Answer :
Let's find the quotient and remainder for the division of the polynomial [tex]\(-2x^3 + 9x^2 - 5x + 4\)[/tex] by [tex]\(x-4\)[/tex] using synthetic division.
Given:
- The polynomial: [tex]\(-2x^3 + 9x^2 - 5x + 4\)[/tex]
- The divisor: [tex]\(x - 4\)[/tex]
Step-by-step solution:
### Step A: Synthetic Division Table
- We will use [tex]\(4\)[/tex] (since [tex]\(x - 4 = 0 \implies x = 4\)[/tex]).
1. Setup the table:
The coefficients of the polynomial are [tex]\([-2, 9, -5, 4]\)[/tex]. Write these on the first row.
[tex]\[ \begin{array}{rrrrr} 4 & | & -2 & 9 & -5 & 4 \\ & | & & & & \\ \end{array} \][/tex]
2. Perform the synthetic division:
- Begin by bringing down the first coefficient. This is assigned to the lower table (the first part of the quotient).
[tex]\[ \begin{array}{rrrrr} 4 & | & -2 & 9 & -5 & 4 \\ & | & -2 & & & \\ \end{array} \][/tex]
- Multiply the divisor [tex]\(4\)[/tex] with the value just written below the line [tex]\(-2\)[/tex] (first quotient part), and write the result under the next coefficient.
[tex]\[ 4 \times -2 = -8 \][/tex]
[tex]\[ \begin{array}{rrrrr} 4 & | & -2 & 9 & -5 & 4 \\ & | & -2 & -8 & & \\ \end{array} \][/tex]
- Add the result to the next coefficient [tex]\(9\)[/tex]:
[tex]\[ 9 + (-8) = 1 \][/tex]
- Repeat this process: Multiply, write under next coefficient, then sum:
[tex]\[ 4 \times 1 = 4 \][/tex]
[tex]\[ \begin{array}{rrrrr} 4 & | & -2 & 9 & -5 & 4 \\ & | & -2 & -8 & 1 & 4 \\ \end{array} \][/tex]
[tex]\[ -5 + 4 = -1 \][/tex]
[tex]\[ 4 \times -1 = -4 \][/tex]
[tex]\[ \begin{array}{rrrrr} 4 & | & -2 & 9 & -5 & 4 \\ & | & -2 & -8 & 1 & 4 \\ & | & & & -1 & -4 \end{array} \][/tex]
The last entry under our coefficients should be:
[tex]\[ 4 + (-4) = 0 \][/tex]
[tex]\[ \begin{array}{rrrrr} 4 & | & -2 & 9 & -5 & 4 \\ & | & -2 & -8 & 1 & 4 \\ & | & & & -1 & 0 \end{array} \][/tex]
So, our completed synthetic division table ends up being:
```
4) -2 9 -5 4
-8 4 -4
------------
-2 1 -1 0
```
### (a) Completed synthetic division table:
[tex]\[ \begin{array}{llll} 4 & | & -2 & 9 & -5 & 4 \\ & | & -8 & 4 & -4 \\ & | -2 & 1 & -1 & 0 \\ \end{array} \][/tex]
### Step B: Quotient and Remainder Form
From the last row of our division table:
- Quotient coefficients: [tex]\([-2, 1, -1]\)[/tex]
- Remainder: [tex]\(0\)[/tex]
Therefore, the quotient is [tex]\(-2x^2 + x - 1\)[/tex] and the remainder is [tex]\(0\)[/tex].
### (b) Write answer in the given form:
[tex]\[ \frac{-2 x^3 + 9 x^2 - 5 x + 4}{x-4} = -2x^2 + x - 1 + \frac{0}{x-4} \][/tex]
Since the remainder is 0, we don't need to include [tex]\(\frac{0}{x-4}\)[/tex] term:
[tex]\[ \frac{-2 x^3 + 9 x^2 - 5 x + 4}{x - 4} = -2x^2 + x - 1 \][/tex]
Given:
- The polynomial: [tex]\(-2x^3 + 9x^2 - 5x + 4\)[/tex]
- The divisor: [tex]\(x - 4\)[/tex]
Step-by-step solution:
### Step A: Synthetic Division Table
- We will use [tex]\(4\)[/tex] (since [tex]\(x - 4 = 0 \implies x = 4\)[/tex]).
1. Setup the table:
The coefficients of the polynomial are [tex]\([-2, 9, -5, 4]\)[/tex]. Write these on the first row.
[tex]\[ \begin{array}{rrrrr} 4 & | & -2 & 9 & -5 & 4 \\ & | & & & & \\ \end{array} \][/tex]
2. Perform the synthetic division:
- Begin by bringing down the first coefficient. This is assigned to the lower table (the first part of the quotient).
[tex]\[ \begin{array}{rrrrr} 4 & | & -2 & 9 & -5 & 4 \\ & | & -2 & & & \\ \end{array} \][/tex]
- Multiply the divisor [tex]\(4\)[/tex] with the value just written below the line [tex]\(-2\)[/tex] (first quotient part), and write the result under the next coefficient.
[tex]\[ 4 \times -2 = -8 \][/tex]
[tex]\[ \begin{array}{rrrrr} 4 & | & -2 & 9 & -5 & 4 \\ & | & -2 & -8 & & \\ \end{array} \][/tex]
- Add the result to the next coefficient [tex]\(9\)[/tex]:
[tex]\[ 9 + (-8) = 1 \][/tex]
- Repeat this process: Multiply, write under next coefficient, then sum:
[tex]\[ 4 \times 1 = 4 \][/tex]
[tex]\[ \begin{array}{rrrrr} 4 & | & -2 & 9 & -5 & 4 \\ & | & -2 & -8 & 1 & 4 \\ \end{array} \][/tex]
[tex]\[ -5 + 4 = -1 \][/tex]
[tex]\[ 4 \times -1 = -4 \][/tex]
[tex]\[ \begin{array}{rrrrr} 4 & | & -2 & 9 & -5 & 4 \\ & | & -2 & -8 & 1 & 4 \\ & | & & & -1 & -4 \end{array} \][/tex]
The last entry under our coefficients should be:
[tex]\[ 4 + (-4) = 0 \][/tex]
[tex]\[ \begin{array}{rrrrr} 4 & | & -2 & 9 & -5 & 4 \\ & | & -2 & -8 & 1 & 4 \\ & | & & & -1 & 0 \end{array} \][/tex]
So, our completed synthetic division table ends up being:
```
4) -2 9 -5 4
-8 4 -4
------------
-2 1 -1 0
```
### (a) Completed synthetic division table:
[tex]\[ \begin{array}{llll} 4 & | & -2 & 9 & -5 & 4 \\ & | & -8 & 4 & -4 \\ & | -2 & 1 & -1 & 0 \\ \end{array} \][/tex]
### Step B: Quotient and Remainder Form
From the last row of our division table:
- Quotient coefficients: [tex]\([-2, 1, -1]\)[/tex]
- Remainder: [tex]\(0\)[/tex]
Therefore, the quotient is [tex]\(-2x^2 + x - 1\)[/tex] and the remainder is [tex]\(0\)[/tex].
### (b) Write answer in the given form:
[tex]\[ \frac{-2 x^3 + 9 x^2 - 5 x + 4}{x-4} = -2x^2 + x - 1 + \frac{0}{x-4} \][/tex]
Since the remainder is 0, we don't need to include [tex]\(\frac{0}{x-4}\)[/tex] term:
[tex]\[ \frac{-2 x^3 + 9 x^2 - 5 x + 4}{x - 4} = -2x^2 + x - 1 \][/tex]
