Question 972920
A polynomial function of degree {{{n=3}}} has {{{3}}} complex zeros.
If a polynomial function w/real coefficients has a non-real complex zero,
then the conjugate complex number is also a zero.
Then, the complex zeros of {{{f(x)}}} are {{{-3}}} , {{{1+5i}}} , and {{{1-5i}}} .
So the factored form of {{{f(x)}}} is
{{{f(x)=a(x-(-3))(x-(1+5i))(x-(1-5i))}}}
{{{f(x)=a(x+3)(x-1-5i)(x-1+5i))}}}
{{{f(x)=a(x+3)((x-1)^2-(5i)^2))}}}
{{{f(x)=a(x+3)(x^2-2x+1-25(i)^2))}}}
{{{f(x)=a(x+3)(x^2-2x+1-25(-1))}}}
{{{f(x)=a(x+3)(x^2-2x+1+25)}}}
{{{f(x)=a(x+3)(x^2-2x+26)}}}
So, {{{130=f(2)=a(2+3)(2^2-2*2+26)}}}-->{{{130=5a(4-4+26)}}}-->{{{130=5a(26)}}}-->{{{130=130a}}}-->{{{a=1}}} .
So, {{{f(x)=(x+3)(x^2-2x+26)}}} in factored form,
and if that's not the required form, we can multiply and get
{{{f(x)=x(x^2-2x+26)+3(x^2-2x+26)=(x^3-2x^2+26x)+(3x^2-6x+78)=x^3+x^2+20x+78}}}