(2y^3-3y^2+3y-4)/(y-2)
To simplify the expression (2y^3 - 3y^2 + 3y - 4)/(y - 2), we can use polynomial long division. Here's how to do it:
1. First, make sure the terms are arranged in descending order of powers of y:
2y^3 - 3y^2 + 3y - 4
2. Divide the first term of the numerator by the first term of the denominator:
(2y^3)/(y) = 2y^2
3. Multiply 2y^2 by the denominator (y - 2) and subtract it from the numerator:
2y^2 * (y - 2) = 2y^3 - 4y^2
Subtracting this from the original numerator:
(2y^3 - 3y^2 + 3y - 4) - (2y^3 - 4y^2) = y^2 + 3y - 4
4. Now, repeat the process with the new numerator (y^2 + 3y - 4):
Divide the first term by the first term of the denominator: (y^2)/(y) = y
5. Multiply y by the denominator (y - 2) and subtract it from the numerator:
y * (y - 2) = y^2 - 2y
Subtracting this from the previous numerator:
(y^2 + 3y - 4) - (y^2 - 2y) = 5y - 4
6. We have a new numerator of 5y - 4. Since this is a linear expression, we can divide it by the linear factor (y - 2) easily:
Divide 5y by y: (5y)/(y) = 5
Multiply 5 by the denominator (y - 2) and subtract it from the numerator:
5 * (y - 2) = 5y - 10
Subtracting this from the previous numerator:
(5y - 4) - (5y - 10) = 6
7. At this point, we have no more terms in the numerator, and we are left with a remainder of 6.
Therefore, the simplified form of (2y^3 - 3y^2 + 3y - 4)/(y - 2) is:
2y^2 + y + 5 + 6/(y - 2)