Factoring Polynomials with GCF worksheet for algebra practice.

Let’s go through each problem one by one. We’re factoring polynomials using the GCF — that means Greatest Common Factor. The goal is to pull out the biggest thing that all terms have in common, and write the expression as a product (multiplication) of that GCF and what’s left.

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Problem 1:

8m³ + 16m²n

Step 1: Look at coefficients: 8 and 16 → GCF is 8
Step 2: Look at variables: m³ and m²n → both have at least m²
So GCF = 8m²

Now divide each term by 8m²:
- 8m³ ÷ 8m² = m
- 16m²n ÷ 8m² = 2n

Answer: 8m²(m + 2n)

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Problem 2:

2x(x + 4) - 3(x + 4)

Notice: Both terms have (x + 4) in them! That’s our GCF.

Factor out (x + 4):
→ (x + 4)(2x - 3)

Answer: (x + 4)(2x - 3)

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Problem 3:

4y(y - 3b) - 2(y - 3b)

Both terms have (y - 3b) → that’s the GCF.

Factor it out:
→ (y - 3b)(4y - 2)

But wait — can we factor more? Yes! In (4y - 2), both 4 and 2 are divisible by 2.

So: (y - 3b) * 2*(2y - 1) → rearrange: 2(y - 3b)(2y - 1)

Answer: 2(y - 3b)(2y - 1)

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Problem 4:

63m⁶ - 49m⁵ - 21m

Coefficients: 63, 49, 21 → GCF is 7
Variables: m⁶, m, m → lowest power is m¹ → so GCF = 7m

Divide each term by 7m:
- 63m⁶ ÷ 7m = 9m⁵
- -49m⁵ ÷ 7m = -7m⁴
- -21m ÷ 7m = -3

Answer: 7m(9m⁵ - 7m⁴ - 3)

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Problem 5:

14a + 21a² + 21a³

Coefficients: 14, 21, 21 → GCF is 7
Variables: a, a², a³ → lowest is a → GCF = 7a

Divide:
- 14a ÷ 7a = 2
- 21a² ÷ 7a = 3a
- 21a³ ÷ 7a = 3a²

Write in order: 7a(2 + 3a + 3a²) or better yet, standard form: 7a(3a² + 3a + 2)

Answer: 7a(3a² + 3a + 2)

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Problem 6:

x(x + 2) + 7(x + 2)

Common factor: (x + 2)

Factor it out: → (x + 2)(x + 7)

Answer: (x + 2)(x + 7)

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Problem 7:

3x⁴ - 21x³ + 10x²

Coefficients: 3, 21, 10 → GCF? 3 and 21 share 3, but 10 doesn’t → so no common number factor except 1? Wait — actually, check again: 3, 21, 10 → GCF is 1? But look at variables: x⁴, x³, x² → all have x² → so GCF = x²

Wait — let me double-check coefficients: 3, 21, 10 → yes, no common factor other than 1. So GCF = x²

Divide:
- 3x⁴ ÷ x² = 3x²
- -21x³ ÷ x² = -21x
- 10x² ÷ x² = 10

Answer: x²(3x² - 21x + 10)

*(Note: This trinomial doesn’t factor nicely with integers, so we stop here.)*

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Problem 8:

(3n + 1)(4n + 1) + (n + 2)(4n + 1)

Both terms have (4n + 1) → that’s the GCF.

Factor it out:
→ (4n + 1)[(3n + 1) + (n + 2)]

Simplify inside brackets:
3n + 1 + n + 2 = 4n + 3

Answer: (4n + 1)(4n + 3)

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Problem 9:

50p³ + 50p² - 20

Coefficients: 50, 50, 20 → GCF is 10
No variable in last term → so GCF = 10

Divide:
- 50p³ ÷ 10 = 5p³
- 50p² ÷ 10 = 5p²
- -20 ÷ 10 = -2

Answer: 10(5p³ + 5p² - 2)

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Problem 10:

6x⁴y⁵ - 2x²y³ + 14x³y⁴

Coefficients: 6, 2, 14 → GCF is 2
Variables:
- x⁴, x², x³ → lowest is x²
- y⁵, y³, y⁴ → lowest is y³
→ GCF = 2x²y³

Divide each term:
- 6x⁴y⁵ ÷ 2x²y³ = 3x²y²
- -2x²y³ ÷ 2x²y³ = -1
- 14x³y⁴ ÷ 2x²y³ = 7xy

Answer: 2x²y³(3x²y² - 1 + 7xy) → rearrange inside: 2x²y³(3x²y² + 7xy - 1)

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Final Answer:

1. 8m²(m + 2n)
2. (x + 4)(2x - 3)
3. 2(y - 3b)(2y - 1)
4. 7m(9m⁵ - 7m⁴ - 3)
5. 7a(3a² + 3a + 2)
6. (x + 2)(x + 7)
7. x²(3x² - 21x + 10)
8. (4n + 1)(4n + 3)
9. 10(5p³ + 5p² - 2)
10. 2x²y³(3x²y² + 7xy - 1)