Find a polynomial p(t) of degree 6 which has a zero of multiplicity 2 at t = 1 and a zero of multiplicity 3 at

t = 2, and also satisfying: p(0) = 2 and p`(0) = 1. What is the other root of p(t)?

Attempt at solution:

zero of multiplicity 2 at t =1 implies (t-1)^2 is a factor or p(1) = 0 and p`(1) = 0

zero of multiplicity 3 at t = 2 implies (t-2)^3 is a factor or p(2) = 0, p`(2) = 0, p``(2) = 0

so now I seem to have 7 pieces of data...

p(0) = 2,

p`(0) = 1,

p(1) = 0,

p`(1) = 0,

p(2) = 0,

p`(2) = 0,

p``(2) = 0

so I thought I would try to put these in a linear system and solve them to get the coefficients.

p(t) = t0 + t1*t + t2*t^2 + t3*t^3 + t4*t^4 + t5*t^5

p`(t) = t1 + 2t*2t + 3t3*t^2 + 4t*t^3 + 5t*t^4

p``(t) = 2t2 + 6t3*t + 12t4*t^2 + 20t5*t^3

p(0) = t0 = 2

p`(0) = t1 = 1

p(1) = t0 + t1*t + t2*t^2 + t3*t^3 + t4*t^4 + t5*t^5 = 0

p`(1) = t1 + 2t*2t + 3t3*t^2 + 4t*t^3 + 5t*t^4 = 0

p(2) = t0 + 2t1 + 4t2 + 8t3 + 16t4 + 32t5 = 0

p`(2) = t1 + 4t2 + 12t3 + 32t4 + 80t5 = 0

p``(2) = 2t2 + 12t3 + 48t4 + 160t5 = 0

so I put these in the calculator and rref them but that doesn't give me any useful information.

I am not sure what to do at this point...

Thank you