Divide $14x^3 - 5x^2 + 9x - 1$ by $2x - 1$.

(N/A) To divide $14x^3 - 5x^2 + 9x - 1$ by $2x - 1$,we use polynomial long division:
$1$. Divide the first term of the dividend $(14x^3)$ by the first term of the divisor $(2x)$: $14x^3 / 2x = 7x^2$. This is the first term of the quotient.
$2$. Multiply $7x^2$ by $(2x - 1)$ to get $14x^3 - 7x^2$. Subtract this from the dividend: $(14x^3 - 5x^2) - (14x^3 - 7x^2) = 2x^2$. Bring down the next term $(9x)$ to get $2x^2 + 9x$.
$3$. Divide the first term of the new expression $(2x^2)$ by the first term of the divisor $(2x)$: $2x^2 / 2x = x$. This is the second term of the quotient.
$4$. Multiply $x$ by $(2x - 1)$ to get $2x^2 - x$. Subtract this from the current expression: $(2x^2 + 9x) - (2x^2 - x) = 10x$. Bring down the last term $(-1)$ to get $10x - 1$.
$5$. Divide the first term $(10x)$ by the first term of the divisor $(2x)$: $10x / 2x = 5$. This is the third term of the quotient.
$6$. Multiply $5$ by $(2x - 1)$ to get $10x - 5$. Subtract this from the current expression: $(10x - 1) - (10x - 5) = 4$.
Thus,the quotient is $7x^2 + x + 5$ and the remainder is $4$.