Answer :
To perform synthetic division with the polynomial and potential root given, let's go through the steps together:
1. Write down the coefficients of the polynomial.
The polynomial is [tex]\(1x^3 + 6x^2 - 7x - 60\)[/tex], which gives us the coefficients: 1, 6, -7, -60.
2. Set up the synthetic division.
The potential root we are testing is -5. So, we place -5 to the left, and the coefficients (1, 6, -7, -60) in a row like this:
```
-5 | 1 6 -7 -60
|
```
3. Bring down the first coefficient.
The first coefficient (1) is brought down unchanged:
```
-5 | 1 6 -7 -60
|
1
```
4. Perform the synthetic division process.
- Step 1: Multiply the root (-5) by the number you just brought down (1), and write the result under the second coefficient (6):
[tex]\(-5 \times 1 = -5\)[/tex].
- Add this result to the second coefficient (6):
[tex]\(6 + (-5) = 1\)[/tex].
Now the setup looks like this:
```
-5 | 1 6 -7 -60
| -5
------------
1 1
```
- Step 2: Repeat the process: multiply -5 by 1, and write the result under the next coefficient (-7):
[tex]\(-5 \times 1 = -5\)[/tex].
- Add this to -7:
[tex]\(-7 + (-5) = -12\)[/tex].
Here's the updated setup:
```
-5 | 1 6 -7 -60
| -5 -5
------------
1 1 -12
```
- Step 3: Once more, multiply -5 by -12, and write the result under the last coefficient (-60):
[tex]\(-5 \times -12 = 60\)[/tex].
- Add this to -60:
[tex]\(-60 + 60 = 0\)[/tex].
The final setup is:
```
-5 | 1 6 -7 -60
| -5 -5 60
------------
1 1 -12 0
```
5. Interpret the result.
The bottom row gives us the coefficients of the quotient polynomial as well as the remainder. So, the synthetic division yields a quotient polynomial with coefficients 1, 1, and -12, and a remainder of 0.
This result means that -5 is indeed a root of the polynomial, with the quotient being [tex]\(1x^2 + 1x - 12\)[/tex].
1. Write down the coefficients of the polynomial.
The polynomial is [tex]\(1x^3 + 6x^2 - 7x - 60\)[/tex], which gives us the coefficients: 1, 6, -7, -60.
2. Set up the synthetic division.
The potential root we are testing is -5. So, we place -5 to the left, and the coefficients (1, 6, -7, -60) in a row like this:
```
-5 | 1 6 -7 -60
|
```
3. Bring down the first coefficient.
The first coefficient (1) is brought down unchanged:
```
-5 | 1 6 -7 -60
|
1
```
4. Perform the synthetic division process.
- Step 1: Multiply the root (-5) by the number you just brought down (1), and write the result under the second coefficient (6):
[tex]\(-5 \times 1 = -5\)[/tex].
- Add this result to the second coefficient (6):
[tex]\(6 + (-5) = 1\)[/tex].
Now the setup looks like this:
```
-5 | 1 6 -7 -60
| -5
------------
1 1
```
- Step 2: Repeat the process: multiply -5 by 1, and write the result under the next coefficient (-7):
[tex]\(-5 \times 1 = -5\)[/tex].
- Add this to -7:
[tex]\(-7 + (-5) = -12\)[/tex].
Here's the updated setup:
```
-5 | 1 6 -7 -60
| -5 -5
------------
1 1 -12
```
- Step 3: Once more, multiply -5 by -12, and write the result under the last coefficient (-60):
[tex]\(-5 \times -12 = 60\)[/tex].
- Add this to -60:
[tex]\(-60 + 60 = 0\)[/tex].
The final setup is:
```
-5 | 1 6 -7 -60
| -5 -5 60
------------
1 1 -12 0
```
5. Interpret the result.
The bottom row gives us the coefficients of the quotient polynomial as well as the remainder. So, the synthetic division yields a quotient polynomial with coefficients 1, 1, and -12, and a remainder of 0.
This result means that -5 is indeed a root of the polynomial, with the quotient being [tex]\(1x^2 + 1x - 12\)[/tex].
