Question

use long division to determine the quotient and remainder when f(x) = 4x³+6x²-2x+5 is divided by (2x-1)

Answer 1

$\begin{matrix} & 2x^2 +4x+1 \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ 2x-1 &)\overline{4x^3+6x^2-2x+5} \\ & - \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ & \underline{4x^3-2x^2} \ \ \ \ \ \ \ \ \ \ \ \ \ \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8x^2-2x+5 \\ & \ \ - \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ & \ \ \ \ \ \ \ \ \underline{8x^2-4x} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2x+5 \\ & \ \ - \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{2x-1} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6 \\ \end{matrix}$

Therefore,

$$\dfrac{4x^3+6x^2-2x+5}{2x-1}=2x^2+4x+1+\dfrac{6}{2x-1}$$

Quotient is $2x^2+4x+1.$

Remainder is $6$