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Divide the following by long division method and check the result:

Let's begin by solving part (a) from the image using the long division method.

We are given:

### (a) Divide \( 6a^2 - a - 5 \) by \( 2a + 1 \)

**Step 1: Set up the division.**

We divide the first term of the dividend \( 6a^2 \) by the first term of the divisor \( 2a \):
\[
\frac{6a^2}{2a} = 3a
\]
So, the first term in the quotient is \( 3a \).

**Step 2: Multiply the divisor by \( 3a \).**
\[
(2a + 1) \times 3a = 6a^2 + 3a
\]

**Step 3: Subtract this from the original polynomial.**
\[
(6a^2 - a - 5) - (6a^2 + 3a) = (-a - 5) - (3a) = -4a - 5
\]

**Step 4: Divide the first term of the new polynomial by the first term of the divisor.**
\[
\frac{-4a}{2a} = -2
\]
So, the next term in the quotient is \( -2 \).

**Step 5: Multiply the divisor by \( -2 \).**
\[
(2a + 1) \times (-2) = -4a - 2
\]

**Step 6: Subtract this from the previous result.**
\[
(-4a - 5) - (-4a - 2) = -5 + 2 = -3
\]

### Final result:
The quotient is \( 3a - 2 \) and the remainder is \( -3 \).

So, we can write:
\[
\frac{6a^2 - a - 5}{2a + 1} = 3a - 2 - \frac{3}{2a + 1}
\]

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Would you like to proceed with the next question or dive deeper into this solution? Here are some related questions:

1. Can you explain the process of synthetic division and when it’s useful?
2. How can you use the Remainder Theorem to verify the result of polynomial division?
3. What are the differences between long division and synthetic division for polynomials?
4. How can polynomial division help solve polynomial equations?
5. What is the role of the remainder in polynomial division?

### Tip:
When dividing polynomials, always arrange both the dividend and the divisor in descending order of their degrees to avoid confusion.

Divide the following by long division method and check the result: (6a^2 - a - 5) by (2a + 1).

Learn how to divide the polynomial (6a^2 - a - 5) by (2a + 1) using the long division method. This detailed, step-by-step solution covers polynomial division concepts and the remainder theorem. Suitable for students in Grades 9-11.

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