Comprehensive algebra worksheet featuring practice problems for factoring expressions using the Greatest Common Factor (GCF).

Let’s solve each problem step by step. We’re factoring using the Greatest Common Factor (GCF) — that means we find the biggest number or variable that divides evenly into all terms, then pull it out front.

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Section A

1) 8x + 24
GCF of 8 and 24 is 8.
→ 8(x + 3)

2) 15 + 25y
GCF of 15 and 25 is 5.
→ 5(3 + 5y)

3) 32 – 40w
GCF of 32 and 40 is 8.
→ 8(4 – 5w)

4) 18c – 36
GCF of 18 and 36 is 18.
→ 18(c – 2)

5) 16d² – 4d
GCF of 16 and 4 is 4; both have at least one d → GCF = 4d
→ 4d(4d – 1)

6) 12s + 60s²
GCF of 12 and 60 is 12; both have at least one s → GCF = 12s
→ 12s(1 + 5s)

7) 21xy + 14x
GCF of 21 and 14 is 7; both have x → GCF = 7x
→ 7x(3y + 2)

8) 27ab – 18a²
GCF of 27 and 18 is 9; both have a → GCF = 9a
→ 9a(3b – 2a)

9) 12s²t + 28s
GCF of 12 and 28 is 4; both have s → GCF = 4s
→ 4s(3st + 7)

10) 72wz + 45w²z
GCF of 72 and 45 is 9; both have w and z → GCF = 9wz
→ 9wz(8 + 5w)

11) 22x²y – 55xy²
GCF of 22 and 55 is 11; both have x and y → GCF = 11xy
→ 11xy(2x – 5y)

12) 16k³ + 24k²
GCF of 16 and 24 is 8; both have k² → GCF = 8k²
→ 8k²(2k + 3)

13) 9h²g – 15h³
GCF of 9 and 15 is 3; both have h² → GCF = 3h²
→ 3h²(3g – 5h)

14) 12c²d² + 20c³
GCF of 12 and 20 is 4; both have c² → GCF = 4c²
→ 4c²(3d² + 5c)

15) 28a³b² – 7a²b
GCF of 28 and 7 is 7; both have a² and b → GCF = 7a²b
→ 7a²b(4ab – 1)

16) 60x²y³ – 35xy²
GCF of 60 and 35 is 5; both have x and y² → GCF = 5xy²
→ 5xy²(12xy – 7)

17) 88s⁴t + 56s³t²
GCF of 88 and 56 is 8; both have s³ and t → GCF = 8s³t
→ 8s³t(11s + 7t)

18) 36p³q⁴ – 48p⁴q²
GCF of 36 and 48 is 12; both have p³ and q² → GCF = 12p³q²
→ 12p³q²(3q² – 4p)

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Section B

1) 6 – 12gh + 3h
GCF of 6, 12, and 3 is 3. No common variables in all terms.
→ 3(2 – 4gh + h)

2) 21st – 7t + 14
GCF of 21, 7, and 14 is 7. Only first two terms have t, last doesn’t → no variable factor.
→ 7(3st – t + 2)

3) 22 – 44vw + 11v
GCF of 22, 44, and 11 is 11. Not all terms have v → no variable factor.
→ 11(2 – 4vw + v)

4) 4ab + 2b – abc
GCF of 4, 2, and 1? Wait — coefficients: 4, 2, 1 → GCF is 1? But wait — all terms have b!
Actually: 4ab, 2b, -abc → all have b. Coefficients: 4, 2, 1 → GCF is 1? But 4 and 2 share 2, but 1 doesn’t. So only common factor is b? Let’s check:

Wait — actually, 4ab, 2b, -abc → yes, all have b. But coefficient GCF is 1? That can’t be right for factoring fully.

Hold on — let’s recheck: 4, 2, and 1 → GCF is 1. But maybe I made a mistake. Actually, 4ab + 2b – abc = b(4a + 2 – ac). Yes! Because every term has b. Even though coefficients don’t share a big number, they still share b.

So: GCF = b
→ b(4a + 2 – ac)

But wait — can we factor more? 4a + 2 – ac → no common factor among those three terms. So yes.

