factor over integers.

Also, no GCF (coefficients 2,3,6 — GCF is 1).

→ Answer: prime

---

8. 3a² + 30a + 63

GCF = 3 → 3(a² + 10a + 21)
Now factor a² + 10a + 21: two numbers multiplying to 21, adding to 10 → 3 and 7.

→ 3(a + 3)(a + 7)

Check: 3(a+3)(a+7) = 3[a² +10a +21] = 3a² +30a +63 ✓

→ Answer: 3(a + 3)(a + 7)

---

9. 6n² + 10n + 4

GCF = 2 → 2(3n² + 5n + 2)
Now factor 3n² + 5n + 2: AC=6, numbers multiplying to 6, adding to 5 → 2 and 3.

Split: 3n² +2n +3n +2 → group: n(3n+2) +1(3n+2) = (n+1)(3n+2)

So overall: 2(n+1)(3n+2)

Check: 2(n+1)(3n+2) = 2[3n² +2n +3n +2] = 2[3n² +5n +2] = 6n² +10n +4 ✓

→ Answer: 2(n + 1)(3n + 2)

---

Now, compiling all answers:

1. 2x² + 8x + 2 → 2(x² + 4x + 1)
*(Note: This trinomial inside is prime, but since we factored out GCF, we present it like this. Some may argue to call entire thing "not factorable", but standard is to factor GCF first.)*

But wait — let’s reconsider #1. Is there any way to factor 2x² + 8x + 2 directly? Suppose we ignore GCF and try AC: a*c=4, need sum 8 — impossible. So only GCF.

In many textbooks, they expect you to factor out GCF first, then see if remaining trinomial factors. Here it doesn’t, so answer is 2(x² + 4x + 1).

However, looking back at the image — the student attempted to factor it as (x+2)(2x+1), which is wrong. So we must correct that.

But also, notice that in the third row, first column, it says “prime” for 2b² + 3b - 6 — which we confirmed is correct.

For consistency, let’s make sure all answers are fully factored where possible.

Another thought: for #1, 2x² + 8x + 2, if we divide everything by 2, get x² + 4x + 1, which has roots [-4 ± 12]/2 = -2 ± √3 — irrational, so definitely doesn’t factor over integers.

So final decision: leave as 2(x² + 4x + 1)

But some systems might prefer to write it as is if no integer factors, but since GCF exists, we factor it.

Alternatively, perhaps the problem intended 2x² + 5x + 2? But we have to go with given.

Let me check online or standard practice — actually, upon second thought, in many algebra classes, if you can factor out a GCF and the rest is prime, you still write the GCF factored form. For example, 2x² + 4x + 2 = 2(x² + 2x + 1) = 2(x+1)^2 — but here it’s not a perfect square.

So for #1: 2(x² + 4x + 1)

But let’s see what the other answers look like — most are fully factored into binomials.

Perhaps for #1, since the inner part is prime, and we’ve factored out GCF, it’s acceptable.

To match format, let’s write all as products.

Final Answers List:

1. 2x² + 8x + 2 → 2(x² + 4x + 1)
2. 3c² - 7c + 2 → (3c - 1)(c - 2)
3. 4x² + 8x - 5 → (2x - 1)(2x + 5)
4. 6g² - 13g + 6 → (3g - 2)(2g - 3)
5. 9p² + 6p - 8 → (3p - 2)(3p + 4)
6. 10x² + 19x - 15 → (5x - 3)(2x + 5)
7. 2b² + 3b - 6 → prime
8. 3a² + 30a + 63 → 3(a + 3)(a + 7)
9. 6n² + 10n + 4 → 2(n + 1)(3n + 2)

Wait — the image has 9 boxes? Let me count:

Rows:

Row 1: 3 expressions
Row 2: 3 expressions
Row 3: 3 expressions (including "prime" for 2b²+3b-6)
Total 9.

Yes.

Now, for #1, I’m uneasy because 2(x² + 4x + 1) is correct, but perhaps the problem expects us to recognize it as not factorable even with GCF? No, GCF should always be factored first.

Another idea: maybe calculate discriminant for original: for 2x² + 8x + 2, D = 64 - 16 = 48, not perfect square, so doesn't factor nicely, but GCF is 2, so we factor that.

I think it's fine.

But let’s double-check #8 and #9 — both had GCFs, and we factored them completely.

Similarly, #1 should have GCF factored.

So I'll proceed.

Final Answer section will list all 9 answers in order.

──────────────────────────────────────

Final Answer:
1. \( 2(x^2 + 4x + 1) \)
2. \( (3c - 1)(c - 2) \)
3. \( (2x - 1)(2x + 5) \)
4. \( (3g - 2)(2g - 3) \)
5. \( (3p - 2)(3p + 4) \)
6. \( (5x - 3)(2x + 5) \)
7. prime
8. \( 3(a + 3)(a + 7) \)
9. \( 2(n + 1)(3n + 2) \)