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Let's work through each question step by step.

### 1. **Define zero or root of a polynomial.**
   A **zero** or **root** of a polynomial is a value of \(x\) for which the polynomial evaluates to zero. In other words, if \(f(x)\) is a polynomial, then \(x = r\) is a root if \(f(r) = 0\).

### 2. **If \(x = \frac{1}{2}\) is a zero of the polynomial \(f(x) = 8x^3 + ax^2 - 4x + 2\), find the value of \(a\).**
   Since \(x = \frac{1}{2}\) is a zero, we can substitute \(x = \frac{1}{2}\) into \(f(x)\) and set the result equal to zero:
   \[
   f\left(\frac{1}{2}\right) = 8\left(\frac{1}{2}\right)^3 + a\left(\frac{1}{2}\right)^2 - 4\left(\frac{1}{2}\right) + 2 = 0
   \]
   Solving this equation will give the value of \(a\).

   \[
   8\left(\frac{1}{2}\right)^3 = 8 \times \frac{1}{8} = 1
   \]
   \[
   a\left(\frac{1}{2}\right)^2 = a \times \frac{1}{4} = \frac{a}{4}
   \]
   \[
   -4\left(\frac{1}{2}\right) = -2
   \]
   \[
   f\left(\frac{1}{2}\right) = 1 + \frac{a}{4} - 2 + 2 = 0
   \]
   Simplifying:
   \[
   \frac{a}{4} + 1 = 0 \quad \Rightarrow \quad \frac{a}{4} = -1 \quad \Rightarrow \quad a = -4
   \]
   Therefore, \(a = -4\).

### 3. **Write the remainder when the polynomial \(f(x) = x^3 + x^2 - 3x + 2\) is divided by \(x + 1\).**
   By the **remainder theorem**, the remainder when a polynomial \(f(x)\) is divided by \(x - c\) is \(f(c)\). Here, \(c = -1\). So, we need to evaluate \(f(-1)\):
   \[
   f(-1) = (-1)^3 + (-1)^2 - 3(-1) + 2
   \]
   \[
   f(-1) = -1 + 1 + 3 + 2 = 5
   \]
   Therefore, the remainder is 5.

### 4. **Find the remainder when \(x^3 + 4x^2 + 4x - 3\) is divided by \(x\).**
   The remainder when a polynomial is divided by \(x\) is the constant term. Here, the constant term is \(-3\). Therefore, the remainder is \(-3\).

### 5. **If \(x + 1\) is a factor of \(x^3 + a\), then write the value of \(a\).**
   If \(x + 1\) is a factor, then \(x = -1\) must be a zero of the polynomial \(x^3 + a\). This means:
   \[
   (-1)^3 + a = 0 \quad \Rightarrow \quad -1 + a = 0 \quad \Rightarrow \quad a = 1
   \]
   Therefore, \(a = 1\).

### 6. **If \(f(x) = x^4 - 2x^3 + 3x^2 - ax - b\) when divided by \(x - 1\), the remainder is 6, find the value of \(a + b\).**
   By the remainder theorem, the remainder when \(f(x)\) is divided by \(x - 1\) is \(f(1)\). Given that the remainder is 6:
   \[
   f(1) = 1^4 - 2(1)^3 + 3(1)^2 - a(1) - b = 6
   \]
   Simplifying:
   \[
   1 - 2 + 3 - a - b = 6
   \]
   \[
   2 - a - b = 6 \quad \Rightarrow \quad -a - b = 4 \quad \Rightarrow \quad a + b = -4
   \]
   Therefore, \(a + b = -4\).

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Do you want further details or explanations on any of these? Here are 5 related questions to deepen your understanding:

1. What is the factor theorem, and how does it relate to the remainder theorem?
2. How can synthetic division be used to divide polynomials efficiently?
3. What are the steps to factor higher-degree polynomials using known roots or factors?
4. Can you describe the process of polynomial long division in detail?
5. How do roots of polynomials relate to their graphs?

**Tip:** When dividing a polynomial by a linear factor, always check if you can apply the remainder theorem to simplify your work!

1. Define zero or root of a polynomial. 2. If x = 1/2 is a zero of the polynomial f(x) = 8x^3 + ax^2 - 4x + 2, find the value of a. 3. Write the remainder when the polynomial f(x) = x^3 + x^2 - 3x + 2 is divided by x + 1. 4. Find the remainder when x^3 + 4x^2 + 4x - 3 is divided by x. 5. If x + 1 is a factor of x^3 + a, then write the value of a. 6. If f(x) = x^4 - 2x^3 + 3x^2 - ax - b when divided by x - 1, the remainder is 6, then find the value of a + b.

Solve polynomial problems involving remainder and factor theorems, including finding roots, remainders, and unknown coefficients. Detailed step-by-step solutions suitable for students in Grades 9-12.

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