Alternatively, sometimes people write it as b(-ac + 4a + 2), but order doesn’t matter.

Actually, let me double-check: Is there a numerical GCF? 4, 2, 1 → no, so just b.

But hold on — 4ab + 2b – abc → factor out b: b(4a + 2 – ac). Correct.

5) 5suv – 10sv + 15su
All terms have s. Coefficients: 5, 10, 15 → GCF = 5. Also, not all have u or v → so GCF = 5s
→ 5s(uv – 2v + 3u)

6) 16xy + 24y – 8xyz
Coefficients: 16, 24, 8 → GCF = 8. All terms have y. Not all have x or z → GCF = 8y
→ 8y(2x + 3 – xz)

7) 9wu – 27wuv + 45w
Coefficients: 9, 27, 45 → GCF = 9. All terms have w. Not all have u or v → GCF = 9w
→ 9w(u – 3uv + 5)

Wait — let’s check: 9wu ÷ 9w = u; -27wuv ÷ 9w = -3uv; 45w ÷ 9w = 5 → yes.

8) 24gh – 12g + 15h
Coefficients: 24, 12, 15 → GCF = 3. Do all terms share a variable? First has g,h; second has g; third has h → no common variable. So GCF = 3
→ 3(8gh – 4g + 5h)

9) 132pqr – 96qr + 108pqrs
Coefficients: 132, 96, 108 → let’s find GCF.

132 = 12×11, 96=12×8, 108=12×9 → GCF at least 12. Can we go higher? 132÷12=11, 96÷12=8, 108÷12=9 → no common factor beyond 12? Actually, 132,96,108 divisible by 12? Yes. Divisible by 24? 132÷24=5.5 → no. So GCF = 12.

Variables: all have q and r? First: p,q,r; second: q,r; third: p,q,r,s → so common variables: q and r → GCF = 12qr

Check: 132pqr ÷ 12qr = 11p; -96qr ÷ 12qr = -8; 108pqrs ÷ 12qr = 9ps

→ 12qr(11p – 8 + 9ps)

We can rearrange inside: 12qr(11p + 9ps – 8)

10) 2x + xy – x²
All terms have x. Coefficients: 2,1,1 → GCF = 1 for numbers, but x is common → GCF = x
→ x(2 + y – x)

11) 5k² – 10jk + k
Coefficients: 5,10,1 → GCF = 1? But all terms have k → GCF = k
→ k(5k – 10j + 1)

12) 9cd – 3c²d + 12c
Coefficients: 9,3,12 → GCF = 3. Variables: all have c? First: c,d; second: c²,d; third: c → so common variable is c. d not in all → GCF = 3c
→ 3c(3d – cd + 4)

Wait: 9cd ÷ 3c = 3d; -3c²d ÷ 3c = -cd; 12c ÷ 3c = 4 → yes.

13) 7xyz + xy² – x²y
Coefficients: 7,1,1 → GCF = 1. Variables: all have x and y? First: x,y,z; second: x,y²; third: x²,y → so common: x and y → GCF = xy
→ xy(7z + y – x)

14) e²f – 5e³f² + e²
Coefficients: 1,5,1 → GCF = 1. Variables: all have e²? First: e²,f; second: e³,f²; third: e² → so common: e². f not in all → GCF = e²
→ e²(f – 5ef² + 1)

15) 8st²u – 32s²t + 64st
Coefficients: 8,32,64 → GCF = 8. Variables: all have s and t? First: s,t²,u; second: s²,t; third: s,t → so common: s and t → GCF = 8st
→ 8st(tu – 4s + 8)

Check: 8st²u ÷ 8st = tu; -32s²t ÷ 8st = -4s; 64st ÷ 8st = 8 → yes.

16) 12g³h – 9g²h² + 18g²h
Coefficients: 12,9,18 → GCF = 3. Variables: all have g² and h? First: g³,h; second: g²,h²; third: g²,h → so common: g² and h → GCF = 3g²h
→ 3g²h(4g – 3h + 6)

17) (1/2)ab + (3/4)a² – a
This has fractions. To factor, find GCF of coefficients: 1/2, 3/4, 1. Better to factor out the smallest fraction or convert to common denominator.

