Question 682329
Find a polynomial function of degree 3 with zeros 1/2, 1-i; where P(4) = 140
<pre>
Since the polynomial is to have real coefficients, since 1-i is a zero,
its conjugate 1+i is also a zero

So the polynomial is

{{{"P(x)"}}}{{{""=""}}}{{{A*(x-1/2)(x-(1-i))(x-(1+i))}}}
{{{"P(x)"}}}{{{""=""}}}{{{A*(x-1/2)(x-1+i)(x-1-i)}}}
{{{"P(x)"}}}{{{""=""}}}{{{A*(x-1/2)((x-1)+i)((x-1)-i)}}}
{{{"P(x)"}}}{{{""=""}}}{{{A*(x-1/2)((x-1)^2-i^2)}}}
{{{"P(x)"}}}{{{""=""}}}{{{A*(x-1/2)((x-1)^2-(-1))}}}
{{{"P(x)"}}}{{{""=""}}}{{{A*(x-1/2)((x-1)^2+1)}}}
{{{"P(x)"}}}{{{""=""}}}{{{A*(2x/2-1/2)((x^2-2x+1)+1)}}}
{{{"P(x)"}}}{{{""=""}}}{{{A*((2x-1)/2)(x^2-2x+1+1)}}}
{{{"P(x)"}}}{{{""=""}}}{{{A*((2x-1)/2)(x^2-2x+2)}}}
{{{2*"P(x)"}}}{{{""=""}}}{{{A(2x-1)(x^2-2x+2)}}}
{{{2*"P(x)"}}}{{{""=""}}}{{{A*(2x^3-4x^2+4x-x^2+2x-2)}}}
{{{2*"P(x)"}}}{{{""=""}}}{{{A*(2x^3-5x^2+6x-2)}}}

Now we substitute {{{"P(4)"}}}{{{""=""}}}{{{140}}}, 
which implies that we also substitute{{{x}}}{{{""=""}}}{{{4}}}.

{{{2*"P(4)"}}}{{{""=""}}}{{{A(2x*4^3-5*4^2+6*4-2)}}}
{{{2*140}}}{{{""=""}}}{{{A*(2*64-5*16+6*4-2)}}}
{{{280}}}{{{""=""}}}{{{A*(128-80+24-2)}}}
{{{280}}}{{{""=""}}}{{{A*(70)}}}
{{{280/70}}}{{{""=""}}}{{{A}}}
{{{4}}}{{{""=""}}}{{{A}}}

So

{{{2*"P(x)"}}}{{{""=""}}}{{{4*(2x^3-5x^2+6x-2)}}}
{{{"P(x)"}}}{{{""=""}}}{{{2*(2x^3-5x^2+6x-2)}}}
{{{"P(x)"}}}{{{""=""}}}{{{4x^3-10x^2+12x-4}}}

Edwin</pre>