Sure, let's solve the given problem using polynomial long division.
We need to divide [tex]\( x^2 + 5x - 1 \)[/tex] by [tex]\( x - 1 \)[/tex].
### Step 1: Setup the division
The dividend is [tex]\( x^2 + 5x - 1 \)[/tex] and the divisor is [tex]\( x - 1 \)[/tex].
### Step 2: Divide the first term
Divide the first term of the dividend, [tex]\( x^2 \)[/tex], by the first term of the divisor, [tex]\( x \)[/tex]:
[tex]\[ \frac{x^2}{x} = x \][/tex]
This gives us the first term of the quotient: [tex]\( x \)[/tex].
### Step 3: Multiply and subtract
Multiply the entire divisor [tex]\( x - 1 \)[/tex] by the first term of the quotient [tex]\( x \)[/tex] and subtract from the dividend:
[tex]\[ (x - 1) \cdot x = x^2 - x \][/tex]
Write this result under the corresponding terms of the dividend and subtract:
[tex]\[
\begin{array}{r}
x^2 + 5x - 1 \\
-(x^2 - x) \\
\hline
6x - 1
\end{array}
\][/tex]
### Step 4: Repeat with the new polynomial [tex]\( 6x - 1 \)[/tex]
Divide the first term of the new polynomial [tex]\( 6x \)[/tex] by the first term of the divisor [tex]\( x \)[/tex]:
[tex]\[ \frac{6x}{x} = 6 \][/tex]
This gives us the next term of the quotient: [tex]\( 6 \)[/tex].
### Step 5: Multiply and subtract again
Multiply the entire divisor [tex]\( x - 1 \)[/tex] by the new term of the quotient [tex]\( 6 \)[/tex] and subtract from the new polynomial:
[tex]\[ (x - 1) \cdot 6 = 6x - 6 \][/tex]
Write this result under the corresponding terms of the new polynomial and subtract:
[tex]\[
\begin{array}{r}
6x - 1 \\
-(6x - 6) \\
\hline
5
\end{array}
\][/tex]
Since [tex]\( 5 \)[/tex] (a constant) is of lower degree than [tex]\( x - 1 \)[/tex], it is the remainder.
### Result
Thus, the quotient is [tex]\( x + 6 \)[/tex]. There's a remainder of [tex]\( 5 \)[/tex], but as the problem specifies not to include the remainder in our final answer, we conclude:
[tex]\[ \boxed{x + 6} \][/tex]
