Question

Use long division to divide the polynomials: x3−x2+4/x+2 .
x2−3x+6+6x+2
x2+3x+6−8x+2
x2−2x+6−6x+2
x2−3x+6−8x+2

Answer 1

To perform long division on the polynomials \( \frac{x^3 - x^2 + 4}{x + 2} \), we follow these steps:

1. **Setup the long division**: Write \( x^3 - x^2 + 4 \) under the long division symbol and \( x + 2 \) outside.

2. **Divide the leading terms**: Divide the leading term of the dividend \( x^3 \) by the leading term of the divisor \( x \):
\[
\frac{x^3}{x} = x^2
\]

3. **Multiply and subtract**: Multiply \( x^2 \) by the entire divisor \( x + 2 \):
\[
x^2(x + 2) = x^3 + 2x^2
\]
Now subtract this result from the original polynomial:
\[
(x^3 - x^2 + 4) - (x^3 + 2x^2) = -x^2 - 2x^2 + 4 = -3x^2 + 4
\]

4. **Bring down the next term (if any)**: In this case, we already have \( -3x^2 + 4 \).

5. **Repeat the process**:
- Now divide the leading term \( -3x^2 \) by \( x \):
\[
\frac{-3x^2}{x} = -3x
\]
- Multiply \( -3x \) by the entire divisor \( x + 2 \):
\[
-3x(x + 2) = -3x^2 - 6x
\]
- Subtract this from \( -3x^2 + 4 \):
\[
(-3x^2 + 4) - (-3x^2 - 6x) = 4 + 6x = 6x + 4
\]

6. **Repeat again**:
- Now divide the leading term \( 6x \) by \( x \):
\[
\frac{6x}{x} = 6
\]
- Multiply \( 6 \) by the entire divisor \( x + 2 \):
\[
6(x + 2) = 6x + 12
\]
- Subtract:
\[
(6x + 4) - (6x + 12) = 4 - 12 = -8
\]

7. **Final result**: We have completed the division. The result is:
\[
x^2 - 3x + 6 \quad \text{with a remainder of } -8
\]
Therefore,
\[
\frac{x^3 - x^2 + 4}{x + 2} = x^2 - 3x + 6 - \frac{8}{x + 2}
\]

So, the answer is:
\[
x^2 - 3x + 6 - \frac{8}{x + 2}
\]