If $$ax^4 + bx^3 + 2x^2 + 4$$ is exactly divisible by $$x^2 - x - 2$$, then $$(a, b) =$$Â
$$x^2 - x - 2$$ = (x+1)(x-2)
Let f(x) =Â $$ax^4 + bx^3 + 2x^2 + 4$$
f(-1) = 0 => a - b + 6 =0
f(2) = 0 => 16a+8b+12 = 0
Solving above two, we get a=-5/2 and b = 7/2
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