Find the remainder (without division) on dividing f(x) by (2x + 1) where

(i) f(x) = 4x² + 5x + 3

(ii) f(x) = 3x³ - 7x² + 4x + 11

(i) By remainder theorem, on dividing f(x) by (x - a) , remainder = f(a)

∴ On dividing, f(x) = 4x² + 5x + 3 by (2x + 1) or 2(x - (-$\dfrac{1}{2}$))

Remainder = f(-$\dfrac{1}{2}$)

$= 4\big(-\dfrac{1}{2}\big)^2 + 5\big(-\dfrac{1}{2}\big) + 3 \\[1em] = 4\big(\dfrac{1}{4}\big) -\dfrac{5}{2} + 3 \\[1em] = 1 - \dfrac{5}{2} + 3 \text{ (On taking L.C.M.)} \\[1em] = \dfrac{2 - 5 + 6}{2} \\[1em] = \dfrac{3}{2} \\[1em] = 1\dfrac{1}{2}.$

Hence, the value of remainder is 1 $\dfrac{1}{2}$.

(ii) By remainder theorem, on dividing f(x) by (x - a) , remainder = f(a)

∴ On dividing, f(x) = 3x³ - 7x² + 4x + 11 by (2x + 1) or 2(x - (-$\dfrac{1}{2}$))

Remainder = f$\big(-\dfrac{1}{2}\big)$

$= 3\big(-\dfrac{1}{2}\big)^3 - 7\big(-\dfrac{1}{2}\big)^2 + 4\big(-\dfrac{1}{2}\big) + 11 \\[1em] = 3\big(-\dfrac{1}{8}\big) - 7\big(\dfrac{1}{4}\big) - 2 + 11 \\[1em] = -\dfrac{3}{8} - \dfrac{7}{4} + 9 \\[1em] = \dfrac{-3 - 14 + 72}{8} \\[1em] = \dfrac{55}{8} \\[1em] = 6\dfrac{7}{8}$

Hence, the value of remainder is $6\dfrac{7}{8}$.