If ax³+bx²+x-6has x+ 2 as factor and leaves a remainder when divided by x-2 find a and b

Given the polynomial \( p(x) = ax^3 + bx^2 + x - 6 \), we know the following:

1. \( x + 2 \) is a factor of the polynomial, meaning that when \( p(x) \) is divided by \( x + 2 \), the remainder is zero. This implies that \( p(-2) = 0 \).
2. The polynomial leaves a remainder \( r \) when divided by \( x - 2 \). This implies that \( p(2) = r \).

We need to find the values of \( a \) and \( b \).

### Step 1: Use the fact that \( x + 2 \) is a factor

Since \( x + 2 \) is a factor, we substitute \( x = -2 \) into the polynomial:

\[
p(-2) = a(-2)^3 + b(-2)^2 + (-2) - 6 = 0
\]

Simplifying:

\[
p(-2) = -8a + 4b - 2 - 6 = 0
\]

\[
-8a + 4b - 8 = 0
\]

\[
-8a + 4b = 8
\]

Dividing the entire equation by 4:

\[
-2a + b = 2 \quad \text{(Equation 1)}
\]

### Step 2: Use the remainder when divided by \( x - 2 \)

We are also given that when divided by \( x - 2 \), the polynomial leaves a remainder \( r \), so \( p(2) = r \). We substitute \( x = 2 \) into the polynomial:

\[
p(2) = a(2)^3 + b(2)^2 + 2 - 6 = r
\]

Simplifying:

\[
p(2) = 8a + 4b + 2 - 6 = r
\]

\[
p(2) = 8a + 4b - 4 = r \quad \text{(Equation 2)}
\]

### Step 3: Solve for \( a \) and \( b \)

From Equation 1:

\[
b = 2a + 2
\]

Substitute \( b = 2a + 2 \) into Equation 2:

\[
8a + 4(2a + 2) - 4 = r
\]

Expand and simplify:

\[
8a + 8a + 8 - 4 = r
\]

\[
16a + 4 = r
\]

Thus, the remainder \( r \) when divided by \( x - 2 \) is:

\[
r = 16a + 4
\]

### Final Answer:
- The value of \( b \) in terms of \( a \) is \( b = 2a + 2 \).
- The remainder when divided by \( x - 2 \) is \( r = 16a + 4 \).

Would you like further details or any clarifications?

### Follow-up Questions:
1. What does it mean when a polynomial has a specific factor like \( x + 2 \)?
2. How can you verify that the values of \( a \) and \( b \) are correct?
3. What would the polynomial look like if \( r = 0 \) when divided by \( x - 2 \)?
4. How does the Remainder Theorem apply to this problem?
5. What happens if you use synthetic division instead of substitution to find the remainder?

### Tip:
Always substitute the root of the factor into the polynomial to simplify calculations when using the Factor Theorem.

Explore how to find coefficients a and b in the polynomial problem where x+2 is a factor and a remainder is given when divided by x-2. This detailed solution covers polynomial factorization, application of the Factor Theorem, and calculating remainders. Suitable for advanced high school students.

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