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...