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The given problem is:

\[
\lim_{{z \to 8}} \frac{2z^2 - 17z + 8}{8 - z}
\]

Let's attempt to solve this limit step by step.

### Step 1: Direct Substitution

If we directly substitute \( z = 8 \) into the expression:

\[
\frac{2(8)^2 - 17(8) + 8}{8 - 8} = \frac{2(64) - 136 + 8}{0} = \frac{128 - 136 + 8}{0} = \frac{0}{0}
\]

This results in an indeterminate form \(\frac{0}{0}\), so we need to simplify the expression further using algebraic techniques like factoring.

### Step 2: Factor the numerator

We factor the quadratic expression \( 2z^2 - 17z + 8 \). To factor it, we look for two numbers that multiply to \( 2 \times 8 = 16 \) and add to \( -17 \).

These numbers are \( -16 \) and \( -1 \). So, we can rewrite the middle term as:

\[
2z^2 - 16z - z + 8
\]

Now, factor by grouping:

\[
2z(z - 8) - 1(z - 8)
\]

This gives:

\[
(2z - 1)(z - 8)
\]

### Step 3: Simplify the expression

Now, we substitute the factored form back into the original limit expression:

\[
\lim_{{z \to 8}} \frac{(2z - 1)(z - 8)}{8 - z}
\]

Notice that \( 8 - z = -(z - 8) \), so we can rewrite the denominator:

\[
\lim_{{z \to 8}} \frac{(2z - 1)(z - 8)}{-(z - 8)}
\]

Cancel the \( (z - 8) \) terms:

\[
\lim_{{z \to 8}} -(2z - 1)
\]

### Step 4: Evaluate the limit

Now, substitute \( z = 8 \) into the simplified expression:

\[
-(2(8) - 1) = -(16 - 1) = -15
\]

Thus, the value of the limit is:

\[
\boxed{-15}
\]

Would you like further details on this solution, or do you have any questions?

### 5 Related Questions:
1. How do you factor quadratic expressions in general?
2. What are some common strategies for resolving indeterminate forms in limits?
3. Can L'Hopital's Rule be used in this case to solve the limit?
4. What is the difference between factoring and using synthetic division for polynomial division?
5. How do you recognize when to cancel terms in limits?

### Tip:
When dealing with limits that result in a \( \frac{0}{0} \) form, factoring the numerator or denominator is a useful technique to simplify the expression.

Calculate \lim_{{z \to 8}} \frac{2z^2 - 17z + 8}{8 - z}

Learn how to solve the limit \lim_{{z \to 8}} \frac{2z^2 - 17z + 8}{8 - z} using factoring techniques and simplifying indeterminate forms. This step-by-step guide is suitable for students in Grades 10-12.

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