Question

Find the quotient of the polynomials using synthetic division:
(𝑥
3 − 9) ÷ (𝑥 − 3

Answer 1

To perform synthetic division on the polynomial \( x^3 - 9 \) by \( x - 3 \), we first need to set up our synthetic division.

Here, the dividend is \( x^3 - 9 \). We can express this polynomial as:

\[
x^3 + 0x^2 + 0x - 9
\]

This means we have the coefficients: \( 1, 0, 0, -9 \).

The divisor \( x - 3 \) implies that we will use \( 3 \) for the synthetic division.

Now, we set up the synthetic division using \( 3 \):

```
3 | 1 0 0 -9
| 3 9 27
-------------------
| 1 3 9 18
```

1. Bring down the first coefficient \( 1 \).
2. Multiply \( 3 \) (from \( x - 3 \)) by \( 1 \) (the value you just brought down), and write the result \( 3 \) under the next coefficient.
3. Add \( 0 + 3 = 3 \).
4. Repeat this process: \( 3 \times 3 = 9 \) under the next coefficient \( 0 \), add \( 0 + 9 = 9 \).
5. Finally, \( 3 \times 9 = 27 \) under \( -9 \), add \( -9 + 27 = 18 \).

So, we have:

```
3 | 1 0 0 -9
| 3 9 27
-------------------
| 1 3 9 18
```

The bottom row gives us the coefficients of the quotient and the remainder. The quotient is:

\[
1x^2 + 3x + 9
\]

And the remainder is \( 18 \).

Therefore, when dividing \( x^3 - 9 \) by \( x - 3 \), the result is:

\[
\text{Quotient: } x^2 + 3x + 9, \quad \text{Remainder: } 18
\]

So we can summarize the result of the division:

\[
\frac{x^3 - 9}{x - 3} = x^2 + 3x + 9 + \frac{18}{x - 3}
\]