Question 1135762


degree {{{3}}} 

zeros of:
{{{x[1]= -2 }}} 
{{{x[2]=5+i}}}-> since complex roots come in pairs, you also have {{{x[3]=5-i}}}


use zero product theorem to find {{{f(x)}}}


{{{f(x)=(x-x[1])(x-x[2])(x-x[3])}}}

{{{f(x)=(x-(-2))(x-(5+i))(x-(5-i))}}}

{{{f(x)=(x+2)(x-5-i)(x-5+i)}}}

{{{f(x)=(x+2)(x^2-5x+ix-5x+25-5i-ix+5i-i^2)}}}

{{{f(x)=(x+2)(x^2-10x+25-(-1))}}}

{{{f(x)=(x+2)(x^2 - 10 x + 26)}}}

{{{f(x)= x^3 - 10 x^2 + 26x+2x^2-20x+52}}}

{{{f(x)= x^3 - 8x^2 + 6x+52}}}