Divide 3x^3 + 3x^2 + 2x – 2 by x + 3 using long division

Answer 1

To divide \(3x^3 + 3x^2 + 2x - 2\) by \(x + 3\) using long division, follow these steps:

First, set up the division in long division format:

```
_______
x + 3 │3x^3 + 3x^2 + 2x - 2
```

Next, divide the first term of the dividend (\(3x^3\)) by the first term of the divisor (\(x\)), which gives \(3x^2\). Write this quotient above the division symbol on the top.

```
3x^2
_______
x + 3 │3x^3 + 3x^2 + 2x - 2
```

Multiply the divisor (\(x + 3\)) by the quotient (\(3x^2\)), which gives \(3x^3 + 9x^2\). Write this product beneath the dividend, and subtract it from the dividend.

```
3x^2
_______
x + 3 │3x^3 + 3x^2 + 2x - 2
- (3x^3 + 9x^2)
```

The subtraction gives \(-6x^2 + 2x - 2\). Bring down the next term, which is \(2x\), and rewrite it below the subtraction result.

```
3x^2
_______
x + 3 │3x^3 + 3x^2 + 2x - 2
- (3x^3 + 9x^2)
- 6x^2 + 2x - 2
```

Repeat the process. Divide the first term of the new result (\(-6x^2\)) by the first term of the divisor (\(x\)), which gives \(-6x\). Write this quotient below the previous quotient.

```
3x^2 - 6x
_______
x + 3 │3x^3 + 3x^2 + 2x - 2
- (3x^3 + 9x^2)
- 6x^2 + 2x - 2
```

Multiply the divisor (\(x + 3\)) by the new quotient (\(-6x\)), which gives \(-6x^2 - 18x\). Write this product beneath the previous subtraction result, and subtract it.

```
3x^2 - 6x
_______
x + 3 │3x^3 + 3x^2 + 2x - 2
- (3x^3 + 9x^2)
- 6x^2 + 2x - 2
- (-6x^2 - 18x)
```

The subtraction gives \(20x - 2\). Bring down the next term, which is \(-2\), and rewrite it below the subtraction result.

```
3x^2 - 6x
_______
x + 3 │3x^3 + 3x^2 + 2x - 2
- (3x^3 + 9x^2)
- 6x^2 + 2x - 2
- (-6x^2 - 18x)
20x - 2
```

Repeat the process. Divide the first term of the new result (\(20x\)) by the first term of the divisor (\(x\)), which gives \(20\). Write this quotient below the previous quotient.

```
3x^2 - 6x + 20
_______
x + 3 │3x^3 + 3x^2 + 2x - 2
- (3x^3 + 9x^2)
- 6x^2 + 2x - 2
- (-6x^2 - 18x)
20x - 2
```

Multiply the divisor (\(x + 3\)) by the new quotient (\(20\)), which gives \(20x + 60\). Write this product beneath the previous subtraction result, and subtract it.

```
3x^2 - 6x + 20
_______
x + 3 │3x^3 + 3x^2 + 2x - 2
- (3x^3 + 9x^2)
- 6x^2 + 2x - 2
- (-6x^2 - 18x)
20x - 2
- (20x + 60)
```

The subtraction gives \(-62\). This is the remainder.

Therefore, the division of \(3x^3 + 3x^2 + 2x - 2\) by \(x + 3\) using long division is:

\[3x^2 - 6x + 20 - \frac{62}{x + 3}\]