Question

using long division determine which of the following polynomials divides evenlt forna diviser of x+3
A. 3x^2-6x+9
B. 5x^2+7x-12
C. X^3 -3x^2 -13x+15
D. X^3-5x^2+10x-15

Answer 1

To determine which of the given polynomials divides evenly for the divisor \(x + 3\), we can use the Remainder Theorem. According to the Remainder Theorem, if a polynomial \(f(x)\) is divided by \(x + c\), then the remainder of that division is \(f(-c)\). If the remainder is zero, then \(x + c\) divides \(f(x)\) evenly.

Since our divisor is \(x + 3\), we will evaluate each polynomial by substituting \(x = -3\) and checking if the value is zero.

1. **For \(A = 3x^2 - 6x + 9\)**:
\[
A(-3) = 3(-3)^2 - 6(-3) + 9 = 3(9) + 18 + 9 = 27 + 18 + 9 = 54 \quad (\text{not divisible})
\]

2. **For \(B = 5x^2 + 7x - 12\)**:
\[
B(-3) = 5(-3)^2 + 7(-3) - 12 = 5(9) - 21 - 12 = 45 - 21 - 12 = 12 \quad (\text{not divisible})
\]

3. **For \(C = x^3 - 3x^2 - 13x + 15\)**:
\[
C(-3) = (-3)^3 - 3(-3)^2 - 13(-3) + 15 = -27 - 3(9) + 39 + 15 = -27 - 27 + 39 + 15 = 0 \quad (\text{divisible})
\]

4. **For \(D = x^3 - 5x^2 + 10x - 15\)**:
\[
D(-3) = (-3)^3 - 5(-3)^2 + 10(-3) - 15 = -27 - 5(9) - 30 - 15 = -27 - 45 - 30 - 15 = -117 \quad (\text{not divisible})
\]

Based on these calculations, the polynomial that divides evenly by \(x + 3\) is:
\[
\boxed{C \, (x^3 - 3x^2 - 13x + 15)}
\]

Answer 2

Are you sure

Answer 3

Let's re-evaluate each polynomial by substituting \(x = -3\) carefully to ensure accuracy, as we want to find any mistakes in my previous calculations.

### 1. For \(A = 3x^2 - 6x + 9\):
\[
A(-3) = 3(-3)^2 - 6(-3) + 9 = 3(9) + 18 + 9 = 27 + 18 + 9 = 54 \quad (\text{not divisible})
\]

### 2. For \(B = 5x^2 + 7x - 12\):
\[
B(-3) = 5(-3)^2 + 7(-3) - 12 = 5(9) - 21 - 12 = 45 - 21 - 12 = 12 \quad (\text{not divisible})
\]

### 3. For \(C = x^3 - 3x^2 - 13x + 15\):
\[
C(-3) = (-3)^3 - 3(-3)^2 - 13(-3) + 15
\]
\[
= -27 - 3(9) + 39 + 15
\]
\[
= -27 - 27 + 39 + 15
\]
\[
= -54 + 39 + 15
\]
\[
= -54 + 54 = 0 \quad (\text{divisible})
\]

### 4. For \(D = x^3 - 5x^2 + 10x - 15\):
\[
D(-3) = (-3)^3 - 5(-3)^2 + 10(-3) - 15
\]
\[
= -27 - 5(9) - 30 - 15
\]
\[
= -27 - 45 - 30 - 15
\]
\[
= -27 - 45 = -72 - 30 = -102 - 15 = -117 \quad (\text{not divisible})
\]

### Summary of Results:
- **For A**: \(A(-3) = 54\) (not divisible)
- **For B**: \(B(-3) = 12\) (not divisible)
- **For C**: \(C(-3) = 0\) (divisible)
- **For D**: \(D(-3) = -117\) (not divisible)

The previous calculations were indeed correct. The polynomial that divides evenly by \(x + 3\) is:

\[
\boxed{C \, (x^3 - 3x^2 - 13x + 15)}
\]

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