Find the remainder when 2x³ - 3x² + 4x + 7 is divided by

(i) x - 2

(ii) x + 3

(iii) 2x + 1

(i) By remainder theorem, on dividing f(x) by (x - a), the remainder left is f(a).

f(x) = 2x³ - 3x² + 4x + 7

∴ On dividing f(x) by x - 2, Remainder = f(2)

$f(2) = 2(2)^3 - 3(2)^2 + 4(2) + 7 \\[0.5em] = 16 - 12 + 8 + 7 \\[0.5em] = 19$

Hence, the value of remainder is 19.

(ii) By remainder theorem, on dividing f(x) by (x - a), the remainder left is f(a).

f(x) = 2x³ - 3x² + 4x + 7

∴ On dividing f(x) by (x + 3) or (x - (-3)), Remainder = f(-3)

$f(-3) = 2(-3)^3 - 3(-3)^2 + 4(-3) + 7 \\[0.5em] = - 54 - 27 - 12 + 7 \\[0.5em] = -86$

Hence, the value of remainder is -86.

(iii) By remainder theorem, on dividing f(x) by (x - a), the remainder left is f(a).

f(x) = 2x³ - 3x² + 4x + 7

∴ On dividing f(x) by (2x + 1) or $2(x - \big(-\dfrac{1}{2}\big))$, Remainder = f$\big(-\dfrac{1}{2}\big)$

$f(-\dfrac{1}{2}) = 2(-\dfrac{1}{2})^3 - 3(-\dfrac{1}{2})^2 + 4(-\dfrac{1}{2}) + 7 \\[1em] = 2(-\dfrac{1}{8}) - 3(\dfrac{1}{4}) - 2 + 7 \\[1em] = -\dfrac{1}{4} - \dfrac{3}{4} + 5 \\[1em] = \dfrac{-1 -3 + 20}{4} \\[1em] = \dfrac{16}{4} \\[1em] = 4$

Hence, the value of remainder is 4.