Answer :
Sure, let's solve the given problem step-by-step using long division.
We need to divide the polynomial [tex]\(-3x^5 - 22x^4 - 13x^3 + 39x^2 + 14x - 6\)[/tex] by the polynomial [tex]\(x^3 + 6x^2 - 3x - 5\)[/tex].
### Step 1: Set up the division
The dividend is [tex]\(-3x^5 - 22x^4 - 13x^3 + 39x^2 + 14x - 6\)[/tex] and the divisor is [tex]\(x^3 + 6x^2 - 3x - 5\)[/tex].
We'll write it in the long division format:
```
________________
x^3 + 6x^2 - 3x - 5 | -3x^5 - 22x^4 - 13x^3 + 39x^2 + 14x - 6
```
### Step 2: Divide the leading terms
First, we divide the leading term of the dividend by the leading term of the divisor:
[tex]\[
\frac{ -3x^5 }{ x^3 } = -3x^2
\][/tex]
### Step 3: Multiply and subtract
Now, we multiply [tex]\( -3x^2 \)[/tex] by the entire divisor:
[tex]\[
-3x^2 \cdot (x^3 + 6x^2 - 3x - 5) = -3x^5 - 18x^4 + 9x^3 + 15x^2
\][/tex]
We subtract this from the dividend:
[tex]\[
\begin{align*}
(-3x^5 - 22x^4 - 13x^3 + 39x^2 + 14x - 6) \\
- (-3x^5 - 18x^4 + 9x^3 + 15x^2) \\
= 0x^5 - 4x^4 - 22x^3 + 24x^2 + 14x - 6
\end{align*}
\][/tex]
### Step 4: Repeat the process
Next, we divide the new leading term by the leading term of the divisor:
[tex]\[
\frac{-4x^4 }{ x^3 } = -4x
\][/tex]
Multiply and subtract:
[tex]\[
\begin{align*}
-4x \cdot (x^3 + 6x^2 - 3x - 5) = -4x^4 - 24x^3 + 12x^2 + 20x \\
0x^5 - 4x^4 - 22x^3 + 24x^2 + 14x - 6 \\
- ( -4x^4 - 24x^3 + 12x^2 + 20x) \\
= 0x^4 + 2x^3 + 12x^2 - 6x - 6
\end{align*}
\][/tex]
### Step 5: Continue the process
Next, we divide:
[tex]\[
\frac{2x^3}{x^3} = 2
\][/tex]
Multiply and subtract:
[tex]\[
\begin{align*}
2 \cdot (x^3 + 6x^2 - 3x - 5) = 2x^3 + 12x^2 - 6x - 10 \\
0x^4 + 2x^3 + 12x^2 - 6x - 6 \\
- ( 2x^3 + 12x^2 - 6x - 10) \\
= 0x^3 + 0x^2 + 0x + 4
\end{align*}
\][/tex]
### Step 6: Write the final quotient and remainder
The division process is complete, and the quotient is:
[tex]\[
-3x^2 - 4x + 2
\][/tex]
And the remainder is:
[tex]\[
4
\][/tex]
Therefore, the final answer is:
[tex]\[
-3x^2 - 4x + 2 \quad \text{with a remainder of} \quad 4
\][/tex]
So, [tex]\(\frac{-3x^5 - 22x^4 - 13x^3 + 39x^2 + 14x - 6}{x^3 + 6x^2 - 3x - 5} = -3x^2 - 4x + 2 + \frac{4}{x^3 + 6x^2 - 3x - 5}\)[/tex].
We need to divide the polynomial [tex]\(-3x^5 - 22x^4 - 13x^3 + 39x^2 + 14x - 6\)[/tex] by the polynomial [tex]\(x^3 + 6x^2 - 3x - 5\)[/tex].
### Step 1: Set up the division
The dividend is [tex]\(-3x^5 - 22x^4 - 13x^3 + 39x^2 + 14x - 6\)[/tex] and the divisor is [tex]\(x^3 + 6x^2 - 3x - 5\)[/tex].
We'll write it in the long division format:
```
________________
x^3 + 6x^2 - 3x - 5 | -3x^5 - 22x^4 - 13x^3 + 39x^2 + 14x - 6
```
### Step 2: Divide the leading terms
First, we divide the leading term of the dividend by the leading term of the divisor:
[tex]\[
\frac{ -3x^5 }{ x^3 } = -3x^2
\][/tex]
### Step 3: Multiply and subtract
Now, we multiply [tex]\( -3x^2 \)[/tex] by the entire divisor:
[tex]\[
-3x^2 \cdot (x^3 + 6x^2 - 3x - 5) = -3x^5 - 18x^4 + 9x^3 + 15x^2
\][/tex]
We subtract this from the dividend:
[tex]\[
\begin{align*}
(-3x^5 - 22x^4 - 13x^3 + 39x^2 + 14x - 6) \\
- (-3x^5 - 18x^4 + 9x^3 + 15x^2) \\
= 0x^5 - 4x^4 - 22x^3 + 24x^2 + 14x - 6
\end{align*}
\][/tex]
### Step 4: Repeat the process
Next, we divide the new leading term by the leading term of the divisor:
[tex]\[
\frac{-4x^4 }{ x^3 } = -4x
\][/tex]
Multiply and subtract:
[tex]\[
\begin{align*}
-4x \cdot (x^3 + 6x^2 - 3x - 5) = -4x^4 - 24x^3 + 12x^2 + 20x \\
0x^5 - 4x^4 - 22x^3 + 24x^2 + 14x - 6 \\
- ( -4x^4 - 24x^3 + 12x^2 + 20x) \\
= 0x^4 + 2x^3 + 12x^2 - 6x - 6
\end{align*}
\][/tex]
### Step 5: Continue the process
Next, we divide:
[tex]\[
\frac{2x^3}{x^3} = 2
\][/tex]
Multiply and subtract:
[tex]\[
\begin{align*}
2 \cdot (x^3 + 6x^2 - 3x - 5) = 2x^3 + 12x^2 - 6x - 10 \\
0x^4 + 2x^3 + 12x^2 - 6x - 6 \\
- ( 2x^3 + 12x^2 - 6x - 10) \\
= 0x^3 + 0x^2 + 0x + 4
\end{align*}
\][/tex]
### Step 6: Write the final quotient and remainder
The division process is complete, and the quotient is:
[tex]\[
-3x^2 - 4x + 2
\][/tex]
And the remainder is:
[tex]\[
4
\][/tex]
Therefore, the final answer is:
[tex]\[
-3x^2 - 4x + 2 \quad \text{with a remainder of} \quad 4
\][/tex]
So, [tex]\(\frac{-3x^5 - 22x^4 - 13x^3 + 39x^2 + 14x - 6}{x^3 + 6x^2 - 3x - 5} = -3x^2 - 4x + 2 + \frac{4}{x^3 + 6x^2 - 3x - 5}\)[/tex].
