To divide (1-x^2-3x+6x^3) by (2x-1), we can use polynomial long division.
1st step: Divide the first term of the dividend by the first term of the divisor. (6x^3) / (2x) = 3x^2
2nd step: Multiply the entire divisor by the result obtained in the previous step and subtract from the dividend. (2x-1) * 3x^2 = 6x^3 - 3x^2 (1-x^2-3x+6x^3) - (6x^3 - 3x^2) = -1 + 3x^2 - 3x
3rd step: Repeat the steps with the result obtained in the previous step. Now we have (-1 + 3x^2 - 3x) as the new dividend and (2x-1) as the divisor.
To divide (1-x^2-3x+6x^3) by (2x-1), we can use polynomial long division.
1st step: Divide the first term of the dividend by the first term of the divisor.
(6x^3) / (2x) = 3x^2
2nd step: Multiply the entire divisor by the result obtained in the previous step and subtract from the dividend.
(2x-1) * 3x^2 = 6x^3 - 3x^2
(1-x^2-3x+6x^3) - (6x^3 - 3x^2) = -1 + 3x^2 - 3x
3rd step: Repeat the steps with the result obtained in the previous step.
Now we have (-1 + 3x^2 - 3x) as the new dividend and (2x-1) as the divisor.
(-3x) / (2x) = -3/2
(2x-1) * (-3/2) = -3x + 3/2
(-1 + 3x^2 - 3x) - (-3x + 3/2) = 2x - 5/2
Since the degree of the new dividend (2x - 5/2) is less than the degree of the divisor (2x - 1), we stop here.
Therefore, the result of dividing (1-x^2-3x+6x^3) by (2x-1) is 3x^2 - 3/2 with a remainder of 2x - 5/2.