Question 174369
Solve:
1) {{{2x^3-5x^2-3x+2 = 2}}} Subtract 2 from both sides.
{{{2x^3-5x^2-3x = 0}}} Factor an x from the left side.
{{{x(2x^2-5x-3) = 0}}} Factor the the parentheses.
{{{x(2x+1)(x-3) = 0}}} Apply the zero product rule.
{{{x = 0}}} or {{{2x+1 = 0}}} or {{{x-3 = 0}}}, so...
{{{highlight(x = 0)}}}
{{{highlight(x = -1/2)}}}
{{{highlight(x = 3)}}}
-------------
2) {{{x^3+64 = 0}}} This is a sum of cubes.
{{{(x)^3+(4)^3 = 0}}} This can be solved thus:{{{A^3+B^3 = (A+B)(A^2-AB+B^2)}}}
{{{(x+4)(x^2-4x+16) = 0}}} Applying the zero product rule:
{{{x+4 = 0}}} or {{{x^2-4x+16 = 0}}}
{{{highlight(x = -4)}}} or 
{{{x = (-(-4)+-sqrt((4)^2-4(1)(16)))/2(1)}}}
{{{x = (4+-sqrt(-48))/2}}}
{{{x = (4+-sqrt(16*(-3)))/2}}}
{{{x = 2+2sqrt(-3)}}} or {{{x = 2-2sqrt(-3)}}} and these can be expressed in terms of i, as:
{{{highlight(x = 2+2sqrt(3)i)}}} or {{{highlight(x = 2-2sqrt(3)i)}}}
--------------------------------------------------------------------
3) {{{x^4-8x^2-9 = 0}}} Rewrite in terms of {{{x^2}}}
{{{(x^2)^2-8(x^2)-9 = 0}}} Factor this.
{{{(x^2+1)(x^2-9) = 0}}} Apply the zero product rule:
{{{x^2+1 = 0}}} or {{{x^2-9 = 0}}}, so...
{{{x^2 = -1}}} Take the square root of both sides.
{{{x = sqrt(-1)}}} or {{{x = -sqrt(-1)}}} or, in terms of i,
{{{highlight(x = i)}}} or {{{highlight(x = -i)}}}
{{{x^2-9 = 0}}}
{{{x^2 = 9}}} Take the square root of both sides.
{{{highlight(x = 3)}}} or {{{highlight(x = -3)}}}