Question 1185932


{{{P(x) = (1/6)(2x^4 + 3x^3-16x- 24)^2}}}

{{{(1/6)(2x^4 + 3x^3-16x- 24)^2=0 }}}will be if

{{{2x^4 + 3x^3-16x- 24=0}}}

{{{(2x^4 + 3x^3)-(16x+ 24)=0}}}

{{{x^3(2x + 3)-8(2x+ 3)=0}}}

{{{(2x + 3) (x^3 - 8) = 0}}}........use {{{8=2^3}}}

{{{(2x + 3) (x^3 - 2^3) = 0}}} .......use the rule for factoring the difference of the cubes

{{{(2x+3)(x-2)(x^2+2x+4)=0}}}......... zeros are


if {{{(2x+3)=0}}} => {{{x=-3/2}}}
if  {{{(x-2) =0}}} =>{{{ x=2}}}

for{{{ x^2+2x+4=0}}} use quadratic formula

{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}} .....in your case {{{a=1}}}, {{{b=2}}}, {{{c=4}}}

{{{x=(-2+-sqrt(2^2-4*1*4))/(2*1)}}} 
{{{x=(-2+-sqrt(4-16))/2}}}
{{{x=(-2+-sqrt(-12))/2}}}
{{{x=(-2+-sqrt(-4*3))/2}}}
{{{x=(-2+-2sqrt(3)*i)/2}}}........simplify
{{{x=(-1+-sqrt(3)*i)}}}

roots: {{{-1+sqrt(3)*i}}},  {{{-1-sqrt(3)*i}}}

all roots will be

real roots:

{{{x=-3/2}}}
{{{x=2}}}

complex roots:

{{{-1+sqrt(3)*i}}}
{{{ -1-sqrt(3)*i}}}