I am trying to learn a very abstract way to finding derivatives using algebraic long division (polynomial division). It goes something like, P(x) = q(x)(x-a) where q(x) is the derivative. Can someone give me an example using regular numbers. The textbook is not really specifying how to use polynomial division and assumes we all just know what to do.
(This is for my basic analysis class)

Question

I am trying to learn a very abstract way to finding derivatives using algebraic long division (polynomial division). It goes something like, P(x) = q(x)(x-a) where q(x) is the derivative. Can someone give me an example using regular numbers. The textbook is not really specifying how to use polynomial division and assumes we all just know what to do. 
(This is for my basic analysis class)

anonymous · Answer

Finding derivatives of polynomials of degree >=2 just to be clear.

aum · Answer

Is q(x) the derivative or the quotient of dividing P(x) by its factor (x-a)?

anonymous · Answer

q(x) is the derivative

aum · Answer

I think you are probably trying to EVALUATE the derivative of a polynomial at x = a by dividing the polynomial and NOT the derivative itself.

aum · Answer

If f(x) = x^3 - 2x^2 + 8x - 5
find f'(2)  without taking the derivative.  Do it by polynomial division.  Is this what you are asking?

anonymous · Answer

yeah thats exactly what i was trying to do. By what do i divide f(x) though? By (x-a)?

anonymous · Answer

or in other words (x-2)

aum · Answer

In my example above, you divide f(x) by (x - 2).
Whatever quotient you get above, divide that again by (x-2).
The remainder will magically be f'(2)!

aum · Answer

Use synthetic division instead of long division.

aum · Answer

f(x) = x^3 - 2x^2 + 8x - 5 find f'(2) without taking the derivative.
Divide x^3 - 2x^2 + 8x - 5  by (x-2).   
Using synthetic division:

2       1    -2    8    -5
                 2    0    16
         1      0    8     11       (remainder is 11, which BTW, is f(2) )

Quotient is: x^2 + 0x + 8  
Divide the quotient by (x-2):

2     1    0    8 
             2    4  
       1    2   12       (remainder is 12, which is f'(2) )

We can verify if it is true:
f(x) = x^3 - 2x^2 + 8x - 5
f'(x) = 3x^2 - 4x + 8
f'(2) = 3(4) - 4(2) + 8 = 12

anonymous · Answer

just did it myself and saw we have the same response. Wow thank you so much this is like magic.

aum · Answer

You are welcome.

To prove it for yourself take some f(x) like the above example, find the derivative, then evaluate it at x = a (instead of a number).
Then try the division method.
Divide f(x) by (x-a).
identify the quotient and divide it again by (x-a).  The remainder will be a function of a.  Compare the remainder to the one you got earlier after actually taking the derivative and evaluating f'(a).  You will find they are the same.