Factoring By Pulling Out The Greatest Common Factor Worksheet ... - Free Printable

You’ve uploaded a worksheet titled “Algebra 2 WS #2: Factoring by pulling out the Greatest Common Factor”, with instructions to factor each polynomial and write PRIME if it does not factor.

Let’s go through each problem one by one, factoring out the Greatest Common Factor (GCF). If no GCF exists (other than 1) and the expression cannot be factored further using basic techniques, we’ll label it PRIME.

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1) 5ax – 5a

✔ GCF = 5a
→ 5a(x – 1)

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2) 5xz + 2xy – 3yz

Check for common factor in all terms:
- 5xz → factors: 5, x, z
- 2xy → 2, x, y
- 3yz → 3, y, z
→ No common variable or constant across all three terms.
→ PRIME

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3) 24ab⁴ + 12ab³ – 18ab²

✔ GCF = 6ab² (smallest power of a, b; largest number dividing 24,12,18 is 6)
→ 6ab²(4b² + 2b – 3)

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4) 3n² + 9

✔ GCF = 3
→ 3(n² + 3)
→ n² + 3 doesn’t factor over reals → Final answer: 3(n² + 3)

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5) x(x + y) – y(x + y)

✔ Common binomial factor: (x + y)
→ (x + y)(x – y)
→ This is also a difference of squares: x² – y², but since we’re factoring by GCF, pulling out (x+y) is correct.
→ (x + y)(x – y)

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6) 25k⁴ + 20k³ + 10k

✔ GCF = 5k
→ 5k(5k³ + 4k² + 2)
→ Check if trinomial factors? 5k³ + 4k² + 2 — no rational roots (try Rational Root Theorem: ±1,2,1/5,2/5 — none work). So leave as is.
→ 5k(5k³ + 4k² + 2)

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7) 8x² + 5x – 7

Check for GCF: coefficients 8,5,7 → no common factor.
Try factoring quadratic:
Need two numbers that multiply to 8×(-7)= -56 and add to 5 → none such integers.
→ PRIME

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8) 7ab³ – 56ab

✔ GCF = 7ab
→ 7ab(b² – 8)
→ b² – 8 doesn’t factor nicely over integers → 7ab(b² – 8)

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9) mnx² – nx² + m²x

Group terms:
= x(mnx – nx + m²)
Factor inside: nx(mx – x) + m² → Not helpful.
Better: look for GCF of all terms:
- mnx², nx², m²x → common factor? Only x
→ x(mnx – nx + m²)
Can we factor further? Group:
= x[ nx(m – 1) + m² ] → no common binomial → can’t factor further
→ x(mnx – nx + m²) — or leave as is.
But technically, this is fully factored after pulling out x.
✔ Final: x(mnx – nx + m²)

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10) x²(x² – 5) – 8(x² – 5)

✔ Common binomial factor: (x² – 5)
→ (x² – 5)(x² – 8)
→ Both are differences of squares?
x² – 5 → irrational roots, not factorable over integers
x² – 8 → same
→ So factored form: (x² – 5)(x² – 8)

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11) 6k³ – 18k²

✔ GCF = 6k²
→ 6k²(k – 3)

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12) 12m³ – 8m² + 20m

✔ GCF = 4m
→ 4m(3m² – 2m + 5)
→ Quadratic: 3m² – 2m + 5 → discriminant = 4 – 60 = -56 → no real roots → doesn’t factor
→ 4m(3m² – 2m + 5)

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13) 6xy – 6xz – 6x

✔ GCF = 6x
→ 6x(y – z – 1)

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14) 3x⁴ + 12x³ – 33

✔ GCF = 3
→ 3(x⁴ + 4x³ – 11)
→ x⁴ + 4x³ – 11 — try factoring? No rational roots (try ±1,11 → none work).
→ 3(x⁴ + 4x³ – 11)

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15) 8a³b⁵ – 28a²b³ + 4a²b²

✔ GCF = 4a²b² (smallest powers: a², b²; 4 divides 8,28,4)
→ 4a²b²(2ab³ – 7b + 1)
→ Inside: 2ab³ – 7b + 1 — no further factoring possible
→ 4a²b²(2ab³ – 7b + 1)

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16) 4k⁵ + 18k³ – 6k⁴

First, rearrange in descending order:
→ 4k⁵ – 6k⁴ + 18k³
✔ GCF = 2k³
→ 2k³(2k² – 3k + 9)
→ Quadratic: 2k² – 3k + 9 → discriminant = 9 – 72 = -63 → no real roots → doesn’t factor
→ 2k³(2k² – 3k + 9)

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## ✔ Final Answers:

1) 5a(x – 1)
2) PRIME
3) 6ab²(4b² + 2b – 3)
4) 3(n² + 3)
5) (x + y)(x – y)
6) 5k(5k³ + 4k² + 2)
7) PRIME
8) 7ab(b² – 8)
9) x(mnx – nx + m²)
10) (x² – 5)(x² – 8)
11) 6k²(k – 3)
12) 4m(3m² – 2m + 5)
13) 6x(y – z – 1)
14) 3(x⁴ + 4x³ – 11)
15) 4a²b²(2ab³ – 7b + 1)
16) 2k³(2k² – 3k + 9)

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📌 Note: “PRIME” here means cannot be factored further using integer coefficients and GCF method. Some expressions may factor over reals or complexes, but per Algebra 2 level and instructions, we use “PRIME” when no GCF and no further factoring is possible with integers.

Let me know if you want step-by-step explanations for any specific problem!