SOLUTION: How do I find the polynomial p(x) with real coefficients of the smallest degree that satisfies the given conditions: p(x) has zeros at x=0, 1/2, 1+ i, and p(1)=2.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: How do I find the polynomial p(x) with real coefficients of the smallest degree that satisfies the given conditions: p(x) has zeros at x=0, 1/2, 1+ i, and p(1)=2.      Log On


   

Question 736102: How do I find the polynomial p(x) with real coefficients of the smallest degree that satisfies the given conditions: p(x) has zeros at x=0, 1/2, 1+ i, and p(1)=2.
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
Start with focusing on the zeros. The complex with imaginary part needs also its conjugate. Your simplest function is x%28x-1%2F2%29%28x-%281%2Bi%29%29%28x-%281-i%29%29
=x%28x-1%2F2%29%28x%5E2-2x%2B2%29.

When you let x=1, you find that the expression evaluation results in value of 1/2. You want p(1)=2, which is an increase by a factor of 4, so you want highlight%28p%28x%29=4x%28x-1%2F2%29%28x%5E2-2x%2B2%29%29
and you could fully multiply the expression into the polynomial if you want it in general form.