Answer :
To perform synthetic division, we start by representing the polynomial [tex]\(8x^3 + x^2 - 5\)[/tex]. We then use the root of the divisor [tex]\(x - 6 = 0\)[/tex], which is [tex]\(x = 6\)[/tex].
To facilitate the division:
1. Write the coefficients of the polynomial [tex]\(8x^3 + x^2 + 0x - 5\)[/tex]. Notice that there is no [tex]\(x\)[/tex] term, so we include a coefficient of [tex]\(0\)[/tex] for completeness. Our list of coefficients is [tex]\([8, 1, 0, -5]\)[/tex].
2. Set up the synthetic division table. Write the root, which is [tex]\(6\)[/tex], to the left and the coefficients to the right:
```
6 | 8 1 0 -5
|
```
3. Bring down the first coefficient directly:
```
6 | 8 1 0 -5
|
| 8
```
4. Now, multiply the root ([tex]\(6\)[/tex]) by the value just written below the line ([tex]\(8\)[/tex]), and write the result under the next coefficient:
```
6 | 8 1 0 -5
| 48
| 8
```
5. Add the result to the next coefficient:
```
6 | 8 1 0 -5
| 48 294
| 8 49
```
6. Repeat steps 4-5 for the remaining coefficients:
[tex]\[ \begin{array}{c|cccc} 6 & 8 & 1 & 0 & -5 \\ \hline & 8 & 49 & 294 & 0 \\ & & 48 & 294 & 0 \\ & & & & 1759 \\ \end{array} \][/tex]
7. The results below the line are the coefficients of the quotient [tex]\([8, 49, 294]\)[/tex], and the last number is the remainder [tex]\(1759\)[/tex].
So, the quotient is [tex]\(8x^2 + 49x + 294\)[/tex] and the remainder is [tex]\(1759\)[/tex].
Therefore, synthetic division of [tex]\((8x^3 + x^2 - 5) \div (x - 6)\)[/tex] gives the quotient [tex]\([8, 49, 294]\)[/tex] and the remainder [tex]\(1759\)[/tex].
To facilitate the division:
1. Write the coefficients of the polynomial [tex]\(8x^3 + x^2 + 0x - 5\)[/tex]. Notice that there is no [tex]\(x\)[/tex] term, so we include a coefficient of [tex]\(0\)[/tex] for completeness. Our list of coefficients is [tex]\([8, 1, 0, -5]\)[/tex].
2. Set up the synthetic division table. Write the root, which is [tex]\(6\)[/tex], to the left and the coefficients to the right:
```
6 | 8 1 0 -5
|
```
3. Bring down the first coefficient directly:
```
6 | 8 1 0 -5
|
| 8
```
4. Now, multiply the root ([tex]\(6\)[/tex]) by the value just written below the line ([tex]\(8\)[/tex]), and write the result under the next coefficient:
```
6 | 8 1 0 -5
| 48
| 8
```
5. Add the result to the next coefficient:
```
6 | 8 1 0 -5
| 48 294
| 8 49
```
6. Repeat steps 4-5 for the remaining coefficients:
[tex]\[ \begin{array}{c|cccc} 6 & 8 & 1 & 0 & -5 \\ \hline & 8 & 49 & 294 & 0 \\ & & 48 & 294 & 0 \\ & & & & 1759 \\ \end{array} \][/tex]
7. The results below the line are the coefficients of the quotient [tex]\([8, 49, 294]\)[/tex], and the last number is the remainder [tex]\(1759\)[/tex].
So, the quotient is [tex]\(8x^2 + 49x + 294\)[/tex] and the remainder is [tex]\(1759\)[/tex].
Therefore, synthetic division of [tex]\((8x^3 + x^2 - 5) \div (x - 6)\)[/tex] gives the quotient [tex]\([8, 49, 294]\)[/tex] and the remainder [tex]\(1759\)[/tex].