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

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

Re: Find a polynomial p(t) of degree 6 which...

Why do you not **use** the information you are given?

Quote:

zero of multiplicity 2 at t =1 implies (t-1)^2 is a factor

zero of multiplicity 3 at t = 2 implies (t-2)^3 is a factor

And since p has degree six, any other factor must be **linear**:

p(t)= (t- 1)^2(t- 2)^3(at+ b)

Now put the other information into that.

Re: Find a polynomial p(t) of degree 6 which...

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?

Re: Find a polynomial p(t) of degree 6 which...

we have, as Plato so graciously hinted at:

p(x) = (x - 1)^{2}(x - 2)^{3}(ax + b)

therefore:

p(0) = (-1)^{2}(-2)^{3}(b) = -8b

since p(0) = 2, this becomes:

2 = -8b

b = -1/4

now let's compute p'(x):

p'(x) = [(x - 1)^{2}(x - 2)^{3}]'(ax + b) + (x - 1)^{2}(x - 2)^{3}(a)

= [(x - 1)^{2}(3(x - 2)^{2}) + 2(x - 1)(x - 2)^{3}](ax + b) + (x - 1)^{2}(x - 2)^{3}(a)

= [(x - 1)(3) + (2)(x - 2)](x - 1)(x - 2)^{2}(ax + b) + (x - 1)^{2}(x - 2)^{3}(a)

= [3x - 3 + 2x - 4](x - 1)(x - 2)^{2}(ax + b) + (x - 1)^{2}(x - 2)^{3}(a)

= (5x - 7)(x - 1)(x - 2)^{2}(ax + b) + (x - 1)^{2}(x - 2)^{3}(a)

= [(5x - 7)(ax + b) + a(x - 1)(x - 2)](x - 1)(x - 2)^{2}

= [5ax^{2} + (5b - 7a)x - 7b + ax^{2} - 3ax + 2a](x - 1)(x - 2)^{2}

= (6ax^{2} + (5b - 10a)x + (2a - 7b))(x - 1)(x - 2)^{2}

therefore:

p'(0) = (2a - 7b)(-1)(-2)^{2} = (2a - 7b)(-4) = 28b - 8a

now we know that b = -1/4, so p'(0) = -7 - 8a.

since p'(0) = 1, we have:

1 = -7 - 8a

8 = -8a

a = -1

so p(x) = -(x - 1)^{2}(x - 2)^{2}(x + 1/4). <---the third root is -1/4. expanding this becomes:

p(x) = -x^{6} + (31/4)x^{5} - 23x^{4} + (127/4)x^{3} - (37/2)x^{2} + x + 2

as a check, we find:

p(0) = 2, and:

p'(x) = -6x^{5} + (155/4)x^{4} - 92x^{3} + (381/4)x^{2} - 37x + 1, so that:

p'(0) = 1.

(i can't vouch for my algebra, i often make mistakes, so i recommend checking my work)