Proving Triple Zero Implies Zero of P, PI, PII, and PIII

[LilTaru]

Homework Statement

The number a is called a triple zero (or a zero of multiplicity 3) of the polynomial P if:
P(x) = (x - a)³q(x) and q(a) [STRIKE]=[/STRIKE] 0.

Prove that if a is a triple zero of P, then a is a zero of P, P^I, P^II, and P^III(a) [STRIKE]=[/STRIKE] 0.

Homework Equations

The Attempt at a Solution

I have no idea about the concept of a triple zero and it is not mentioned in the section of the textbook these questions are in... I have no idea how to even prove this... My first thought was to differentiate the equation P(x) three times and see if the requirements fit? Is this on the right track or no?

 

[micromass]

Yes, just differentiate P(x) three times. That should do...

 

[LilTaru]

Okay I got all the requirements, but was wondering for the P^III(a) [STRIKE]=[/STRIKE] 0... I got it to equal 0 + 0 + 0 + 0 + undefined + 0 + 0 + 0, which I think equals undefined which would not equal zero... is this correct?

 

[micromass]

undefined? How did you get an undefined term? I think that you made a small mistake

If you can't find it, then post the derivatives and I'll try to spot the mistake...

 

[LilTaru]

Hmmm... maybe... I used the product rule since P(x) = (x-a)³ * q(x)

P^I(x) = (x-a)³ * q^I(x) + q(x) * 3(x-a)²

P^II(x) = [(x-a)³ * q^II(x) + q^I(x) * 3(x-a)²] + [q(x) * 6(x-a) + 3(x-a)² * q^I(x)]

P^III(x) = [(x-a)³ * q^III(x) + q^II(x) * 3(x-a)²] + [q^I(x) * 6(x-a) + 3(x-a)² * q^II(x)] + [q(x) * 6(x-a)^-1 + 6(x-a) * q^I(x)] + [3(x-a)² * q^II(x) + q^I(x) * 6(x-a)]

 

[LilTaru]

Therefore, by plugging x = a into P^III(x) it equals
0 + 0 + 0 + 0 + undefined + 0 + 0 + 0
The undefined term comes from 6(x-a)^-1 since it equals 6/(x-a) = 6/(a-a) = 6/0

 

[micromass]

How did you arrive at $$(x-a)^{-1}$$. The derivative of x-a is just 1... I'm not sure why you have an exponent of -1...

 

[LilTaru]

Oh I was thinking of the formula xⁿ becomes nx^n-1 when finding the derivative... so 6(x-a) would be 6(1) and become 6 when derived? It still matched because it would still not equal zero...

 

[micromass]

Yes, I know it doesn't match. I don't think the third derivative equals zero anyway. So I don't think it is true what you are trying to show...

 

[LilTaru]

The third derivative is not supposed to equal zero... the question asks to prove that... so I think your suggestion works because P^III(a) = 6q(a) and it states in the question that q(a) does not equal zero, which proves all three requirements and answers the question!

 

[micromass]

Ah yes! It seems you've proved it then

 

[LilTaru]

Yes thanks to you! Thank you very much! The help was much appreciated!