The remainder when $2{\mathrm{x}}^5–3{\mathrm{x}}^4+5{\mathrm{x}}^3–3{\mathrm{x}}^2+7\mathrm{x}–9$ is divisible by ${\mathrm{x}}^2 – \mathrm{x} – 3$ is :

$\begin{array}{l}GivenEquation2x^5-3x^4+5x^3-3x^2+7x-9isdivisibleby{x}^2-x-3\\ Bysyntheticdivision\end{array}$

$\therefore \mathrm{Re}mainder=41x+44$