Sure, let's work through the division of the polynomial step-by-step. We need to divide:
[tex]\[
\frac{-14 z^2 + 17 + 4 z^3}{-1 + 2 z}
\][/tex]
### Step 1: Rewrite the division in a more familiar form
This can be expressed as:
[tex]\[
\frac{4z^3 - 14z^2 + 17}{2z - 1}
\][/tex]
### Step 2: Perform polynomial long division
We will divide the terms one by one. We want to see how many times the denominator [tex]\( (2z - 1) \)[/tex] fits into each term of the numerator.
1. First Division:
- Divide the leading term [tex]\( 4z^3 \)[/tex] by [tex]\( 2z \)[/tex]:
[tex]\[
\frac{4z^3}{2z} = 2z^2
\][/tex]
- Multiply [tex]\( 2z^2 \)[/tex] by [tex]\( 2z - 1 \)[/tex] and subtract from the original polynomial:
[tex]\[
(2z^2) \cdot (2z - 1) = 4z^3 - 2z^2
\][/tex]
Subtract to get the new polynomial:
[tex]\[
(4z^3 - 14z^2 + 17) - (4z^3 - 2z^2) = -12z^2 + 17
\][/tex]
2. Second Division:
- Divide the leading term [tex]\( -12z^2 \)[/tex] by [tex]\( 2z \)[/tex]:
[tex]\[
\frac{-12z^2}{2z} = -6z
\][/tex]
- Multiply [tex]\( -6z \)[/tex] by [tex]\( 2z - 1 \)[/tex] and subtract:
[tex]\[
(-6z) \cdot (2z - 1) = -12z^2 + 6z
\][/tex]
Subtract to get the new polynomial:
[tex]\[
(-12z^2 + 17) - (-12z^2 + 6z) = -6z + 17
\][/tex]
3. Third Division:
- Divide the leading term [tex]\( -6z \)[/tex] by [tex]\( 2z \)[/tex]:
[tex]\[
\frac{-6z}{2z} = -3
\][/tex]
- Multiply [tex]\( -3 \)[/tex] by [tex]\( 2z - 1 \)[/tex] and subtract:
[tex]\[
(-3) \cdot (2z - 1) = -6z + 3
\][/tex]
Subtract to get the new polynomial:
[tex]\[
(-6z + 17) - (-6z + 3) = 14
\][/tex]
### Step 3: Combine Quotients and Remainder
The result of the division of [tex]\( 4z^3 - 14z^2 + 17 \)[/tex] by [tex]\( 2z - 1 \)[/tex] is:
[tex]\[
2z^2 - 6z - 3 + \frac{14}{2z - 1}
\][/tex]
Therefore, we can express:
[tex]\[
\frac{4z^3 - 14z^2 + 17}{2z - 1} = 2z^2 - 6z - 3 + \frac{14}{2z - 1}
\][/tex]
The numerator and the denominator terms are accurate, and the fraction part is the final remainder over the original divisor. The solution is indeed:
[tex]\[
\boxed{\frac{4z^3 - 14z^2 + 17}{2z - 1}}
\][/tex]
