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Let's solve each problem on the "Factor Perfect Square Trinomials" worksheet. We'll go through each polynomial and determine if it is a perfect square trinomial, then factor it accordingly.

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🔍 What is a Perfect Square Trinomial?

A perfect square trinomial is of the form:

$$
(a - b)^2 = a^2 - 2ab + b^2 \quad \text{or} \quad (a + b)^2 = a^2 + 2ab + b^2
$$

So, for a trinomial $ ax^2 + bx + c $ to be a perfect square:
- The first term ($ a^2 $) must be a perfect square.
- The last term ($ c $) must be a perfect square.
- The middle term must be twice the product of the square roots of the first and last terms.

We’ll apply this logic to each problem.

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✔ Problem 1: $ 27b^2 - 144b + 192 $

First, check if it’s a perfect square trinomial.

But notice: 27 is not a perfect square, so let’s try factoring out the GCF first.

GCF of 27, 144, and 192:

- 27 = $ 3^3 $
- 144 = $ 12^2 = 2^4 \cdot 3^2 $
- 192 = $ 64 \cdot 3 = 2^6 \cdot 3 $

So GCF = 3

Factor out 3:

$$
27b^2 - 144b + 192 = 3(9b^2 - 48b + 64)
$$

Now look at $ 9b^2 - 48b + 64 $

Check if this is a perfect square trinomial:

- $ 9b^2 = (3b)^2 $
- $ 64 = 8^2 $
- Middle term should be $ 2 \cdot 3b \cdot 8 = 48b $ → Yes! But we have –48b

So:

$$
9b^2 - 48b + 64 = (3b - 8)^2
$$

✔ So full factorization:

$$
\boxed{3(3b - 8)^2}
$$

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✔ Problem 2: $ 75m^2 - 240m + 192 $

Factor out GCF:

GCF of 75, 240, 192:

- 75 = $ 3 \cdot 5^2 $
- 240 = $ 16 \cdot 15 = 2^4 \cdot 3 \cdot 5 $
- 192 = $ 64 \cdot 3 = 2^6 \cdot 3 $

So GCF = 3

$$
75m^2 - 240m + 192 = 3(25m^2 - 80m + 64)
$$

Now check $ 25m^2 - 80m + 64 $:

- $ 25m^2 = (5m)^2 $
- $ 64 = 8^2 $
- Middle term: $ 2 \cdot 5m \cdot 8 = 80m $ → yes, but we have –80m

So:

$$
25m^2 - 80m + 64 = (5m - 8)^2
$$

✔ Final answer:

$$
\boxed{3(5m - 8)^2}
$$

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✔ Problem 3: $ 9b^2 - 48b + 64 $

This one is similar to #1.

- $ 9b^2 = (3b)^2 $
- $ 64 = 8^2 $
- Middle term: $ 2 \cdot 3b \cdot 8 = 48b $, and we have –48b

So:

$$
9b^2 - 48b + 64 = (3b - 8)^2
$$

✔ Answer:

$$
\boxed{(3b - 8)^2}
$$

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✔ Problem 4: $ 100c^2 - 320c + 256 $

Check for GCF:

- 100, 320, 256 → all divisible by 4?

100 ÷ 4 = 25
320 ÷ 4 = 80
256 ÷ 4 = 64

So GCF = 4

$$
100c^2 - 320c + 256 = 4(25c^2 - 80c + 64)
$$

Now check $ 25c^2 - 80c + 64 $:

- $ 25c^2 = (5c)^2 $
- $ 64 = 8^2 $
- $ 2 \cdot 5c \cdot 8 = 80c $, and we have –80c

So:

$$
25c^2 - 80c + 64 = (5c - 8)^2
$$

✔ Final answer:

$$
\boxed{4(5c - 8)^2}
$$

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✔ Problem 5: $ 50a^2 - 160a + 128 $

Find GCF:

