Without actual division, prove that x⁴ + 2x³ - 2x² + 2x - 3 is exactly divisible by x² + 2x - 3.

Let, f(x) = x⁴ + 2x³ - 2x² + 2x - 3

g(x) = x² + 2x - 3
= x² + 3x - x - 3
= x(x + 3) - 1(x + 3)
= (x - 1)(x + 3)

$\Rightarrow$ (x - 1) and (x + 3) are factors of g(x).

In order to prove that f(x) is exactly divisible by g(x), it is sufficient to prove that x - 1 and x + 3 are factors of f(x) i.e. it is sufficient to show that f(1) = 0 and f(-3) = 0.

Now,
f(1) = (1)⁴ + 2(1)³ - 2(1)² + 2(1) - 3

= 1 + 2 - 2 + 2 - 3 = 0

f(-3) = (-3)⁴ + 2(-3)³ - 2(-3)² + 2(-3) - 3

= 81 - 54 - 18 - 6 - 3 = 0

∴ f(x) is divisible by (x - 1) and (x + 3)

Hence, f(x) is exactly divisible by g(x).