(1)/(4p)(x-h)^(2)+k=0 multiply the equation by 4p. explain how different values of k affect the number of zeros of the polynomial. consider k > 0, k = 0, and k < 0. assume p > 0.

Assume that and multiply the equation by 4p. Then you obtain the equation .

1) If k>0, then 4pk>0 and the equation doesn't have solutions, because and this is unreal.

2) If k=0, then 4pk=0 and . There is one solution x=h.

2) If k<0, then 4pk<0 and the equation
has two different solutions and .