$\displaystyle{10}={1}\cdot{10}={2}\cdot{5}=-{1}\cdot{\left(-{10}\right)}=-{2}\cdot{\left(-{5}\right)}\usepackage{cmbright}\Huge\$ . The factors whose sum is $\displaystyle-{7}\usepackage{cmbright}\Huge\$ are $\displaystyle-{2}\usepackage{cmbright}\Huge\$ and $\displaystyle-{5}\usepackage{cmbright}\Huge\$ . Now rewrite the trinomial and use grouping:

$\displaystyle{{x}}^{{2}}-{2}{x}-{5}{x}+{10}={x}{\left({x}-{2}\right)}-{5}{\left({x}-{2}\right)}={\left({x}-{2}\right)}{\left({x}-{5}\right)}\usepackage{cmbright}\Huge\$