Write the function in the form f(x) = (x − k)q(x) + r for the given value of k.

Write the function in the form f(x) = (x − k)q(x) + r for the given value of k.

f(x) = x³ + 5x² − 3x − 22,

k = \(\sqrt3\)

We need to divide:

\(\begin{array}{|rcll|} \hline \dfrac{ x^3 + 5x^2 - 3x - 22 } {(x - \sqrt3)} &=& x^2 + (5+\sqrt3)x+5\sqrt3 + \dfrac{-7}{x-\sqrt3} \qquad | \qquad \cdot (x - \sqrt3)\\\\ x^3 + 5x^2 - 3x - 22 &=& (x - \sqrt3)\cdot \left[~ x^2 + (5+\sqrt3)x+5\sqrt3 ~\right] + (x - \sqrt3)\cdot \left[~ \dfrac{-7}{x-\sqrt3} ~\right] \\\\ x^3 + 5x^2 - 3x - 22 &=& (x - \sqrt3)\cdot \underbrace{ \left[~ x^2 + (5+\sqrt3)x+5\sqrt3 ~\right] } _{=q(x)} \quad \underbrace{ -\quad 7 }_{= r} \\\\ \hline \end{array} \)

\(f(x)\div (x-\sqrt3)\\=f(x)\div ((x^2-3)\div(x+\sqrt3))\\=f(x)\div (x^2-3)\times (x+\sqrt 3)\)

\(\text{We need to get f(x) into the form }(x-\sqrt{3})q(x)+r\\ \text{ so we need to find q(x) and r}\)

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