Solving for Factors in a Polynomial Equation | Step-by-Step Guide

[Mentallic]

Homework Statement

A polynomial $$P(x)=(x-b)^7Q(x)$$
a) Show that $$P(b)=P ' (b)=0$$
b) Hence find a and b, if $$(x-1)^7$$ is a factor of: $$P(x)=x^7+3x^6+ax^5+x^4+3x^3+bx^2-x-1$$

Homework Equations

If $$P(x)=Q(x)R(x)$$
Then $$P ' (x)=Q ' (x)R(x)+Q(x)R ' (x)$$

I can't think of anything for the factoring aspect of the question.

The Attempt at a Solution

For a)
$$P(b)=(b-b)^7Q(b)=0$$
$$P ' (x)=7(x-b)^6Q(x)+(x-b)^7Q'(x)$$
$$P'(b)=7(b-b)^6Q(b)+(b-b)^7Q'(b)=0$$

But for b) I have no idea how to apply anything from a) to answer the question. Any ideas?

 

[gabbagabbahey]

Well if $(x-1)^7$ is a factor, then you can write P(x) in the form: $P(x)=(x-1)^7Q(x)$...So P(1)=__? and P'(1)=___?

But what are P(1) and P'(1) for $P(x)=x^7+3x^6+ax^5+x^4+3x^3+bx^2-x-1$?

 

[icystrike]

Well if $(x-1)^7$ is a factor, then you can write P(x) in the form: $P(x)=(x-1)^7Q(x)$...So P(1)=__? and P'(1)=___?

But what are P(1) and P'(1) for $P(x)=x^7+3x^6+ax^5+x^4+3x^3+bx^2-x-1$?

From P(1) and P'(1) you will get a simultaneous equation whereby a+b=? and 5a+2b=?

 

[Mentallic]

Ahh since $$P(1)=P'(1)=0$$ and by finding $$P(1)=6+a+b$$ and $$P'(1)=37+5a+2b$$ from substituting into the equation, I find a and b through simultaneous equations. Thus, $$a=-8\frac{1}{3},b=2\frac{1}{3}$$
I really hope I can pick these ideas up in the test...