The polynomial of degree 5, P(x) has leading coefficient 1, has roots of multiplicity 2 at x=1 and x=0, and a root of multiplicity 1 at x=-1 Find a possible formula for P(x)?

Question

25150 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-08T23:04:49

# P(x) = x^2(x-1)^2(x+1) #

Given that we have a root of multiplicity #2# #at x=1#, we know that #P(x)# has a factor #(x-1)^2#

Given that we have a root of multiplicity #2# at #x=0#, we know that #P(x)# has a factor #x^2#

Given that we have a root of multiplicity #1# at #x=-1#, we know that #P(x)# has a factor #x+1#

We are given that #P(x)# is a polynomial of degree #5#, and we have therefore identified all five roots, and factors, so we can write

# P(x) = 0 => x^2(x-1)^2(x+1) = 0 #

And we can therefore write

# P(x) = Ax^2(x-1)^2(x+1) #

We also know that the leading coefficient is #1 => A=1#

Hence,

# P(x) = x^2(x-1)^2(x+1) #