\(f(x)\div (x-\sqrt3)\\=f(x)\div ((x^2-3)\div(x+\sqrt3))\\=f(x)\div (x^2-3)\times (x+\sqrt 3)\)
\(\text{We need to get f(x) into the form }(x-\sqrt{3})q(x)+r\\ \text{ so we need to find q(x) and r}\)
First we need to divide x^3 + 5x^2 - 3x - 22 by x^2-3
Quotient = x Remainder = -22
So that \(\dfrac{f(x)}{x^2-3}= x - \dfrac{22}{x^2-3}\)
Then multiply this to x+sqrt3
\((x-\dfrac{22}{x^2-3})(x+\sqrt3)\)
\(= x^2 + \sqrt3x - \dfrac{22(x+\sqrt3)}{x^2-3}\)
\(= x^2+\sqrt3x-\dfrac{22}{x-\sqrt3}\)
There we have found q(x) and r
q(x) = x^2 + sqrt3 x r = -22
