# Grade 12 Calculus Problem - Differentiation and Division

1. Aug 13, 2011

### G-S

1. The problem statement, all variables and given/known data

Let p be a polynomial function with p(a)= 0 =p'(a) for some real a. Which of the following must be true?

A) p(x) is divisible by x+a
B) p(x) is divisible by x^2+a^2
C) p(x) is divisible by x^2-a^2
D) p(x) is divisible by x^2+2ax+a^2
E) p(x) is divisible by x^2-2ax+a^2

2. Relevant equations

3. The attempt at a solution

I'm honestly beyond stumped for this one. Having a hard time finding out where to start on this problem.

2. Aug 13, 2011

### SammyS

Staff Emeritus
What does it mean for a polynomial (or any function) if p(x) = 0 at x=a ?

What does it mean for a polynomial, if p'(x) = 0 at x=a ?

3. Aug 13, 2011

### vela

Staff Emeritus
Hint: p(x) is a polynomial and p(a)=0 means a is a root of the polynomial. If you were to factor p(x), what factor do you know has to be there if a is a root?

4. Aug 13, 2011

### G-S

Thanks for the replies.
So this is the conclusion I've come to so far,

p(x)=0 at x=a
p'(x)=0 at x=a
a is a root of the polynomial

So,
x2-a2=0
x2=a2
x=a

Therefore, C) must be true.

Am I in the right direction?

5. Aug 13, 2011

### SammyS

Staff Emeritus
No, unless you know that -a is also a root.

What is true about a function if it's first derivative is zero at some value of x ?

6. Aug 13, 2011

### G-S

Sorry SammyS but the only thing that comes to mind is the function is a constant. Either that or the first step to find critical points.

After looking at the problem again I've come to the conclusion that in order to fulfill the condition p(a)=0 only C) and E) can be true.

Considering that we only know that a is a root (and not -a) E) seems like the only one that fulfills the conditions.

0=x2-2ax+a2
0=(x-a)2
0=x-a
x=a

Taking stabs in the dark here.

7. Aug 13, 2011

### e(ho0n3

You have narrowed down the list of choices to just E, so E must be the answer. Is there anything more you need to know?

8. Aug 13, 2011

### G-S

I'm trying to get confirmation as to whether the answer and my thought process is correct and if not, how one would approach this question.

9. Aug 13, 2011

### SammyS

Staff Emeritus
And, in general, why is it that you look for critical points?

Also, what is the slope of a constant function? (I'm assuming that this is not a constant polynomial.)

10. Aug 13, 2011

### G-S

To find max/min values. And the slope of a constant function is 0.

11. Aug 13, 2011

### SammyS

Staff Emeritus
More to the point, what is the slope of the line tangent to the polynomial at x = a, if p'(a) = 0 ?

This is the question I should have asked in my previous post.

12. Aug 13, 2011

### G-S

The slope of the tangent line is 0 if p'(a)=0.

13. Aug 13, 2011

### SammyS

Staff Emeritus
The behavior of a polynomial in the neighborhood of one of its zeros is due mainly to the factor which 'causes' the zero, and the multiplicity of that zero.

So, in the neighborhood of x=a, the polynomial, p(x), behaves like ±(x - a)n, where n ≥ 2 .

How do we know it's not like ±(x - a)1 ?

14. Aug 13, 2011

### G-S

Because it's not one of the options? Haha, I haven't got a clue.

15. Aug 13, 2011

### SammyS

Staff Emeritus
The derivative of (x - a) ≠ 0 at x = 1 .

16. Aug 14, 2011

### HallsofIvy

I presume you mean at x= a.

17. Aug 14, 2011

### SammyS

Staff Emeritus
Yes. Thanks for the correction.