Question 1157330


{{{h(x)=2x^4+3x^3+48x^2+75x-50}}}

if given zero {{{x[1]= -5i}}}, then you also have {{{x[2]= 5i}}}

so, find a product {{{(x-x[1])(x-x[2])}}} and divide given equation using long division

{{{(x-(-5i))(x-5i)}}} 
{{{(x+5i)(x-5i)}}}
{{{x^2-cross(5i*x)+cross(5i*x)-(5i)^2}}}
{{{x^2-25(i)^2}}}
{{{x^2-25(-1)}}}
{{{x^2+25}}}


.......................({{{2x^2+3x-2}}}
{{{x^2+25}}}|{{{2x^4+3x^3+48x^2+75x-50}}}
.....................{{{2x^4+0*x^3+50x^2}}}..........subtract
....................................{{{3x^3-2x^2}}}......bring down {{{75x}}}
....................................{{{3x^3-2x^2+75x}}}
....................................{{{3x^3+0*x^2+75x}}} .........subtract 
................................................{{{-2x^2}}}......bring down {{{-50}}}
................................................{{{-2x^2-50}}}.
................................................{{{-2x^2-50}}}.......subtract 
.............................................................{{{0}}}

so  {{{2x^4+3x^3+48x^2+75x-50}}}=  {{{(x^2+25)(2x^2+3x-2)}}}

now factor {{{2x^2+3x-2=0}}}

{{{2x^2-x+4x-2=0}}}
{{{(2x^2-x)+(4x-2)=0}}}
{{{x(2x-1)+2(2x-1)=0}}}
{{{(x + 2) (2 x - 1) = 0}}}

remaining zeros are:

{{{x=-2}}}
{{{x=1/2}}}