Answer :
Sure, let's go through the problem step by step.
### Step 1: Verify the divisor form
The divisor provided is [tex]\( y - 3 \)[/tex]. This is already in the form [tex]\( (x - r) \)[/tex] where [tex]\( r = 3 \)[/tex].
So, the answer to the first part is:
Yes, the divisor is in the [tex]\((x-r)\)[/tex] form.
### Step 2: Identify the number of terms in the dividend
The dividend polynomial is [tex]\( -3y^3 + 11y^2 - 6y \)[/tex].
To count the number of terms in the dividend, we look at each term with a non-zero coefficient:
- [tex]\(-3y^3\)[/tex]
- [tex]\(11y^2\)[/tex]
- [tex]\(-6y\)[/tex]
Therefore, there are 3 terms in the dividend.
So, the answer to the second part is:
There are 3 terms in the dividend.
### Step 3: Perform synthetic division
To perform synthetic division, we use the coefficients of the dividend [tex]\(-3y^3 + 11y^2 - 6y\)[/tex] and the root [tex]\( r = 3 \)[/tex] from the divisor [tex]\( y - 3 \)[/tex].
1. Write down the coefficients of the dividend including the constant term (which is 0 in this case):
- Coefficients: [tex]\([-3, 11, -6, 0\)[/tex]]
2. Set up the synthetic division table:
- Initial array: [tex]\([-3]\)[/tex]
3. Perform the synthetic division steps:
- Multiply the first coefficient by [tex]\( r \)[/tex] (3), and add to the next coefficient.
1. [tex]\(-3 \times 3 + 11 = -9 + 11 = 2\)[/tex]
- Synthetic array: [tex]\([-3, 2]\)[/tex]
2. Multiply the result by [tex]\( r \)[/tex] (3) again, and add to the next coefficient.
- [tex]\(2 \times 3 + (-6) = 6 - 6 = 0\)[/tex]
- Synthetic array: [tex]\([-3, 2, 0]\)[/tex]
3. Multiply the result by [tex]\( r \)[/tex] (3) again, and add to the next coefficient.
- [tex]\(0 \times 3 + 0 = 0 + 0 = 0\)[/tex]
- Synthetic array: [tex]\([-3, 2, 0, 0]\)[/tex]
4. Interpret the synthetic array:
- The first part [tex]\([-3, 2, 0]\)[/tex] represents the coefficients of the quotient.
- The last value (0) represents the remainder.
So, the quotient and remainder are:
Quotient: [tex]\([-3, 2, 0]\)[/tex]
Remainder: 0
### Final Answers:
1. Is the divisor given in [tex]\((x-r)\)[/tex] form?
Yes.
2. How many terms are in the dividend?
3 terms.
3. Enter the quotient and remainder:
Quotient: [tex]\([-3, 2, 0]\)[/tex]
Remainder: 0
### Step 1: Verify the divisor form
The divisor provided is [tex]\( y - 3 \)[/tex]. This is already in the form [tex]\( (x - r) \)[/tex] where [tex]\( r = 3 \)[/tex].
So, the answer to the first part is:
Yes, the divisor is in the [tex]\((x-r)\)[/tex] form.
### Step 2: Identify the number of terms in the dividend
The dividend polynomial is [tex]\( -3y^3 + 11y^2 - 6y \)[/tex].
To count the number of terms in the dividend, we look at each term with a non-zero coefficient:
- [tex]\(-3y^3\)[/tex]
- [tex]\(11y^2\)[/tex]
- [tex]\(-6y\)[/tex]
Therefore, there are 3 terms in the dividend.
So, the answer to the second part is:
There are 3 terms in the dividend.
### Step 3: Perform synthetic division
To perform synthetic division, we use the coefficients of the dividend [tex]\(-3y^3 + 11y^2 - 6y\)[/tex] and the root [tex]\( r = 3 \)[/tex] from the divisor [tex]\( y - 3 \)[/tex].
1. Write down the coefficients of the dividend including the constant term (which is 0 in this case):
- Coefficients: [tex]\([-3, 11, -6, 0\)[/tex]]
2. Set up the synthetic division table:
- Initial array: [tex]\([-3]\)[/tex]
3. Perform the synthetic division steps:
- Multiply the first coefficient by [tex]\( r \)[/tex] (3), and add to the next coefficient.
1. [tex]\(-3 \times 3 + 11 = -9 + 11 = 2\)[/tex]
- Synthetic array: [tex]\([-3, 2]\)[/tex]
2. Multiply the result by [tex]\( r \)[/tex] (3) again, and add to the next coefficient.
- [tex]\(2 \times 3 + (-6) = 6 - 6 = 0\)[/tex]
- Synthetic array: [tex]\([-3, 2, 0]\)[/tex]
3. Multiply the result by [tex]\( r \)[/tex] (3) again, and add to the next coefficient.
- [tex]\(0 \times 3 + 0 = 0 + 0 = 0\)[/tex]
- Synthetic array: [tex]\([-3, 2, 0, 0]\)[/tex]
4. Interpret the synthetic array:
- The first part [tex]\([-3, 2, 0]\)[/tex] represents the coefficients of the quotient.
- The last value (0) represents the remainder.
So, the quotient and remainder are:
Quotient: [tex]\([-3, 2, 0]\)[/tex]
Remainder: 0
### Final Answers:
1. Is the divisor given in [tex]\((x-r)\)[/tex] form?
Yes.
2. How many terms are in the dividend?
3 terms.
3. Enter the quotient and remainder:
Quotient: [tex]\([-3, 2, 0]\)[/tex]
Remainder: 0