Question 995119
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The polynomial of degree 4, P(x)  has a root of multiplicity 2 at x=2  and roots of multiplicity 1 at x=0  and x=−2. It goes through the point (5,252). 
Find a formula for P(x).
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According to the given information about the roots,  the polynomial is 


P(x) =  {{{a*(x-2)^2*(x-0)*(x-(-2))}}} = {{{a*(x-2)^2*x*(x+2)}}}           (1)


with some unknown real coefficient  {{{a}}}.


To find  {{{a}}},  use the fact that the plot of the polynomial goes through the point  (5,252).  It means that  P(5) = 252. 

So,  simply substitute  x=5  into the polynomial  (1). You will get the equation


{{{a*(5-2)^2*5*(5+2)}}} = {{{252}}},     or


{{{a*3^2*5*7}}} = {{{252}}},       or


{{{315a}}} = {{{252}}}.


Hence,  {{{a}}} = {{{252/315}}} = {{{4/5}}}.


Therefore,  P(x) = {{{4/5}}}.{{{(x-2)^2*x*(x+2)}}}.


You can open the parentheses if you need.