Answer
$-15n^{4}-18n^{3}+6n^{2}$
Work Step by Step
First, we multiply each term inside the parenthesis by $3n^{2}$. Then, we use the rule $a^{m}\times a^{n}=a^{m+n}$ to simplify,
$(-3n^{2})(5n^{2}+6n-2)$
=$(-3n^{2})(5n^{2})+(-3n^{2})(6n)+(-3n^{2})(-2)$
=$(-3\times5\times n^{2}\times n^{2})+(-3\times6\times n^{2}\times n)+(-3\times-2\times n^{2})$
=$(-15\times n^{2+2})+(-18\times n^{2+1})+(6\times n^{2})$
=$(-15\times n^{4})+(-18\times n^{3})+(6\times n^{2})$
=$(-15n^{4})+(-18n^{3})+(6n^{2})$
=$-15n^{4}-18n^{3}+6n^{2}$
