Factor the trinomial. $$ w^{2}-18 w+80 $$

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To factor the trinomial $$ w^{2} - 18w + 80 $$, we need to find two numbers that satisfy the following conditions:

1. **Product:** The two numbers must multiply to $$ +80 $$.
2. **Sum:** The two numbers must add up to $$ -18 $$.

Let's consider the possible pairs of factors for 80:

- $$ 1 \times 80 = 80 $$ (Sum: $$ 1 + 80 = 81 $$)
- $$ 2 \times 40 = 80 $$ (Sum: $$ 2 + 40 = 42 $$)
- $$ 4 \times 20 = 80 $$ (Sum: $$ 4 + 20 = 24 $$)
- $$ 5 \times 16 = 80 $$ (Sum: $$ 5 + 16 = 21 $$)
- $$ 8 \times 10 = 80 $$ (Sum: $$ 8 + 10 = 18 $$)

The pair $$ 8 $$ and $$ 10 $$ adds up to $$ 18 $$. Since the middle term in the trinomial is $$ -18w $$, both numbers must be negative to satisfy the sum condition:

- $$ (-8) \times (-10) = 80 $$
- $$ (-8) + (-10) = -18 $$

Therefore, the factored form of the trinomial is:

$$
(w - 8)(w - 10)
$$

**Answer:** $$\boxed{\, (w - 8)(w - 10)\,}$$
The trinomial $$ w^{2} - 18w + 80 $$ factors to $$ (w - 8)(w - 10) $$.

Factor the trinomial.

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