WAEC Further Mathematics 2020 Paper

$\frac{(\sqrt{5}+3)(4+\sqrt{10})}{(4-\sqrt{10})(4+\sqrt{10})}$

= $\frac{4\sqrt{5}+\sqrt{50}+12+3\sqrt{10}}{4^2-(\sqrt{10})^2}$

= $\frac{4\sqrt{5}+5\sqrt{2}+12+3\sqrt{10}}{16-10}$

= $\frac{4\sqrt{5}}{6}+\frac{5\sqrt{2}}{6}+\frac{12}{6}+\frac{3\sqrt{10}}{6}$

= $\frac{2}{3}$$\sqrt{5}$ + $\frac{5}{6}\sqrt{2}$ + $\frac{1}{2}\sqrt{10}$ + 2