To factorize the quadratic expression \( 2y^2 - 11y + 5 \), we can use the following steps:

### Step 1: Multiply the coefficient of \( y^2 \) by the constant term
Multiply the coefficient of \( y^2 \) (which is 2) by the constant term (which is 5):
\[
2 \times 5 = 10
\]

### Step 2: Find two numbers that multiply to 10 and add to -11
We need to find two numbers that multiply to 10 and add to -11. These numbers are -10 and -1, because:
\[
-10 \times -1 = 10 \quad \text{and} \quad -10 + (-1) = -11
\]

### Step 3: Rewrite the middle term using these two numbers
Rewrite the middle term \(-11y\) as \(-10y - y\):
\[
2y^2 - 10y - y + 5
\]

### Step 4: Factor by grouping
Group the terms in pairs and factor each group:
\[
(2y^2 - 10y) - (y - 5)
\]
Factor out the common factors from each group:
\[
2y(y - 5) - 1(y - 5)
\]

### Step 5: Factor out the common binomial
Since \( y - 5 \) is a common factor, factor it out:
\[
(2y - 1)(y - 5)
\]

So, the factorized form of \( 2y^2 - 11y + 5 \) is:
\[
\boxed{(2y - 1)(y - 5)}
\]

Would you like any more details or have any questions? Here are some related questions:

1. How do you solve quadratic equations by factoring?
2. What if the middle term cannot be easily split into factors?
3. Can the quadratic formula be used to factor quadratic expressions?
4. How can you verify if the factorization is correct?
5. What is the significance of the discriminant in factorization?

**Tip:** Always check your factorization by expanding the factors to ensure you get back the original expression.

Learn how to factorize the quadratic expression 2y²-11y+5 step-by-step with detailed explanations. This problem involves quadratic equations, polynomial factorization, and application of the quadratic formula. Suitable for high school students.

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