In the following problem, divide using long division. State the quotient, $q(x)$, and the remainder, $r(x)$. \[ \frac{2 x^{4}-3 x^{2}+4 x}{x-4} \]

Step 1 ：Write the problem in long division format: \(\frac{2x^4 - 3x^2 + 4x}{x - 4}\)

Step 2 ：Divide the first term in the dividend \(2x^4\) by the first term in the divisor \(x\) to get \(2x^3\)

Step 3 ：Multiply the divisor \(x - 4\) by the term we just found \(2x^3\) to get \(2x^4 - 8x^3\), and subtract this from the dividend to get \(8x^3 - 3x^2 + 4x\)

Step 4 ：Bring down the next term from the original dividend \(-3x^2\) to get \(8x^3 - 3x^2 + 4x\)

Step 5 ：Divide the first term of the new dividend \(8x^3\) by the first term of the divisor \(x\) to get \(8x^2\), and subtract \(8x^3 - 32x^2\) from the dividend to get \(29x^2 + 4x\)

Step 6 ：Bring down the next term from the original dividend \(4x\) to get \(29x^2 + 4x\)

Step 7 ：Divide the first term of the new dividend \(29x^2\) by the first term of the divisor \(x\) to get \(29x\), and subtract \(29x^2 - 116x\) from the dividend to get \(120x\)

Step 8 ：Divide the first term of the new dividend \(120x\) by the first term of the divisor \(x\) to get \(120\), and subtract \(120x - 480\) from the dividend to get \(480\)

Step 9 ：Subtract the last line from the dividend to find the remainder \(480\)

Step 10 ：\(\boxed{2x^3 + 8x^2 + 29x + 120}\) is the quotient and \(\boxed{480}\) is the remainder