Question 1161068

given:

 {{{p(x)}}} is a degree{{{ 5}}}


has zeros :
{{{x[1]=i }}}-> complex zeros always come in pair, so you also have
{{{x[2]=-i }}}
 {{{x[3]=4-i }}}, and also
{{{x[4]=4+i }}}

passes through the origin: point ({{{0}}},{{{0}}})-> 5th root is {{{x[5]=0}}}

{{{p(5)=520}}}


use zero product rule:

{{{p(x)=a(x-x[1])(x-x[2])(x-x[3])(x-x[4])(x-x[5])}}}....substitute given zeros


{{{p(x)=a(x-i)(x-(-i))(x-(4-i ))(x-(4+i ))(x-0)}}}

{{{p(x)=a(x^5 - 8x^4 + 18x^3 - 8x^2 + 17x)}}}......since we are given{{{p(5)=520}}}, use it to find   {{{a}}}

{{{520=a(5^5 - 8*5^4 + 18*5^3 - 8*5^2 + 17*5)}}}

{{{520=a(260)}}}

{{{a=520/260}}}

{{{a=2}}}

and your equation is

{{{p(x)=2(x^5 - 8x^4 + 18x^3 - 8x^2 + 17x)}}}

{{{p(x)=2x^5 - 16x^4 + 36x^3 - 16x^2 + 34x }}}

{{{ graph( 600, 600, -5, 5, -1000, 1000, 2x^5 - 16x^4 + 36x^3 - 16x^2 + 34x) }}}