If a polynomial f(x) = x⁴ - 2x³ + 3x² - ax - b leaves remainders 5 and 19 when divided by (x - 1) and (x + 1) respectively, find the values of a and b. Hence, determine the remainder when f(x) is divided by (x - 2).

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

f(x) = x⁴ - 2x³ + 3x² - ax - b

∴ On dividing f(x) by (x + 1) or (x - (-1)), Remainder = f(-1)

Given, on dividing by (x + 1) remainder = 19,

∴ f(-1) = 19

$\Rightarrow (-1)^4 - 2(-1)^3 + 3(-1)^2 - a(-1) - b = 19 \\[0.5em] \Rightarrow 1 + 2 + 3 + a - b = 19 \\[0.5em] \Rightarrow a - b + 6 = 19 \\[0.5em] \Rightarrow a - b = 13 \\[0.5em] a = 13 + b \text{ \space (Equation 1) }$

∴ On dividing f(x) by (x - 1), Remainder = f(1)

Given, on dividing by (x - 1) remainder = 5,

∴ f(1) = 5

$\Rightarrow (1)^4 - 2(1)^3 + 3(1)^2 - a(1) - b = 5 \\[0.5em] \Rightarrow 1 - 2 + 3 - a - b = 5 \\[0.5em] \Rightarrow 2 - a - b = 5 \\[0.5em] \Rightarrow -a - b = 3$

Putting value of a = 13 + b from equation 1,

$\Rightarrow -(13 + b) - b = 3 \\[0.5em] \Rightarrow -13 - b - b = 3 \\[0.5em] \Rightarrow -13 - 2b = 3 \\[0.5em] \Rightarrow 2b = -16 \\[0.5em] \Rightarrow b = -\dfrac{16}{2} \\[0.5em] \Rightarrow b = -8 \\[0.5em] \text{ and } a = 13 + b = 13 - 8 = 5$

Putting a = 5 and b = -8 in f(x) we get,

f(x) = x⁴ - 2x³ + 3x² - 5x + 8.

On dividing f(x) by (x - 2), remainder = f(2) by remainder theorem

$f(2) = (2)^4 - 2(2)^3 + 3(2)^2 - 5(2) + 8 \\[0.5em] = 16 - 16 + 12 - 10 + 8 \\[0.5em] = 10.$

Hence, the value of a is 5 and b is -8.
On dividing x⁴ - 2x³ + 3x² - 5x + 8 by (x - 2) the value of remainder is 10.