Question 392824
Solve for x:
{{{3x+15 = x^3+5x^2}}} Reformat to look like:
{{{x^3+5x^2-3x-15 = 0}}} Factor.
{{{(x^2-3)(x+5) = 0}}} Apply the zero product rule.
{{{x^2-3 = 0}}} or {{{x+5 = 0}}} so...
{{{x^2 =3}}} then {{{x = 3}}} and {{{x = -3}}} also...
{{{x+5 = 0}}} then {{{x = -5}}} so...The values of x are:
{{{x = 3}}}, {{{x = -3}}}, {{{x = -5}}}