Find the remainder when $2x^3-3x^2+4x-5$ is divided by $x-2$.

Step 1 ：Step 1: Arrange the dividend and divisor in descending order of powers. So, we have $2x^3-3x^2+4x-5$ divided by $x-2$.

Step 2 ：Step 2: Divide the highest degree term in the dividend by the highest degree term in the divisor. Here, $2x^3$ divided by $x$ gives us $2x^2$.

Step 3 ：Step 3: Multiply the entire divisor by the result from step 2 and subtract the result from the original dividend. $(x-2)\ast 2x^2=2x^3-4x^2$. Subtracting this from the original dividend gives us $x^2+4x-5$.

Step 4 ：Step 4: Repeat the process starting from step 2 with the new dividend of $x^2+4x-5$. Divide $x^2$ by $x$ to get $x$. Multiply the entire divisor by $x$ to get $x^2-2x$. Subtract this from the new dividend to get $6x-5$.

Step 5 ：Step 5: Repeat the process again with the new dividend of $6x-5$. Divide $6x$ by $x$ to get $6$. Multiply the entire divisor by $6$ to get $6x-12$. Subtract this from the new dividend to get $7$, which is the remainder.