Cube Root Word Problems Form - Fill Out and Sign Printable PDF - Free Printable

It looks like you've uploaded a worksheet titled "Prime Factorization Method" that asks students to find the square root of numbers using prime factorization. However, I can't view or access images directly. But based on your description and the text visible in the image (which you've partially transcribed), I can help solve the problems listed.

Let me go through the problems one by one and explain how to solve them using prime factorization to find the square root.

---

🔍 Problem 1: √144

We need to find the square root of 144 using prime factorization.

#### Step 1: Prime factorize 144
Divide 144 by the smallest prime number (2):

- 144 ÷ 2 = 72
- 72 ÷ 2 = 36
- 36 ÷ 2 = 18
- 18 ÷ 2 = 9
- 9 ÷ 3 = 3
- 3 ÷ 3 = 1

So,
144 = 2 × 2 × 2 × 2 × 3 × 3

Group the primes into pairs:

- (2 × 2) × (2 × 2) × (3 × 3)

Now take one from each pair:

- 2 × 2 × 3 = 12

✔ So, √144 = 12

---

🔍 Problem 2: √484

Factorize 484:

- 484 ÷ 2 = 242
- 242 ÷ 2 = 121
- 121 ÷ 11 = 11
- 11 ÷ 11 = 1

So,
484 = 2 × 2 × 11 × 11

Group into pairs:

- (2 × 2) × (11 × 11)

Take one from each pair:

- 2 × 11 = 22

✔ So, √484 = 22

---

🔍 Problem 3: √729

Factorize 729:

- 729 is odd → not divisible by 2
Try dividing by 3:

- 729 ÷ 3 = 243
- 243 ÷ 3 = 81
- 81 ÷ 3 = 27
- 27 ÷ 3 = 9
- 9 ÷ 3 = 3
- 3 ÷ 3 = 1

So,
729 = 3 × 3 × 3 × 3 × 3 × 3 = 3⁶

Group into pairs:

- (3×3) × (3×3) × (3×3)

Take one from each pair:

- 3 × 3 × 3 = 27

✔ So, √729 = 27

---

🔍 Problem 4: √3744

Let’s factor 3744:

Start with 2:

- 3744 ÷ 2 = 1872
- 1872 ÷ 2 = 936
- 936 ÷ 2 = 468
- 468 ÷ 2 = 234
- 234 ÷ 2 = 117
- 117 ÷ 3 = 39
- 39 ÷ 3 = 13
- 13 ÷ 13 = 1

So,
3744 = 2⁵ × 3² × 13

Now group into pairs:

- (2×2) × (2×2) × 2 × (3×3) × 13

We have:
- Two pairs of 2s → 2×2 = 4
- One pair of 3s → 3
- Leftover: 2 and 13 → unpaired

Since there are leftover factors (2 and 13), 3744 is not a perfect square, so its square root is not an integer.

But we can still write it as:

√3744 = √(2⁴ × 3² × 2 × 13) = 2² × 3 × √(2×13) = 4×3×√26 = 12√26

✔ So, √3744 = 12√26 (irrational)

---

🔍 Problem 5: √10035

Check if 10035 is a perfect square.

First, factor it:

- It's odd → not divisible by 2
- Sum of digits: 1+0+0+3+5 = 9 → divisible by 3

- 10035 ÷ 3 = 3345
- 3345 ÷ 3 = 1115
- 1115 ÷ 5 = 223
- 223 is a prime number (check divisibility: not divisible by 2,3,5,7,11,13 → yes, prime)

So,
10035 = 3² × 5 × 223

We have:
- One pair of 3s → √3² = 3
- Leftover: 5 and 223 → not paired

So,
√10035 = 3 × √(5×223) = 3√1115

✔ Not a perfect square → √10035 = 3√1115

---

🔍 Problem 6: √1936

Factor 1936:

- Even → divide by 2:
- 1936 ÷ 2 = 968
- 968 ÷ 2 = 484
- 484 ÷ 2 = 242
- 242 ÷ 2 = 121
- 121 ÷ 11 = 11
- 11 ÷ 11 = 1

So,
1936 = 2⁴ × 11²

Group into pairs:

- (2×2) × (2×2) × (11×11)

Take one from each pair:

- 2 × 2 × 11 = 44

✔ So, √1936 = 44

---

🔍 Problem 7: √10000

This one is easy.

10000 = 100 × 100 = (10²)² = 10⁴

So,
√10000 = 10² = 100

Or via prime factorization:

- 10000 = 10⁴ = (2×5)⁴ = 2⁴ × 5⁴

Group into pairs:

- (2×2) × (2×2) × (5×5) × (5×5)

Take one from each pair:

- 2×2×5×5 = 4×25 = 100

✔ So, √10000 = 100

---

🔍 Problem 8: √242

Factor 242:

- Even → ÷2 = 121
- 121 = 11×11

So,
242 = 2 × 11²

Group:

- (11×11) × 2

Only one pair → √242 = 11√2

✔ So, √242 = 11√2

---

🔍 Problem 9: √600

Factor 600:

- 600 ÷ 2 = 300
- 300 ÷ 2 = 150
- 150 ÷ 2 = 75
- 75 ÷ 3 = 25
- 25 ÷ 5 = 5
- 5 ÷ 5 = 1

So,
600 = 2³ × 3 × 5²

Group:

- (2×2) × 2 × 3 × (5×5)

Pairs: 2² and 5² → take one from each: 2×5 = 10

Leftover: 2 and 3 → √(2×3) = √6

So, √600 = 10√6

✔ So, √600 = 10√6

---

## ✔ Final Answers:

| Number | Square Root |
|--------|-------------|
| 1. √144 | 12 |
| 2. √484 | 22 |
| 3. √729 | 27 |
| 4. √3744 | 12√26 |
| 5. √10035 | 3√1115 |
| 6. √1936 | 44 |
| 7. √10000 | 100 |
| 8. √242 | 11√2 |
| 9. √600 | 10√6 |

---

📌 Summary of Steps for Prime Factorization Method:

1. Factor the number completely into primes.
2. Group the prime factors into pairs.
3. Take one factor from each pair and multiply them.
4. If all primes are paired → result is a perfect square.
5. If some primes are left unpaired → include them under a square root.

Let me know if you'd like this explained visually or want a printable version!