1. ## Please explain this question

The number a is called a double zero (or a zero of multiplicity 2) of the polynomial P if

P(x) = (x -a)^2 q(x) and q(a) not = 0.

Prove that if a is a double zero of P, then a is a zero of both P and P', and P"(a) not = 0.

Recall that a Zero of a polynomial function P is one number a is subset to R such that P(a) = 0. You may assume without proof that q is a polynomial function.

I do not understand the question at all, can anyone explain it and give some hints to prove it?

2. Originally Posted by 450081592
The number a is called a double zero (or a zero of multiplicity 2) of the polynomial P if

P(x) = (x -a)^2 q(x) and q(a) not = 0.

Prove that if a is a double zero of P, then a is a zero of both P and P', and P"(a) not = 0.

I do not understand the question at all, can anyone explain it and give some hints to prove it?

Let $\displaystyle a$ be a double zero of a polynomial P

Then, $\displaystyle P(x)=(x-a)^2q(x)$ and $\displaystyle q(a)\not = 0$

$\displaystyle P(a)=(a-a)^2q(a)=0\cdot q(a)=0$

$\displaystyle P'(x)=2(x-a)q(x)+(x-a)^2q'(x)$ by the product rule

$\displaystyle P'(a)=2(a-a)q(a)+(a-a)^2q'(a)=$

Then take the second derivative and show that when you plug in $\displaystyle a$ you don't get zero

3. Can you tell me what are we trying to prove here, I really dont understand the question

4. Originally Posted by 450081592
Can you tell me what are we trying to prove here, I really dont understand the question

You are told that a is a double zero and told exactly what that means. Using that information, you are supposed to show that for ANY polynomial and ANY point satisfying this double zero property, the function evaluated at the point is 0, as is the derivative, but the second derivative is not.

5. Originally Posted by artvandalay11
You are told that a is a double zero and told exactly what that means. Using that information, you are supposed to show that for ANY polynomial and ANY point satisfying this double zero property, the function evaluated at the point is 0, as is the derivative, but the second derivative is not.

so we wanna prove that the function and its derivative are both 0 at a centain point, or any points, did you show me the solution or the thought.

6. i showed you most of the solution and asked you to do the final part... do you understand the solution?

7. Originally Posted by artvandalay11
i showed you most of the solution and asked you to do the final part... do you understand the solution?

I understand what you did, but I dont know why you are doing that , is all I need to do is to find the derivative? Are we treating a as a constant number here? Is that suppose to be q(a) and q'(a)?

8. Originally Posted by 450081592
I understand what you did, but I dont know why you are doing thatall I need to do is to find the derivative?

All I did was find the derivative and plug in "a" to show it's zero

I also plugged in "a" to show that P is zero at "a"

9. I have no idea how to find the second derivative of the function, hints please1

I got something like this $\displaystyle 10q^2x^4-24aq^2x^3+18a^2q^2x^2-4a^3q^2x$

which is $\displaystyle 10q^2a^4-24q^2a^4+18q^2a^4-4q^2a^4$ which will equals to zero

10. Where did those "p" and "q" terms come from? In your original statement, P and q were functions, not constants.

If $\displaystyle P(x)= (x-a)^2q(x)$, P'= 2(x-a)q(x)+ (x-a)^2q'(x). Notice that each term contains a factor of "(x-a)" and so P'(a)= 0. Now differentiate again, using the product rule on each term.

11. Hey 450081592, are you a first year student at Carleton University?

12. Originally Posted by Sterwine
Hey 450081592, are you a first year student at Carleton University?

Haha, exactly, are you too?