Which second-degree polynomial function has a leading coefficient of –1 and root 5 with multiplicity 2?

A second degree polynomial function has the general form:

                  , where .

The leading coefficient is a, so we have a=-1.

5 is a double root means that :

i) f(5)=0,
ii) the discriminant D is 0, where .

Substituting x=5, we have

                                        f(5)=a(5)^2+b(5)+c,

and since f(5)=0, and a is -1 we have:

                                          0=-25+5b+c
thus c=25-5b.

By ii) .

Substituting a with -1 and c with 25-5b we have:
                                     
          
          
          
          
          
           

Finally we find c: c=25-5b=25-50=-25

Thus the function is        

Remark: It is also possible to solve the problem by considering the form

directly.

In general, if a quadratic function has leading coefficient a, and has a root r of multiplicity 2, then its form is