Question #bfb76

1 Answer
Feb 2, 2018

#P(x)=x^2(x+4)^4(x-7)#

Explanation:

If we write P(x) as a general factorial formular, we get:
#P(x)=k(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)(x-x_6)(x-x_7)#
#x_n="root"_n#
#k# has to be equal to one, because the leading coefficient is equal to k and equal to one, wherefore:
#k=1#
If we have roots of a multiplicity n, we well get:
#(x-x_n)^n#
as one of the factors. Therefore, we get the formular:
#P(x)=x^2(x+4)^4(x-7)#