Find the quotient and remainder using long division.
\[
\frac{10 x^{3}+14 x^{2}-18 x-12}{2 x+4}
\]

The quotient is

The remainder is

Question Help:
Video

Step 1 ：First, divide the leading term of the dividend, \(10x^3\), by the leading term of the divisor, \(2x\), to get the first term of the quotient, \(5x^2\).

Step 2 ：Next, multiply the divisor, \(2x + 4\), by the first term of the quotient, \(5x^2\), to get \(10x^3 + 20x^2\).

Step 3 ：Subtract this from the original dividend, \(10x^3 + 14x^2 - 18x - 12\), to get a new dividend of \(-6x^2 - 18x - 12\).

Step 4 ：Repeat the process: divide the leading term of the new dividend, \(-6x^2\), by the leading term of the divisor, \(2x\), to get the next term of the quotient, \(-3x\).

Step 5 ：Then, multiply the divisor, \(2x + 4\), by the new term of the quotient, \(-3x\), to get \(-6x^2 - 12x\).

Step 6 ：Subtract this from the new dividend, \(-6x^2 - 18x - 12\), to get a new dividend of \(-6x - 12\).

Step 7 ：Repeat the process one more time: divide the leading term of the new dividend, \(-6x\), by the leading term of the divisor, \(2x\), to get the final term of the quotient, \(3\).

Step 8 ：Then, multiply the divisor, \(2x + 4\), by the final term of the quotient, \(3\), to get \(6x + 12\).

Step 9 ：Subtract this from the new dividend, \(-6x - 12\), to get a remainder of \(0\).

Step 10 ：So, the quotient is \(5x^2 - 3x + 3\) and the remainder is \(0\).

Step 11 ：Final Answer: \(\boxed{5x^2 - 3x + 3, 0}\)