Dividing Polynomials with Long and Synthetic Division: Practice Problems Video - Free Printable

Problem Analysis:

The image shows a polynomial long division problem. The dividend is \( 6y^3 + 7y^2 - 5y + 5 \), and the divisor is \( 3y - 1 \). The quotient is partially shown as \( 2y^2 + 3y \), and the remainder is \( -2y + 5 \).

We are tasked with understanding and explaining the steps of this polynomial long division.

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Step-by-Step Solution:

#### 1. Setup the Division:
We are dividing \( 6y^3 + 7y^2 - 5y + 5 \) by \( 3y - 1 \).

#### 2. Divide the Leading Terms:
- The leading term of the dividend is \( 6y^3 \).
- The leading term of the divisor is \( 3y \).
- Divide \( 6y^3 \) by \( 3y \):
\[
\frac{6y^3}{3y} = 2y^2
\]
So, the first term of the quotient is \( 2y^2 \).

#### 3. Multiply and Subtract:
- Multiply the entire divisor \( 3y - 1 \) by \( 2y^2 \):
\[
2y^2 \cdot (3y - 1) = 6y^3 - 2y^2
\]
- Subtract this result from the original dividend:
\[
(6y^3 + 7y^2 - 5y + 5) - (6y^3 - 2y^2) = 9y^2 - 5y + 5
\]

#### 4. Repeat the Process:
- The new dividend is \( 9y^2 - 5y + 5 \).
- Divide the leading term \( 9y^2 \) by the leading term of the divisor \( 3y \):
\[
\frac{9y^2}{3y} = 3y
\]
So, the next term of the quotient is \( 3y \).

#### 5. Multiply and Subtract Again:
- Multiply the entire divisor \( 3y - 1 \) by \( 3y \):
\[
3y \cdot (3y - 1) = 9y^2 - 3y
\]
- Subtract this result from the current dividend:
\[
(9y^2 - 5y + 5) - (9y^2 - 3y) = -2y + 5
\]

#### 6. Final Remainder:
- The degree of the remainder \( -2y + 5 \) is less than the degree of the divisor \( 3y - 1 \), so we stop here.
- The quotient is \( 2y^2 + 3y \), and the remainder is \( -2y + 5 \).

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Final Answer:

The result of the polynomial division is:
\[
\boxed{2y^2 + 3y \text{ with a remainder of } -2y + 5}
\]

This can also be written in the form:
\[
\frac{6y^3 + 7y^2 - 5y + 5}{3y - 1} = 2y^2 + 3y + \frac{-2y + 5}{3y - 1}
\]