### Polynomials : Multiplicity of roots Homework problem

Posted:

**Sat Feb 21, 2015 1:27 pm UTC**P(x) = ax

P(x) has a double root of x = 1

show that a = n, and b = -(n+1)

What I have done so far:

P(x) has a double root of x = 1, so P'(x) has a root of x = 1

P'(x) = (n+1)ax

P'(1) = (n+1)a + nb = 0

(n+1)a = -nb

n ≠ n+1

So it follows that a = -n and b = n+1 or a = n and b = -(n+1)

I don't know where to go from here, and how I can eliminate the other case.

Either I'm stupid or this question is wrong.

^{n+1}+ bx^{n}+ rP(x) has a double root of x = 1

show that a = n, and b = -(n+1)

What I have done so far:

P(x) has a double root of x = 1, so P'(x) has a root of x = 1

P'(x) = (n+1)ax

^{n}+ nbx^{n-1}P'(1) = (n+1)a + nb = 0

(n+1)a = -nb

n ≠ n+1

So it follows that a = -n and b = n+1 or a = n and b = -(n+1)

I don't know where to go from here, and how I can eliminate the other case.

Either I'm stupid or this question is wrong.