Note: all terms have a. Coefficients: 1/2, 3/4, 1. The GCF of fractions: think of them as multiples. Factor out a, and also factor out 1/4? Let’s see:

Write as: a [ (1/2)b + (3/4)a – 1 ]

To make it nicer, factor out 1/4? But 1/2 = 2/4, 3/4=3/4, 1=4/4 → so inside: (2/4)b + (3/4)a – 4/4 = (1/4)(2b + 3a – 4)

So overall: a * (1/4)(2b + 3a – 4) = (a/4)(2b + 3a – 4)

But usually we prefer integer coefficients inside. Alternatively, factor out a/4:

Original: (1/2)ab + (3/4)a² – a = a [ (1/2)b + (3/4)a – 1 ]

Multiply inside by 4/4: = a/4 * [2b + 3a – 4]

Yes → (a/4)(3a + 2b – 4)

But some might leave it as a( (1/2)b + (3/4)a - 1 ), but "factor fully" usually means no fractions inside if possible.

Standard way: factor out the greatest common factor including fractions. Since 1/4 is a common factor of the coefficients when written with denominator 4.

Coefficients: 1/2 = 2/4, 3/4, 1 = 4/4 → GCF of numerators 2,3,4 is 1, so GCF is 1/4? But also variable a.

So GCF = (1/4)a? Let's test:

(1/4)a * ? = (1/2)ab → ? = 2b

(1/4)a * ? = (3/4)a² → ? = 3a

(1/4)a * ? = -a → ? = -4

Yes! So → (1/4)a (2b + 3a - 4)

We can write as \frac{a}{4}(3a + 2b - 4)

18) (3/4)x⁴y – x²y³ + (1/2)x³y²

Again, fractions. Coefficients: 3/4, 1, 1/2. Write with denominator 4: 3/4, 4/4, 2/4 → GCF of numerators 3,4,2 is 1, so GCF coefficient is 1/4.

Variables: all have x² and y? First: x⁴,y; second: x²,y³; third: x³,y² → so min powers: x², y¹ → GCF = (1/4)x²y

Check:

(1/4)x²y * ? = (3/4)x⁴y → ? = 3x²

(1/4)x²y * ? = -x²y³ → ? = -4y²

(1/4)x²y * ? = (1/2)x³y² → ? = 2xy

So → (1/4)x²y (3x² - 4y² + 2xy)

Or rearrange: (1/4)x²y (3x² + 2xy - 4y²)

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Now, compiling all answers neatly.

Final Answer:

Section A:
1) 8(x + 3)
2) 5(3 + 5y)
3) 8(4 - 5w)
4) 18(c - 2)
5) 4d(4d - 1)
6) 12s(1 + 5s)
7) 7x(3y + 2)
8) 9a(3b - 2a)
9) 4s(3st + 7)
10) 9wz(8 + 5w)
11) 11xy(2x - 5y)
12) 8k²(2k + 3)
13) 3h²(3g - 5h)
14) 4c²(3d² + 5c)
15) 7a²b(4ab - 1)
16) 5xy²(12xy - 7)
17) 8s³t(11s + 7t)
18) 12p³q²(3q² - 4p)

Section B:
1) 3(2 - 4gh + h)
2) 7(3st - t + 2)
3) 11(2 - 4vw + v)
4) b(4a + 2 - ac)
5) 5s(uv - 2v + 3u)
6) 8y(2x + 3 - xz)
7) 9w(u - 3uv + 5)
8) 3(8gh - 4g + 5h)
9) 12qr(11p + 9ps - 8)
10) x(2 + y - x)
11) k(5k - 10j + 1)
12) 3c(3d - cd + 4)
13) xy(7z + y - x)
14) e²(f - 5ef² + 1)
15) 8st(tu - 4s + 8)
16) 3g²h(4g - 3h + 6)
17) \frac{a}{4}(3a + 2b - 4)
18) \frac{1}{4}x^2y(3x^2 + 2xy - 4y^2)