Find the polynomial in xof degree three that has a value zero when x = -1 and x = 2, a value 8

when x = 0, and a remainder 16

when divided by 3x + 2.

To find the polynomial \( P(x) \) of degree 3 that meets the given conditions, we can follow these steps:

### Step 1: Use the given roots
The polynomial has roots at \( x = -1 \) and \( x = 2 \). Therefore, \( P(x) \) can be expressed as:
\[
P(x) = a(x + 1)(x - 2)(x - r)
\]
where \( a \) is a constant and \( r \) is the remaining root to be determined.

### Step 2: Use the given value when \( x = 0 \)
We are also given that \( P(0) = 8 \). Substituting \( x = 0 \) in the polynomial, we get:
\[
P(0) = a(0 + 1)(0 - 2)(0 - r) = a(-2)(-r) = 2ar
\]
Setting \( 2ar = 8 \) gives us:
\[
ar = 4
\]

### Step 3: Use the remainder when divided by \( 3x + 2 \)
The polynomial leaves a remainder of 16 when divided by \( 3x + 2 \). This means that:
\[
P\left(-\frac{2}{3}\right) = 16
\]
Substituting \( x = -\frac{2}{3} \) in the polynomial, we get:
\[
P\left(-\frac{2}{3}\right) = a\left(-\frac{2}{3} + 1\right)\left(-\frac{2}{3} - 2\right)\left(-\frac{2}{3} - r\right)
\]
Simplifying the expression, we find:
\[
P\left(-\frac{2}{3}\right) = a\left(-\frac{2}{3} + \frac{3}{3}\right)\left(-\frac{2}{3} - \frac{6}{3}\right)\left(-\frac{2}{3} - r\right) = a\left(\frac{1}{3}\right)\left(-\frac{8}{3}\right)\left(-\frac{2 + 3r}{3}\right)
\]
We know \( P\left(-\frac{2}{3}\right) = 16 \), so:
\[
a\left(\frac{1}{3}\right)\left(-\frac{8}{3}\right)\left(-\frac{2 + 3r}{3}\right) = 16
\]
This simplifies to:
\[
\frac{8a(2 + 3r)}{27} = 16
\]
\[
8a(2 + 3r) = 432
\]
\[
a(2 + 3r) = 54
\]

### Step 4: Solve for \( r \) and \( a \)
We now have two equations:
1. \( ar = 4 \)
2. \( a(2 + 3r) = 54 \)

Solving these equations simultaneously:
- From equation 1: \( a = \frac{4}{r} \)
- Substitute into equation 2: 
\[
\frac{4}{r}(2 + 3r) = 54
\]
\[
8 + 12r = 54r
\]
\[
46r = 8
\]
\[
r = \frac{4}{23}
\]
Substitute back to find \( a \):
\[
a = \frac{4}{\frac{4}{23}} = 23
\]

### Final Polynomial
The polynomial is then:
\[
P(x) = 23(x + 1)\left(x - 2\right)\left(x - \frac{4}{23}\right)
\]
This can be expanded if needed. 

If you have any further questions or need more details, feel free to ask!

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**Related Questions:**

1. What are the steps to expand the polynomial \(P(x)\) fully?
2. How do we determine the roots of the polynomial after it is expanded?
3. Can the process be simplified if we have more roots or other conditions?
4. What would the polynomial look like if it had a different degree, say 4?
5. How would you factorize a polynomial given similar conditions?

**Tip:** When solving for polynomial roots, checking the remainder with synthetic division can be a quick way to verify your results.

Learn how to find a polynomial of degree three that satisfies given conditions: roots at x = -1 and x = 2, P(0) = 8, and remainder 16 when divided by 3x + 2. Detailed steps include polynomial expansion and application of the remainder theorem.

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