Answer :
Sure! Let's go through the process of synthetic division step-by-step for the given problem.
We are dividing the polynomial [tex]\( P(x) = x^3 + 6x^2 - 7x - 60 \)[/tex] by [tex]\( x + 5 \)[/tex]. In synthetic division, we use the opposite of the root being tested, so we use -5 in this case.
Here’s the step-by-step setup and process:
1. Write down the coefficients of the polynomial: 1 (for [tex]\( x^3 \)[/tex]), 6 (for [tex]\( x^2 \)[/tex]), -7 (for [tex]\( x \)[/tex]), and -60 (the constant term). So, we start with the array [tex]\( [1, 6, -7, -60] \)[/tex].
2. Set up the synthetic division with -5 being the number on the left side. The first number from our coefficients is always brought down unchanged.
```
-5 | 1 6 -7 -60
|
-----------------
1
```
3. Multiply the number you just brought down by -5, and add it to the next coefficient.
- Multiply 1 by -5 and add to 6:
- Result: [tex]\( -5 + 6 = 1 \)[/tex]
```
-5 | 1 6 -7 -60
| -5
-----------------
1 1
```
4. Repeat the process: Multiply the result by -5 and add to the next coefficient:
- Multiply 1 by -5 and add to -7:
- Result: [tex]\( -5 + (-7) = -12 \)[/tex]
```
-5 | 1 6 -7 -60
| -5 -5
-----------------
1 1 -12
```
5. Continue the process with the last coefficient:
- Multiply -12 by -5 and add to -60:
- Result: [tex]\( 60 + (-60) = 0 \)[/tex]
```
-5 | 1 6 -7 -60
| -5 -5 60
-----------------
1 1 -12 0
```
6. Here, the last number 0 is the remainder. Since the remainder is 0, -5 is a root of the polynomial.
So, the numbers that complete the division problem are [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], [tex]\( c = -12 \)[/tex], and [tex]\( d = 0 \)[/tex]. This means our quotient is [tex]\( x^2 + 1x - 12 \)[/tex], with no remainder.
We are dividing the polynomial [tex]\( P(x) = x^3 + 6x^2 - 7x - 60 \)[/tex] by [tex]\( x + 5 \)[/tex]. In synthetic division, we use the opposite of the root being tested, so we use -5 in this case.
Here’s the step-by-step setup and process:
1. Write down the coefficients of the polynomial: 1 (for [tex]\( x^3 \)[/tex]), 6 (for [tex]\( x^2 \)[/tex]), -7 (for [tex]\( x \)[/tex]), and -60 (the constant term). So, we start with the array [tex]\( [1, 6, -7, -60] \)[/tex].
2. Set up the synthetic division with -5 being the number on the left side. The first number from our coefficients is always brought down unchanged.
```
-5 | 1 6 -7 -60
|
-----------------
1
```
3. Multiply the number you just brought down by -5, and add it to the next coefficient.
- Multiply 1 by -5 and add to 6:
- Result: [tex]\( -5 + 6 = 1 \)[/tex]
```
-5 | 1 6 -7 -60
| -5
-----------------
1 1
```
4. Repeat the process: Multiply the result by -5 and add to the next coefficient:
- Multiply 1 by -5 and add to -7:
- Result: [tex]\( -5 + (-7) = -12 \)[/tex]
```
-5 | 1 6 -7 -60
| -5 -5
-----------------
1 1 -12
```
5. Continue the process with the last coefficient:
- Multiply -12 by -5 and add to -60:
- Result: [tex]\( 60 + (-60) = 0 \)[/tex]
```
-5 | 1 6 -7 -60
| -5 -5 60
-----------------
1 1 -12 0
```
6. Here, the last number 0 is the remainder. Since the remainder is 0, -5 is a root of the polynomial.
So, the numbers that complete the division problem are [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], [tex]\( c = -12 \)[/tex], and [tex]\( d = 0 \)[/tex]. This means our quotient is [tex]\( x^2 + 1x - 12 \)[/tex], with no remainder.