50, 160, 128

- 50 = $ 2 \cdot 5^2 $
- 160 = $ 16 \cdot 10 = 2^5 \cdot 5 $
- 128 = $ 2^7 $

GCF = 2

$$
50a^2 - 160a + 128 = 2(25a^2 - 80a + 64)
$$

Now check $ 25a^2 - 80a + 64 $:

- $ 25a^2 = (5a)^2 $
- $ 64 = 8^2 $
- $ 2 \cdot 5a \cdot 8 = 80a $, and we have –80a

So:

$$
25a^2 - 80a + 64 = (5a - 8)^2
$$

✔ Final answer:

$$
\boxed{2(5a - 8)^2}
$$

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✔ Problem 6: $ 27z^2 - 144z + 192 $

Find GCF:

27, 144, 192

- 27 = $ 3^3 $
- 144 = $ 12^2 = 2^4 \cdot 3^2 $
- 192 = $ 2^6 \cdot 3 $

GCF = 3

$$
27z^2 - 144z + 192 = 3(9z^2 - 48z + 64)
$$

Now check $ 9z^2 - 48z + 64 $:

- $ 9z^2 = (3z)^2 $
- $ 64 = 8^2 $
- $ 2 \cdot 3z \cdot 8 = 48z $, and we have –48z

So:

$$
9z^2 - 48z + 64 = (3z - 8)^2
$$

✔ Final answer:

$$
\boxed{3(3z - 8)^2}
$$

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✔ Problem 7: $ 75k^2 - 240k + 192 $

GCF of 75, 240, 192:

Same as earlier — GCF = 3

$$
75k^2 - 240k + 192 = 3(25k^2 - 80k + 64)
$$

Now check $ 25k^2 - 80k + 64 $:

- $ 25k^2 = (5k)^2 $
- $ 64 = 8^2 $
- $ 2 \cdot 5k \cdot 8 = 80k $, and we have –80k

So:

$$
25k^2 - 80k + 64 = (5k - 8)^2
$$

✔ Final answer:

$$
\boxed{3(5k - 8)^2}
$$

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✔ Problem 8: $ 18k^2 - 96k + 128 $

GCF of 18, 96, 128:

- 18 = $ 2 \cdot 3^2 $
- 96 = $ 2^5 \cdot 3 $
- 128 = $ 2^7 $

GCF = 2

$$
18k^2 - 96k + 128 = 2(9k^2 - 48k + 64)
$$

Now check $ 9k^2 - 48k + 64 $:

- $ 9k^2 = (3k)^2 $
- $ 64 = 8^2 $
- $ 2 \cdot 3k \cdot 8 = 48k $, and we have –48k

So:

$$
9k^2 - 48k + 64 = (3k - 8)^2
$$

✔ Final answer:

$$
\boxed{2(3k - 8)^2}
$$

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✔ Summary of All Answers:

| Problem | Factored Form |
|--------|----------------|
| 1 | $ \boxed{3(3b - 8)^2} $ |
| 2 | $ \boxed{3(5m - 8)^2} $ |
| 3 | $ \boxed{(3b - 8)^2} $ |
| 4 | $ \boxed{4(5c - 8)^2} $ |
| 5 | $ \boxed{2(5a - 8)^2} $ |
| 6 | $ \boxed{3(3z - 8)^2} $ |
| 7 | $ \boxed{3(5k - 8)^2} $ |
| 8 | $ \boxed{2(3k - 8)^2} $ |

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📌 Key Pattern:

All these trinomials are perfect square trinomials after factoring out the GCF. They follow the pattern:

$$
(\text{first term})^2 - 2 \cdot (\text{first}) \cdot (\text{last}) + (\text{last})^2 = (\text{first} - \text{last})^2
$$

And the common structure is:

- Coefficients like 9, 25, 100, etc., which are squares.
- Constant terms like 64, 192, 128 — often multiples of 64.
- Middle term is always $ -2 \times \sqrt{\text{first}} \times \sqrt{\text{last}} $

You can recognize them by checking whether the middle term is twice the product of the square roots of the first and last terms (after factoring out GCF).

Let me know if you’d like a printable version or explanation in another format!