# Find a polynomial p(t) of degree 6 which

1. Oct 14, 2012

### zodiacbrave

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

2. Oct 14, 2012

### Ray Vickson

If the other root is at t = a (where you do not yet know a), what is the form of p(t)?

RGV

3. Oct 14, 2012

### SammyS

Staff Emeritus
Write p(t) as a product of its factors:

$\displaystyle p(t)=a(t-1)^2(t-2)^3(t-b)\,,$ where b is the sixth root.

Use the extended product rule to find p'(t).

4. Oct 14, 2012

### zodiacbrave

Okay, so that makes sense, the factor thing, but now it seems we have two unknowns...How would I get the polynomial from its factors, given two unknowns?

Unless taking the derivative is the key?

5. Oct 14, 2012

### Ray Vickson

You are given two other restrictions on p; what do they give you?

RGV